Chapter 3

 

Analyzing motion of systems of particles

 

In this chapter, we shall discuss

  1. The concept of a particle
  2. Position/velocity/acceleration relations for a particle
  3. Newton’s laws of motion for a particle
  4. How to use Newton’s laws to calculate the forces needed to make a particle move in a particular way
  5. How to use Newton’s laws to derive `equations of motion’ for a system of particles
  6. How to solve equations of motion for particles by hand or using a computer.

 

The focus of this chapter is on setting up and solving equations of motion MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  we will not discuss in detail the behavior of the various examples that are solved. 

 

 

3.1 Equations of motion for a particle

 

We start with some basic definitions and physical laws.

 

3.1.1 Definition of a particle

 

A `Particle’ is a point mass at some position in space. It can move about, but has no characteristic orientation or rotational inertia.  It is characterized by its mass.

 

Examples of applications where you might choose to idealize part of a system as a particle include:

1.      Calculating the orbit of a satellite MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  for this application, you don’t need to know the orientation of the satellite, and you know that the satellite is very small compared with the dimensions of its orbit.

2.      A molecular dynamic simulation, where you wish to calculate the motion of individual atoms in a material.  Most of the mass of an atom is usually concentrated in a very small region (the nucleus) in comparison to inter-atomic spacing. It has negligible rotational inertia.   This approach is also sometimes used to model entire molecules, but rotational inertia can be important in this case.

 

Obviously, if you choose to idealize an object as a particle, you will only be able to calculate its position.  Its orientation or rotation cannot be computed.

 

 

3.1.2 Position, velocity, acceleration relations for a particle (Cartesian coordinates)

 

In most practical applications we are interested in the position or the velocity (or speed) of the particle as a function of time.   But Newton’s laws will only tell us its acceleration.   We therefore need equations that relate the position, velocity and acceleration.

 

Position vector: In Newtonian physics we have to measure position and motion in a so-called ‘Inertial Frame’.  This concept will be discussed in more detail in Section 3.2.  For now, suppose that we can identify

1.      Three, mutually perpendicular, fixed directions in space: the three directions are described by unit vectors { i,j,k } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaaca WHPbGaaiilaiaahQgacaGGSaGaaC4AaaGaay5Eaiaaw2haaaaa@3C60@

2.      We choose a convenient stationary (or non-accelerating) point to use as origin.

 

The position vector (relative to the origin) is then specified by the three distances (x,y,z) shown in the figure.

r=x(t)i+y(t)j+z(t)k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadIhacaGGOaGaamiDaiaacMcacaWHPbGaey4kaSIaamyEaiaa cIcacaWG0bGaaiykaiaahQgacqGHRaWkcaWG6bGaaiikaiaadshaca GGPaGaaC4Aaaaa@4684@

 

In dynamics problems, x,y,z can all be functions of time, but { i,j,k } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaaca WHPbGaaiilaiaahQgacaGGSaGaaC4AaaGaay5Eaiaaw2haaaaa@3C60@  are fixed.

 

 

Velocity vector: By definition, the velocity is the derivative of the position vector with respect to time (following the usual machinery of calculus)

v= lim δt0 r(t+δt)r(t) δt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9maaxababaGaciiBaiaacMgacaGGTbaaleaacqaH0oazcaWG0bGa eyOKH4QaaGimaaqabaGcdaWcaaqaaiaahkhacaGGOaGaamiDaiabgU caRiabes7aKjaadshacaGGPaGaeyOeI0IaaCOCaiaacIcacaWG0bGa aiykaaqaaiabes7aKjaadshaaaaaaa@4E08@

Velocity is a vector, and can therefore be expressed in terms of its Cartesian components

v= v x i+ v y j+ v z k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9iaadAhadaWgaaWcbaGaamiEaaqabaGccaWHPbGaey4kaSIaamOD amaaBaaaleaacaWG5baabeaakiaahQgacqGHRaWkcaWG2bWaaSbaaS qaaiaadQhaaeqaaOGaaC4Aaaaa@442C@

You can visualize a velocity vector as follows

·         The direction of the vector is parallel to the direction of motion

·         The magnitude of the vector v=| v |= v x 2 + v y 2 + v z 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODaiabg2 da9maaemaabaGaaCODaaGaay5bSlaawIa7aiabg2da9maakaaabaGa amODamaaDaaaleaacaWG4baabaGaaGOmaaaakiabgUcaRiaadAhada qhaaWcbaGaamyEaaqaaiaaikdaaaGccqGHRaWkcaWG2bWaa0baaSqa aiaadQhaaeaacaaIYaaaaaqabaaaaa@48B3@  is the speed of the particle (in meters/sec, for example). 

 

When both position and velocity vectors are expressed in terms Cartesian components, it is simple to calculate the velocity from the position vector.   For this case, the basis vectors { i,j,k } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaaca WHPbGaaiilaiaahQgacaGGSaGaaC4AaaGaay5Eaiaaw2haaaaa@3C60@  are constant (independent of time) and so

v x i+ v y j+ v z k= d dt ( x(t)i+y(t)j+z(t)k )= dx dt i+ dy dt j+ dz dt k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaWG4baabeaakiaahMgacqGHRaWkcaWG2bWaaSbaaSqaaiaa dMhaaeqaaOGaaCOAaiabgUcaRiaadAhadaWgaaWcbaGaamOEaaqaba GccaWHRbGaeyypa0ZaaSaaaeaacaWGKbaabaGaamizaiaadshaaaWa aeWaaeaacaWG4bGaaiikaiaadshacaGGPaGaaCyAaiabgUcaRiaadM hacaGGOaGaamiDaiaacMcacaWHQbGaey4kaSIaamOEaiaacIcacaWG 0bGaaiykaiaahUgaaiaawIcacaGLPaaacqGH9aqpdaWcaaqaaiaads gacaWG4baabaGaamizaiaadshaaaGaaCyAaiabgUcaRmaalaaabaGa amizaiaadMhaaeaacaWGKbGaamiDaaaacaWHQbGaey4kaSYaaSaaae aacaWGKbGaamOEaaqaaiaadsgacaWG0baaaiaahUgaaaa@6645@

This is really three equations MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  one for each velocity component, i.e.

v x = dx dt v y = dy dt v z = dz dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaWG4baabeaakiabg2da9maalaaabaGaamizaiaadIhaaeaa caWGKbGaamiDaaaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaadAhadaWgaaWcbaGaamyEaaqabaGccqGH9aqp daWcaaqaaiaadsgacaWG5baabaGaamizaiaadshaaaGaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadAhadaWgaaWcbaGa amOEaaqabaGccqGH9aqpdaWcaaqaaiaadsgacaWG6baabaGaamizai aadshaaaaaaa@6245@

 

 

Acceleration vector: The acceleration is the derivative of the velocity vector with respect to time; or, equivalently, the second derivative of the position vector with respect to time.

a= lim δt0 v(t+δt)v(t) δt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabg2 da9maaxababaGaciiBaiaacMgacaGGTbaaleaacqaH0oazcaWG0bGa eyOKH4QaaGimaaqabaGcdaWcaaqaaiaahAhacaGGOaGaamiDaiabgU caRiabes7aKjaadshacaGGPaGaeyOeI0IaaCODaiaacIcacaWG0bGa aiykaaqaaiabes7aKjaadshaaaaaaa@4DFB@

The acceleration is a vector, with Cartesian representation a= a x i+ a y j+ a z k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabg2 da9iaadggadaWgaaWcbaGaamiEaaqabaGccaWHPbGaey4kaSIaamyy amaaBaaaleaacaWG5baabeaakiaahQgacqGHRaWkcaWGHbWaaSbaaS qaaiaadQhaaeqaaOGaaC4Aaaaa@43D8@ .

 

Like velocity, acceleration has magnitude and direction. Sometimes it may be possible to visualize an acceleration vector MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  for example, if you know your particle is moving in a straight line, the acceleration vector must be parallel to the direction of motion; or if the particle moves around a circle at constant speed, its acceleration is towards the center of the circle.  But sometimes you can’t trust your intuition regarding the magnitude and direction of acceleration, and it can be best to simply work through the math.

 

The relations between Cartesian components of position, velocity and acceleration are

a x = d v x dt = d 2 x d t 2 a y = d v y dt = d 2 y d t 2 a z = d v z dt = d 2 z d t 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyamaaBa aaleaacaWG4baabeaakiabg2da9maalaaabaGaamizaiaadAhadaWg aaWcbaGaamiEaaqabaaakeaacaWGKbGaamiDaaaacqGH9aqpcaaMc8 +aaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaqaaiaa dsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaakiaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caWGHbWaaSbaaSqaaiaadMha aeqaaOGaeyypa0ZaaSaaaeaacaWGKbGaamODamaaBaaaleaacaWG5b aabeaaaOqaaiaadsgacaWG0baaaiaaykW7cqGH9aqpdaWcaaqaaiaa dsgadaahaaWcbeqaaiaaikdaaaGccaWG5baabaGaamizaiaadshada ahaaWcbeqaaiaaikdaaaaaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caWGHbWaaSbaaSqaaiaadQhaaeqaaOGaaGPaVlabg2da9maala aabaGaamizaiaadAhadaWgaaWcbaGaamOEaaqabaaakeaacaWGKbGa amiDaaaacqGH9aqpdaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaa GccaWG6baabaGaamizaiaadshadaahaaWcbeqaaiaaikdaaaaaaaaa @7AE5@

 

 

3.1.3 Review of some ideas from calculus

 

To analyze motion, we often have to calculate accelerations given position vectors as a function of time, or (more commonly) calculate velocity and position by integrating accelerations.

 

You should have been beaten into submission with this sort of problem in calculus courses (and maybe some physics courses as well) so we’ll just review the most important procedures here as a reminder.   We will focus on the generic one-dimensional problem: given a MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadggaaaa@3720@ , calculate v and x; or vice-versa.

 

Calculating positions, velocities and accelerations graphically

 

You will remember that:

·         Speed is the slope of the distance-v-time curve

·         Distance is the area under the speed-v-time curve

Or alternatively

·         Acceleration is the slope of the speed-v-time curve

·         Speed is the area under the acceleration-v-time curve

 

These ideas are illustrated in the figure below: if you can sketch a graph of acceleration, velocity or position, you can often use geometry to calculate all the other quantities of interest.

 

Rules for integrating accelerations and speeds

 

·         Acceleration given as a function of time

dv dt =a(t) v0 v dv = 0 t a(t)dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiaadsgacaWG2baabaGaam izaiaadshaaaGaeyypa0JaamyyaiaacIcacaWG0bGaaiykaiaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqGHshI3caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8+aa8qCaeaacaWG KbGaamODaaWcbaGaamODaiaaicdaaeaacaWG2baaniabgUIiYdGccq GH9aqpdaWdXbqaaiaadggacaGGOaGaamiDaiaacMcacaWGKbGaamiD aaWcbaGaaGimaaqaaiaadshaa0Gaey4kIipaaaa@6413@

Example:

dv dt = t v= v 0 + 2 3 t 3/2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiaadsgacaWG2baabaGaam izaiaadshaaaGaeyypa0ZaaOaaaeaacaWG0baaleqaaOGaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaeyO0H4TaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamODaiab g2da9iaadAhadaWgaaWcbaGaaGimaaqabaGccqGHRaWkdaWcaaqaai aaikdaaeaacaaIZaaaaiaadshadaahaaWcbeqaaiaaiodacaGGVaGa aGOmaaaaaaa@5BEA@

 

 

·         Acceleration given as a function of speed

dv dt =g(t)f(v) v0 v dv f(v) = 0 t g(t)dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiaadsgacaWG2baabaGaam izaiaadshaaaGaeyypa0Jaam4zaiaacIcacaWG0bGaaiykaiaadAga caGGOaGaamODaiaacMcacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaeyO0H4TaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVpaapehabaWaaSaaaeaacaWGKbGaamODaaqaaiaadA gacaGGOaGaamODaiaacMcaaaaaleaacaWG2bGaaGimaaqaaiaadAha a0Gaey4kIipakiabg2da9maapehabaGaam4zaiaacIcacaWG0bGaai ykaiaadsgacaWG0baaleaacaaIWaaabaGaamiDaaqdcqGHRiI8aaaa @6AAD@

Example:

dv dt =cv v0 v dv v = 0 t cdt log( v v 0 )=ct MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiaadsgacaWG2baabaGaam izaiaadshaaaGaeyypa0JaeyOeI0Iaam4yaiaaykW7caWG2bGaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7cqGHshI3caaMc8UaaGPaVlaayk W7caaMc8+aa8qCaeaadaWcaaqaaiaadsgacaWG2baabaGaamODaaaa aSqaaiaadAhacaaIWaaabaGaamODaaqdcqGHRiI8aOGaeyypa0Jaey OeI0Yaa8qCaeaacaWGJbGaamizaiaadshaaSqaaiaaicdaaeaacaWG 0baaniabgUIiYdGccaaMc8UaaGPaVlaaykW7cqGHshI3caaMc8UaaG PaVlaaykW7ciGGSbGaai4BaiaacEgacaGGOaWaaSaaaeaacaWG2baa baGaamODamaaBaaaleaacaaIWaaabeaaaaGccaGGPaGaeyypa0Jaey OeI0Iaam4yaiaadshaaaa@73C3@

 

·         Acceleration given as a function of distance

dv dt =f(x) dv dx dx dt =f(x) dv dx v=f(x) v0 v vdv = 0 x f(x)dx MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiaadsgacaWG2baabaGaam izaiaadshaaaGaeyypa0JaamOzaiaacIcacaWG4bGaaiykaiaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabgkDiElaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8+aaSaaaeaacaWG KbGaamODaaqaaiaadsgacaWG4baaamaalaaabaGaamizaiaadIhaae aacaWGKbGaamiDaaaacqGH9aqpcaWGMbGaaiikaiaadIhacaGGPaGa aGPaVlaaykW7caaMc8UaeyO0H49aaSaaaeaacaWGKbGaamODaaqaai aadsgacaWG4baaaiaadAhacqGH9aqpcaWGMbGaaiikaiaadIhacaGG PaGaeyO0H4TaaGPaVlaaykW7daWdXbqaaiaadAhacaWGKbGaamODaa WcbaGaamODaiaaicdaaeaacaWG2baaniabgUIiYdGccqGH9aqpdaWd XbqaaiaadAgacaGGOaGaamiEaiaacMcacaWGKbGaamiEaaWcbaGaaG imaaqaaiaadIhaa0Gaey4kIipaaaa@86AE@

Example

dv dt =kx v0 v vdv = 0 x kxdx 1 2 v 2 1 2 v 0 2 = 1 2 k x 2 v= v 0 2 k x 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiaadsgacaWG2baabaGaam izaiaadshaaaGaeyypa0JaeyOeI0Iaam4AaiaadIhacaaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7cqGHshI3caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7daWdXbqaaiaa dAhacaWGKbGaamODaaWcbaGaamODaiaaicdaaeaacaWG2baaniabgU IiYdGccqGH9aqpdaWdXbqaaiabgkHiTiaadUgacaWG4bGaamizaiaa dIhaaSqaaiaaicdaaeaacaWG4baaniabgUIiYdGccaaMc8UaaGPaVl aaykW7caaMc8UaeyO0H4TaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7daWcaaqaaiaaigdaaeaacaaIYaaaaiaadA hadaahaaWcbeqaaiaaikdaaaGccqGHsisldaWcaaqaaiaaigdaaeaa caaIYaaaaiaadAhadaqhaaWcbaGaaGimaaqaaiaaikdaaaGccqGH9a qpcqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaaiaadUgacaWG4bWa aWbaaSqabeaacaaIYaaaaOGaaGPaVlaaykW7caaMc8UaeyO0H4Taam ODaiabg2da9maakaaabaGaamODamaaDaaaleaacaaIWaaabaGaaGOm aaaakiabgkHiTiaadUgacaWG4bWaaWbaaSqabeaacaaIYaaaaaqaba aaaa@9935@

 

·         Velocity given as a function of time

dx dt =v(t) x0 x dx = 0 t v(t)dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiaadsgacaWG4baabaGaam izaiaadshaaaGaeyypa0JaamODaiaacIcacaWG0bGaaiykaiaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqGHshI3caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8+aa8qCaeaacaWG KbGaamiEaaWcbaGaamiEaiaaicdaaeaacaWG4baaniabgUIiYdGccq GH9aqpdaWdXbqaaiaadAhacaGGOaGaamiDaiaacMcacaWGKbGaamiD aaWcbaGaaGimaaqaaiaadshaa0Gaey4kIipaaaa@6445@

Example

v= v 0 + 2 3 t 3/2 x= x 0 + v 0 t+ 4 15 t 5/2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaaMc8UaamODaiabg2da9iaadAhada WgaaWcbaGaaGimaaqabaGccqGHRaWkdaWcaaqaaiaaikdaaeaacaaI ZaaaaiaadshadaahaaWcbeqaaiaaiodacaGGVaGaaGOmaaaakiabgk DiElaadIhacqGH9aqpcaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaey4k aSIaamODamaaBaaaleaacaaIWaaabeaakiaadshacqGHRaWkdaWcaa qaaiaaisdaaeaacaaIXaGaaGynaaaacaWG0bWaaWbaaSqabeaacaaI 1aGaai4laiaaikdaaaaaaa@4E94@

 

·         Velocity given as a separable function of position and time

dx dt =g(t)f(x) v0 v dx f(x) = 0 t g(t)dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiaadsgacaWG4baabaGaam izaiaadshaaaGaeyypa0Jaam4zaiaacIcacaWG0bGaaiykaiaadAga caGGOaGaamiEaiaacMcacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaeyO0H4TaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVpaapehabaWaaSaaaeaacaWGKbGaamiEaaqaaiaadA gacaGGOaGaamiEaiaacMcaaaaaleaacaWG2bGaaGimaaqaaiaadAha a0Gaey4kIipakiabg2da9maapehabaGaam4zaiaacIcacaWG0bGaai ykaiaadsgacaWG0baaleaacaaIWaaabaGaamiDaaqdcqGHRiI8aaaa @6AB5@

Example

dx dt = v 0 2 k x 2 x 0 x dx v 0 2 k x 2 = 0 t dt 1 k { sin 1 ( k x v 0 ) sin 1 ( k x 0 v 0 ) }=t x= v 0 k sin( k t+ sin 1 ( k x 0 v 0 ) ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaamaalaaabaGaamizaiaadIhaae aacaWGKbGaamiDaaaacqGH9aqpdaGcaaqaaiaadAhadaqhaaWcbaGa aGimaaqaaiaaikdaaaGccqGHsislcaWGRbGaamiEamaaCaaaleqaba GaaGOmaaaaaeqaaOGaaGPaVlaaykW7caaMc8UaaGPaVlabgkDiElaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVpaapehabaWaaSaaae aacaWGKbGaamiEaaqaamaakaaabaGaamODamaaDaaaleaacaaIWaaa baGaaGOmaaaakiabgkHiTiaadUgacaWG4bWaaWbaaSqabeaacaaIYa aaaaqabaaaaaqaaiaadIhadaWgaaadbaGaaGimaaqabaaaleaacaWG 4baaniabgUIiYdGccqGH9aqpdaWdXbqaaiaadsgacaWG0baaleaaca aIWaaabaGaamiDaaqdcqGHRiI8aaGcbaGaeyO0H49aaSaaaeaacaaI XaaabaWaaOaaaeaacaWGRbaaleqaaaaakmaacmaabaGaci4CaiaacM gacaGGUbWaaWbaaSqabeaacqGHsislcaaIXaaaaOWaaeWaaeaadaWc aaqaamaakaaabaGaam4AaaWcbeaakiaadIhaaeaacaWG2bWaaSbaaS qaaiaaicdaaeqaaaaaaOGaayjkaiaawMcaaiabgkHiTiGacohacaGG PbGaaiOBamaaCaaaleqabaGaeyOeI0IaaGymaaaakmaabmaabaWaaS aaaeaadaGcaaqaaiaadUgaaSqabaGccaWG4bWaaSbaaSqaaiaaicda aeqaaaGcbaGaamODamaaBaaaleaacaaIWaaabeaaaaaakiaawIcaca GLPaaaaiaawUhacaGL9baacqGH9aqpcaWG0baabaGaeyO0H4TaamiE aiabg2da9maalaaabaGaamODamaaBaaaleaacaaIWaaabeaaaOqaam aakaaabaGaam4AaaWcbeaaaaGcciGGZbGaaiyAaiaac6gadaqadaqa amaakaaabaGaam4AaaWcbeaakiaadshacqGHRaWkciGGZbGaaiyAai aac6gadaahaaWcbeqaaiabgkHiTiaaigdaaaGcdaqadaqaamaalaaa baWaaOaaaeaacaWGRbaaleqaaOGaamiEamaaBaaaleaacaaIWaaabe aaaOqaaiaadAhadaWgaaWcbaGaaGimaaqabaaaaaGccaGLOaGaayzk aaaacaGLOaGaayzkaaaaaaa@9DB0@

 

 

 

 

3.1.4 Examples using position-velocity-acceleration relations

 

It is important for you to be comfortable with calculating velocity and acceleration from the position vector of a particle.   You will need to do this in nearly every problem we solve.  In this section we provide a few examples.  Each example gives a set of formulas that will be useful in practical applications.

 

 

Example 1: Constant acceleration along a straight lineThere are many examples where an object moves along a straight line, with constant acceleration.   Examples include free fall near the surface of a planet (without air resistance), the initial stages of the acceleration of a car, or and aircraft during takeoff roll, or a spacecraft during blastoff.

 

Suppose that

The particle moves parallel to a unit vector i

The particle has constant acceleration, with magnitude a

At time t= t 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 da9iaadshadaWgaaWcbaGaaGimaaqabaaaaa@3ADB@  the particle has speed v 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaaIWaaabeaaaaa@38DE@

At time t= t 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 da9iaadshadaWgaaWcbaGaaGimaaqabaaaaa@3ADB@  the particle has position vector r= x 0 i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadIhadaWgaaWcbaGaaGimaaqabaGccaWHPbaaaa@3BDD@

The position, velocity acceleration vectors are then

r=( x 0 + v 0 (t t 0 )+ 1 2 a (t t 0 ) 2 )i v=( v 0 +at )i a=ai MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWHYb Gaeyypa0ZaaeWaaeaacaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaey4k aSIaamODamaaBaaaleaacaaIWaaabeaakiaacIcacaWG0bGaeyOeI0 IaamiDamaaBaaaleaacaaIWaaabeaakiaacMcacqGHRaWkdaWcaaqa aiaaigdaaeaacaaIYaaaaiaadggacaGGOaGaamiDaiabgkHiTiaads hadaWgaaWcbaGaaGimaaqabaGccaGGPaWaaWbaaSqabeaacaaIYaaa aaGccaGLOaGaayzkaaGaaCyAaaqaaiaahAhacqGH9aqpdaqadaqaai aadAhadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcaWGHbGaamiDaaGa ayjkaiaawMcaaiaahMgaaeaacaWHHbGaeyypa0JaamyyaiaahMgaaa aa@5BC1@

 

Verify for yourself that the position, velocity and acceleration (i) have the correct values at t=0 and (ii) are related by the correct expressions (i.e. differentiate the position and show that you get the correct expression for the velocity, and differentiate the velocity to show that you get the correct expression for the acceleration).

 

HEALTH WARNING: These results can only be used if the acceleration is constant.  In many problems acceleration is a function of time, or position MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  in this case these formulas cannot be used. People who have taken high school physics classes have used these formulas to solve so many problems that they automatically apply them to everything MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  this works for high school problems but not always in real life!

 

Example 2: Simple Harmonic MotionThe vibration of a very simple spring-mass system is an example of simple harmonic motion.

 

In simple harmonic motion (i) the particle moves along a straight line; and (ii) the position, velocity and acceleration are all trigonometric functions of time.

 

For example, the position vector of the mass might be given by

r=x(t)i=( X 0 +ΔXsin(2πt/T) )i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadIhacaGGOaGaamiDaiaacMcacaWHPbGaeyypa0ZaaeWaaeaa caWGybWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSIaeuiLdqKaamiwai GacohacaGGPbGaaiOBaiaacIcacaaIYaGaeqiWdaNaamiDaiaac+ca caWGubGaaiykaaGaayjkaiaawMcaaiaahMgaaaa@4EE1@

Here X 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiwamaaBa aaleaacaaIWaaabeaaaaa@38C0@  is the average length of the spring, X 0 +ΔX MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiwamaaBa aaleaacaaIWaaabeaakiabgUcaRiabfs5aejaadIfaaaa@3BEF@  is the maximum length of the spring, and T is the time for the mass to complete one complete cycle of oscillation (this is called the `period’ of oscillation). 

 

Harmonic vibrations are also often characterized by the frequency of vibration:

·         The frequency in cycles per second (or Hertz) is related to the period by f=1/T

·         The angular frequency is related to the period by ω=2π/T MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyYdCNaey ypa0JaaGOmaiabec8aWjaac+cacaWGubaaaa@3DD5@

 

The motion is plotted in the figure on the right.

 

The velocity and acceleration can be calculated by differentiating the position, as follows

v= dx(t) dt i=( 2πΔX T cos(2πt/T) )i a= d 2 x(t) d t 2 i=( 4 π 2 ΔX T 2 sin(2πt/T) )i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWH2b Gaeyypa0ZaaSaaaeaacaWGKbGaamiEaiaacIcacaWG0bGaaiykaaqa aiaadsgacaWG0baaaiaahMgacqGH9aqpdaqadaqaamaalaaabaGaaG Omaiabec8aWjabfs5aejaadIfaaeaacaWGubaaaiGacogacaGGVbGa ai4CaiaacIcacaaIYaGaeqiWdaNaamiDaiaac+cacaWGubGaaiykaa GaayjkaiaawMcaaiaahMgaaeaacaWHHbGaeyypa0ZaaSaaaeaacaWG KbWaaWbaaSqabeaacaaIYaaaaOGaamiEaiaacIcacaWG0bGaaiykaa qaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaakiaahMgacqGH 9aqpdaqadaqaaiabgkHiTmaalaaabaGaaGinaiabec8aWnaaCaaale qabaGaaGOmaaaakiabfs5aejaadIfaaeaacaWGubWaaWbaaSqabeaa caaIYaaaaaaakiGacohacaGGPbGaaiOBaiaacIcacaaIYaGaeqiWda NaamiDaiaac+cacaWGubGaaiykaaGaayjkaiaawMcaaiaahMgaaaaa @7291@

 

Note that:

·         The velocity and acceleration are also harmonic, and have the same period and frequency as the displacement.

·         If you know the frequency, and amplitude and of either the displacement, velocity, or acceleration, you can immediately calculate the amplitudes of the other two.  For example, if ΔX MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam iwaaaa@3940@ , ΔV MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam Ovaaaa@393E@ , ΔA MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam yqaaaa@3929@  denote the amplitudes of the displacement, velocity and acceleration, we have that

ΔV= 2π T ΔXΔA= ( 2π T ) 2 ΔX= 2π T ΔV MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam Ovaiabg2da9maalaaabaGaaGOmaiabec8aWbqaaiaadsfaaaGaeuiL dqKaamiwaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaeuiLdqKaamyqaiabg2da9maa bmaabaWaaSaaaeaacaaIYaGaeqiWdahabaGaamivaaaaaiaawIcaca GLPaaadaahaaWcbeqaaiaaikdaaaGccqqHuoarcaWGybGaeyypa0Za aSaaaeaacaaIYaGaeqiWdahabaGaamivaaaacqqHuoarcaWGwbaaaa@62DE@

 

 

Example 3: Motion at constant speed around a circular path  Circular motion is also very common MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  examples include any rotating machinery, vehicles traveling around a circular path, and so on.

 

The simplest way to make an object move at constant speed along a circular path is to attach it to the end of a shaft (see the figure), and then rotate the shaft at a constant angular rate.  Then, notice that

·       The angle θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUdehaaa@38B4@  increases at constant rate.  We can write θ=ωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUdeNaey ypa0JaeqyYdCNaamiDaaaa@3C7F@ , where ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyYdChaaa@38CA@  is the (constant) angular speed of the shaft, in radians/seconds

·       The speed of the particle is related to ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyYdChaaa@38CA@  by V=Rω MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 da9iaadkfacqaHjpWDaaa@3B83@ .   To see this, notice that the circumferential distance traveled by the particle is s=Rθ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Caiabg2 da9iaadkfacqaH4oqCaaa@3B89@ .  Therefore, V=ds/dt=Rdθ/dt=Rω MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 da9iaadsgacaWGZbGaai4laiaadsgacaWG0bGaeyypa0JaamOuaiaa dsgacqaH4oqCcaGGVaGaamizaiaadshacqGH9aqpcaWGsbGaeqyYdC haaa@4810@ .

 

For this example the position vector is

r=Rcosθi+Rsinθj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadkfaciGGJbGaai4BaiaacohacqaH4oqCcaWHPbGaey4kaSIa amOuaiGacohacaGGPbGaaiOBaiabeI7aXjaahQgaaaa@468A@

The velocity can be calculated by differentiating the position vector. 

v= dr dt =R dθ dt sinθi+R dθ dt cosθj=Rω(sinθi+cosθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9maalaaabaGaamizaiaahkhaaeaacaWGKbGaamiDaaaacqGH9aqp cqGHsislcaWGsbWaaSaaaeaacaWGKbGaeqiUdehabaGaamizaiaads haaaGaci4CaiaacMgacaGGUbGaeqiUdeNaaCyAaiabgUcaRiaadkfa daWcaaqaaiaadsgacqaH4oqCaeaacaWGKbGaamiDaaaaciGGJbGaai 4BaiaacohacqaH4oqCcaWHQbGaeyypa0JaamOuaiabeM8a3jaacIca cqGHsislciGGZbGaaiyAaiaac6gacqaH4oqCcaWHPbGaey4kaSIaci 4yaiaac+gacaGGZbGaeqiUdeNaaCOAaiaacMcaaaa@6747@

Here, we have used the chain rule of differentiation, and noted that dθ/dt=ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiabeI 7aXjaac+cacaWGKbGaamiDaiabg2da9iabeM8a3baa@3F04@ .

 

The acceleration vector follows as

a= dv dt =Rω( dθ dt cosθi dθ dt sinθj)=R ω 2 (cosθi+sinθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabg2 da9maalaaabaGaamizaiaahAhaaeaacaWGKbGaamiDaaaacqGH9aqp caWGsbGaeqyYdCNaaiikaiabgkHiTmaalaaabaGaamizaiabeI7aXb qaaiaadsgacaWG0baaaiGacogacaGGVbGaai4CaiabeI7aXjaahMga cqGHsisldaWcaaqaaiaadsgacqaH4oqCaeaacaWGKbGaamiDaaaaci GGZbGaaiyAaiaac6gacqaH4oqCcaWHQbGaaiykaiabg2da9iabgkHi TiaadkfacqaHjpWDdaahaaWcbeqaaiaaikdaaaGccaGGOaGaci4yai aac+gacaGGZbGaeqiUdeNaaCyAaiabgUcaRiGacohacaGGPbGaaiOB aiabeI7aXjaahQgacaGGPaaaaa@6A83@

Note that

 (i) The magnitude of the velocity is V=Rω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 da9iaadkfacqaHjpWDaaa@3B82@ , and its direction is (obviously!) tangent to the path (to see this, visualize (using trig) the direction of the unit vector t=(sinθi+cosθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiDaiabg2 da9iaacIcacqGHsislciGGZbGaaiyAaiaac6gacqaH4oqCcaWHPbGa ey4kaSIaci4yaiaac+gacaGGZbGaeqiUdeNaaCOAaiaacMcaaaa@4724@

(ii) The magnitude of the acceleration is R ω 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabeM 8a3naaCaaaleqabaGaaGOmaaaaaaa@3A8A@  and its direction is towards the center of the circle. To see this, visualize (using trig) the direction of the unit vector n=(cosθi+sinθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOBaiabg2 da9iabgkHiTiaacIcaciGGJbGaai4BaiaacohacqaH4oqCcaWHPbGa ey4kaSIaci4CaiaacMgacaGGUbGaeqiUdeNaaCOAaiaacMcaaaa@471E@  

 

We can write these mathematically as

v= dr dt =Rωt=Vta= dv dt =R ω 2 n= V 2 R n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9maalaaabaGaamizaiaahkhaaeaacaWGKbGaamiDaaaacqGH9aqp caWGsbGaeqyYdCNaaCiDaiabg2da9iaadAfacaWH0bGaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaCyyaiabg2da9maalaaabaGaamizaiaahAhaae aacaWGKbGaamiDaaaacqGH9aqpcaWGsbGaeqyYdC3aaWbaaSqabeaa caaIYaaaaOGaaCOBaiabg2da9maalaaabaGaamOvamaaCaaaleqaba GaaGOmaaaaaOqaaiaadkfaaaGaaCOBaaaa@66F1@

 

Example 4: More general motion around a circular path 

 

We next look at more general circular motion, where the particle still moves around a circular path, but does not move at constant speed.  The angle θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUdehaaa@38B4@  is now a general function of time. 

 

We can write down some useful scalar relations:

·       Angular rate: ω= dθ dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyYdCNaey ypa0ZaaSaaaeaacaWGKbGaeqiUdehabaGaamizaiaadshaaaaaaa@3E62@

·       Angular acceleration α= dω dt = d 2 θ d t 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySdeMaey ypa0ZaaSaaaeaacaWGKbGaeqyYdChabaGaamizaiaadshaaaGaeyyp a0ZaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaeqiUdehaba GaamizaiaadshadaahaaWcbeqaaiaaikdaaaaaaaaa@45BE@

·       Speed V=R dθ dt =Rω MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 da9iaadkfadaWcaaqaaiaadsgacqaH4oqCaeaacaWGKbGaamiDaaaa cqGH9aqpcaWGsbGaeqyYdChaaa@41F1@

·       Rate of change of speed dV dt =R d 2 θ d t 2 =R dω dt =Rα MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbGaamOvaaqaaiaadsgacaWG0baaaiabg2da9iaadkfadaWcaaqa aiaadsgadaahaaWcbeqaaiaaikdaaaGccqaH4oqCaeaacaWGKbGaam iDamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpcaWGsbWaaSaaaeaa caWGKbGaeqyYdChabaGaamizaiaadshaaaGaeyypa0JaamOuaiabeg 7aHbaa@4D09@

 

We can now calculate vector velocities and accelerations

r=Rcosθi+Rsinθj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadkfaciGGJbGaai4BaiaacohacqaH4oqCcaWHPbGaey4kaSIa amOuaiGacohacaGGPbGaaiOBaiabeI7aXjaahQgaaaa@468A@

The velocity can be calculated by differentiating the position vector. 

v= dr dt =R dθ dt sinθi+R dθ dt cosθj=Rω(sinθi+cosθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9maalaaabaGaamizaiaahkhaaeaacaWGKbGaamiDaaaacqGH9aqp cqGHsislcaWGsbWaaSaaaeaacaWGKbGaeqiUdehabaGaamizaiaads haaaGaci4CaiaacMgacaGGUbGaeqiUdeNaaCyAaiabgUcaRiaadkfa daWcaaqaaiaadsgacqaH4oqCaeaacaWGKbGaamiDaaaaciGGJbGaai 4BaiaacohacqaH4oqCcaWHQbGaeyypa0JaamOuaiabeM8a3jaacIca cqGHsislciGGZbGaaiyAaiaac6gacqaH4oqCcaWHPbGaey4kaSIaci 4yaiaac+gacaGGZbGaeqiUdeNaaCOAaiaacMcaaaa@6747@

The acceleration vector follows as

a= dv dt =R dω dt (sinθi+cosθj)+Rω( dθ dt cosθi dθ dt sinθj) =Rα(sinθi+cosθj)R ω 2 (cosθi+sinθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWHHb Gaeyypa0ZaaSaaaeaacaWGKbGaaCODaaqaaiaadsgacaWG0baaaiab g2da9iaadkfadaWcaaqaaiaadsgacqaHjpWDaeaacaWGKbGaamiDaa aacaGGOaGaeyOeI0Iaci4CaiaacMgacaGGUbGaeqiUdeNaaCyAaiab gUcaRiGacogacaGGVbGaai4CaiabeI7aXjaahQgacaGGPaGaey4kaS IaamOuaiabeM8a3jaacIcacqGHsisldaWcaaqaaiaadsgacqaH4oqC aeaacaWGKbGaamiDaaaaciGGJbGaai4BaiaacohacqaH4oqCcaWHPb GaeyOeI0YaaSaaaeaacaWGKbGaeqiUdehabaGaamizaiaadshaaaGa ci4CaiaacMgacaGGUbGaeqiUdeNaaCOAaiaacMcaaeaacaaMc8UaaG PaVlaaykW7caaMc8Uaeyypa0JaamOuaiabeg7aHjaacIcacqGHsisl ciGGZbGaaiyAaiaac6gacqaH4oqCcaWHPbGaey4kaSIaci4yaiaac+ gacaGGZbGaeqiUdeNaaCOAaiaacMcacqGHsislcaWGsbGaeqyYdC3a aWbaaSqabeaacaaIYaaaaOGaaiikaiGacogacaGGVbGaai4CaiabeI 7aXjaahMgacqGHRaWkciGGZbGaaiyAaiaac6gacqaH4oqCcaWHQbGa aiykaaaaaa@95D5@

 

It is often more convenient to re-write these in terms of the unit vectors n and t normal and tangent to the circular path, noting that t=(sinθi+cosθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiDaiabg2 da9iaacIcacqGHsislciGGZbGaaiyAaiaac6gacqaH4oqCcaWHPbGa ey4kaSIaci4yaiaac+gacaGGZbGaeqiUdeNaaCOAaiaacMcaaaa@4724@ , n=(cosθi+sinθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOBaiabg2 da9iabgkHiTiaacIcaciGGJbGaai4BaiaacohacqaH4oqCcaWHPbGa ey4kaSIaci4CaiaacMgacaGGUbGaeqiUdeNaaCOAaiaacMcaaaa@471E@ .  Then

v=Rωt=Vta=Rαt+R ω 2 n= dV dt t+ V 2 R n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9iaadkfacqaHjpWDcaWH0bGaeyypa0JaamOvaiaahshacaaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaahggacqGH9aqpcaWGsbGa eqySdeMaaCiDaiabgUcaRiaadkfacqaHjpWDdaahaaWcbeqaaiaaik daaaGccaWHUbGaeyypa0ZaaSaaaeaacaWGKbGaamOvaaqaaiaadsga caWG0baaaiaahshacqGHRaWkdaWcaaqaaiaadAfadaahaaWcbeqaai aaikdaaaaakeaacaWGsbaaaiaah6gaaaa@6A35@

 

These are the famous circular motion formulas that you might have seen in physics class. 

Using MATLAB ‘live scripts’

 

If you find that your calculus is a bit rusty you can use MATLAB to do the tedious work for you.  For example, to differentiate the vector

r=x(t)i+y(t)j+z(t)k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadIhacaGGOaGaamiDaiaacMcacaWHPbGaey4kaSIaamyEaiaa cIcacaWG0bGaaiykaiaahQgacqGHRaWkcaWG6bGaaiikaiaadshaca GGPaGaaC4Aaaaa@4684@

you would type

syms x(t) y(t) z(t) r(t) v(t) a(t)
r(t) = [x(t),y(t),z(t)]
v(t) = diff(r(t),t)
a(t) = diff(v(t),t)

 

It is essential to type in the (t) after x,y,and z MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqacKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E8@  if you don’t do this, MAPLE assumes that these variables are constants, and takes their derivative to be zero.  You must enter (t) after _any_ variable that changes with time.

 

Here’s how you would do the circular motion calculation if you only know that the angle θ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUdehaaa@38B3@  is some arbitrary function of time, but don’t know what the function is

syms R theta(t) r(t) v(t) a(t)
r(t) = [R*cos(theta(t)),R*sin(theta(t))]
v(t) = simplify(diff(r(t),t))
a(t) = simplify(diff(v(t),t))

 

MATLAB can make very long and complicated calculations fairly painless.  It is a godsend to engineers, who generally find that every real-world problem they need to solve is long and complicated.  But of course it’s important to know what the program is doing MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  so keep taking those math classes…

 

 

 

 

3.1.5 Velocity and acceleration in normal-tangential coordinates.

 

In some cases it is helpful to use special basis vectors to write down velocity and acceleration vectors, instead of a fixed {i,j,k} basis.  If you see that this approach can be used to quickly solve a problem MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  go ahead and use it.  If not, just use Cartesian coordinates MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  this will always work, and with Matlab is not very hard. 

 

Normal-tangential coordinates for particles moving along a prescribed planar path

                         

If a particle moves along some known path, the formulas for velocity and acceleration look much simpler if they are expressed as components using a basis oriented parallel and perpendicular to the path.  These basis vectors are called ‘normal-tangential’ coordinates.

 

For example, normal-tangential coordinates are nearly always used in vehicle dynamics problems, because the {t,n,k} MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaacUhacaWH0b Gaaiilaiaah6gacaGGSaGaaC4Aaiaac2haaaa@3C82@  directions point ‘forwards’ ‘sideways’ and ‘vertically’, so it is easy to understand the significance of accelerations along {t,n,k} MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaacUhacaWH0b Gaaiilaiaah6gacaGGSaGaaC4Aaiaac2haaaa@3C82@ , while {i,j,k} components are generally very difficult to interpret.

 

To use normal-tangential coordinates we

·         Specify the path by writing down the position vector of a point on the path in terms of the distance s travelled along the path.   For a 2D curve:

r(s)=x(s)i+y(s)j MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhacaGGOa Gaam4CaiaacMcacqGH9aqpcaWG4bGaaiikaiaadohacaGGPaGaaCyA aiabgUcaRiaadMhacaGGOaGaam4CaiaacMcacaWHQbaaaa@43F0@

·       Introduce two unit vectors n and t, with t pointing tangent to the path and n pointing normal to the path, towards the center of curvature (this sounds a bit scary, but n is just perpendicular to t, and if the path curves to the right n points to the right; if it curves to the left, it points to the left).

·       Introduce the radius of curvature of the path R (in most problems we solve R is given, but we’ll give some formulas later).

·       Denote the speed of the particle by V.    The speed can vary with time.

 

We then use the following formulas to calculate speed, velocity and acceleration

v=Vt a= dV dt t+ V 2 R n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWH2b Gaeyypa0JaamOvaiaahshaaeaacaWHHbGaeyypa0ZaaSaaaeaacaWG KbGaamOvaaqaaiaadsgacaWG0baaaiaahshacqGHRaWkdaWcaaqaai aadAfadaahaaWcbeqaaiaaikdaaaaakeaacaWGsbaaaiaah6gaaaaa @4613@

We can also use the formula V=ds/dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGwbGaeyypa0 JaamizaiaadohacaGGVaGaamizaiaadshaaaa@3BCD@  to write velocity and acceleration in terms of distance traveled along the path and its time derivatives

v= ds dt t a= d 2 s d t 2 t+ 1 R ( ds dt ) 2 n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWH2b Gaeyypa0ZaaSaaaeaacaWGKbGaam4CaaqaaiaadsgacaWG0baaaiaa hshaaeaacaWHHbGaeyypa0ZaaSaaaeaacaWGKbWaaWbaaSqabeaaca aIYaaaaOGaam4CaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaa aaaakiaahshacqGHRaWkdaWcaaqaaiaaigdaaeaacaWGsbaaamaabm aabaWaaSaaaeaacaWGKbGaam4CaaqaaiaadsgacaWG0baaaaGaayjk aiaawMcaamaaCaaaleqabaGaaGOmaaaakiaah6gaaaaa@504A@

 

In words, these equations tell us:

(1)   The direction of the velocity vector of a particle is tangent to its path. 

(2)   The magnitude of the velocity vector is equal to the speed.

(3)   The speed is the derivative of distance traveled with respect to time

(4)   The acceleration vector can be constructed by adding two components:

·         the component of acceleration tangent to the particle’s path is equal to dV/dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsgacaWGwb Gaai4laiaadsgacaWG0baaaa@3A93@ .  We call this the ‘tangential component’ of acceleration.

·         The component of acceleration perpendicular to the path (towards the center of curvature) is equal to V 2 /R MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfadaahaa WcbeqaaiaaikdaaaGccaGGVaGaamOuaaaa@3992@ .   We call this the ‘normal component’ of acceleration.

 

 

Deriving the normal-tangential formulas

In Newtonian physics, we have to start by defining an ‘inertial frame’, which (mathematically) is always a non-accelerating and non-rotating Cartesian {i,j,k} basis.   The x,y,z coordinates of a particle and their derivatives tell us the Newtonian definitions of position, velocity and acceleration.

 

As we will see with specific examples below, some problems can be simplified by replacing the x,y,z coordinates with some simpler set of coordinates (these might be angles, or distances, or some combination of both that describe our system).  Whenever we use new coordinated, we proceed by writing down the x,y,z coordinates in terms of our new ones.  We usually also replace the {i,j,k} basis with a new set of directions that are defined by our new coordinate system.  As we make small changes to our new coordinates, we will move around in space MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzagaeaa aaaaaaa8qacaWFtacaaa@3806@  the directions we move usually have some special significance, so it is helpful to write down all vector quantities of interest as components in the basis defined by these new directions.

 

This sounds very abstract, so let’s see how it works for normal-tangential coordinates.

 

For problems where a particle moves along a known path, we can always write down the position of a point on the path in terms of the distance travelled along the path.   For a 2D curve:

r(s)=x(s)i+y(s)j MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhacaGGOa Gaam4CaiaacMcacqGH9aqpcaWG4bGaaiikaiaadohacaGGPaGaaCyA aiabgUcaRiaadMhacaGGOaGaam4CaiaacMcacaWHQbaaaa@43F0@

Here s is the arc-length traveled along the path.   Since the path is known (x,y are given functions) we only need one coordinate (s) to specify where we are.

 

We now generate some basis vectors by working out how position vector changes if we make small changes to our new coordinate s.  It is convenient to define

t= dr ds n=R dt ds R= 1 dt ds dt ds = 1 ( d 2 x d s 2 ) 2 + ( d 2 y d s 2 ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahshacqGH9a qpdaWcaaqaaiaadsgacaWHYbaabaGaamizaiaadohaaaGaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaCOBaiabg2da9iaa dkfadaWcaaqaaiaadsgacaWH0baabaGaamizaiaadohaaaGaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaamOuaiabg2da9maalaaabaGaaGymaaqaam aakaaabaWaaSaaaeaacaWGKbGaaCiDaaqaaiaadsgacaWGZbaaaiaa ykW7cqGHflY1daWcaaqaaiaadsgacaWH0baabaGaamizaiaadohaaa aaleqaaOGaaGPaVdaacqGH9aqpdaWcaaqaaiaaigdaaeaadaGcaaqa amaabmaabaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaam iEaaqaaiaadsgacaWGZbWaaWbaaSqabeaacaaIYaaaaaaaaOGaayjk aiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgUcaRmaabmaabaWaaS aaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaamyEaaqaaiaadsga caWGZbWaaWbaaSqabeaacaaIYaaaaaaaaOGaayjkaiaawMcaamaaCa aaleqabaGaaGOmaaaaaeqaaaaaaaa@8F6D@

The vectors {t,n,k} MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaacUhacaWH0b Gaaiilaiaah6gacaGGSaGaaC4Aaiaac2haaaa@3C82@  have unit length, are mutually perpendicular, and therefore define a new Cartesian basis.   We can define any vector in this basis in the usual way.  

 

With these definitions we are now ready to derive our formulas for velocity and acceleration:

v= dr(s) dt = dr(s) ds ds dt =Vt a= dv dt = dV dt t+V dt dt = dV dt t+V dt ds ds dt = dV dt t+V n R V= dV dt t+ V 2 R n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaaCODai abg2da9maalaaabaGaamizaiaahkhacaGGOaGaam4CaiaacMcaaeaa caWGKbGaamiDaaaacqGH9aqpdaWcaaqaaiaadsgacaWHYbGaaiikai aadohacaGGPaaabaGaamizaiaadohaaaWaaSaaaeaacaWGKbGaam4C aaqaaiaadsgacaWG0baaaiabg2da9iaadAfacaWH0baabaGaaCyyai abg2da9maalaaabaGaamizaiaahAhaaeaacaWGKbGaamiDaaaacqGH 9aqpdaWcaaqaaiaadsgacaWGwbaabaGaamizaiaadshaaaGaaCiDai abgUcaRiaadAfadaWcaaqaaiaadsgacaWH0baabaGaamizaiaadsha aaGaeyypa0ZaaSaaaeaacaWGKbGaamOvaaqaaiaadsgacaWG0baaai aahshacqGHRaWkcaWGwbWaaSaaaeaacaWGKbGaaCiDaaqaaiaadsga caWGZbaaamaalaaabaGaamizaiaadohaaeaacaWGKbGaamiDaaaacq GH9aqpdaWcaaqaaiaadsgacaWGwbaabaGaamizaiaadshaaaGaaCiD aiabgUcaRiaadAfadaWcaaqaaiaah6gaaeaacaWGsbaaaiaadAfacq GH9aqpdaWcaaqaaiaadsgacaWGwbaabaGaamizaiaadshaaaGaaCiD aiabgUcaRmaalaaabaGaamOvamaaCaaaleqabaGaaGOmaaaaaOqaai aadkfaaaGaaCOBaaaaaa@80F8@

These are all just repeated applications of the chain rule….

 

 

Examples using normal-tangential coordinates

 

We’ll work through a few examples below that show how to work with normal-tangential coordinates.   These are all quite hard: you need to be able to use lots of basic ideas from vectors and calculus to be able to answer them.

 

Example: Circular motion in normal-tangential coordinates

 

We already analyzed circular motion as a special case.  We’ll revisit that example as an example of motion along a general curved path.  

 

The distance traveled around the circle is the arc length s.  

 

The arc-length formula gives θ=s/R MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXjabg2 da9iaadohacaGGVaGaamOuaaaa@3B78@  

 

Therefore r=Rcos s R i+Rsin s R j MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhacqGH9a qpcaWGsbGaci4yaiaac+gacaGGZbWaaSaaaeaacaWGZbaabaGaamOu aaaacaWHPbGaey4kaSIaamOuaiGacohacaGGPbGaaiOBamaalaaaba Gaam4CaaqaaiaadkfaaaGaaCOAaaaa@4619@   (just use trig)

 

The tangent vector is t= dr ds =sin s R i+cos s R j MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahshacqGH9a qpdaWcaaqaaiaadsgacaWHYbaabaGaamizaiaadohaaaGaeyypa0Ja eyOeI0Iaci4CaiaacMgacaGGUbWaaSaaaeaacaWGZbaabaGaamOuaa aacaWHPbGaey4kaSIaci4yaiaac+gacaGGZbWaaSaaaeaacaWGZbaa baGaamOuaaaacaWHQbaaaa@4A35@   (this agrees with our earlier formula)

 

The normal vector is n=R dt ds =cos s R isin s R j MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaah6gacqGH9a qpcaWGsbWaaSaaaeaacaWGKbGaaCiDaaqaaiaadsgacaWGZbaaaiab g2da9iabgkHiTiGacogacaGGVbGaai4CamaalaaabaGaam4Caaqaai aadkfaaaGaaCyAaiabgkHiTiGacohacaGGPbGaaiOBamaalaaabaGa am4CaaqaaiaadkfaaaGaaCOAaaaa@4B13@  (also the same as our earlier formula)

 

The formula says velocity vector is v= ds dt t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAhacqGH9a qpdaWcaaqaaiaadsgacaWGZbaabaGaamizaiaadshaaaGaaCiDaaaa @3D0F@   (the same as our earlier formula)

The formula for acceleration vector is a= d 2 s d t 2 t+ 1 R ( ds dt ) 2 n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahggacqGH9a qpdaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWGZbaabaGa amizaiaadshadaahaaWcbeqaaiaaikdaaaaaaOGaaCiDaiabgUcaRm aalaaabaGaaGymaaqaaiaadkfaaaWaaeWaaeaadaWcaaqaaiaadsga caWGZbaabaGaamizaiaadshaaaaacaGLOaGaayzkaaWaaWbaaSqabe aacaaIYaaaaOGaaCOBaaaa@48AA@   (also agrees with the earlier result)

 

 

Example (a simple exam problem):

 

A vehicle starts at rest at A and travels with constant tangential acceleration a t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadggadaWgaa WcbaGaamiDaaqabaaaaa@3845@  around a circular path.   To avoid skidding, the magnitude of the acceleration must not exceed μg MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeY7aTjaadE gaaaa@38DC@  , where μ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeY7aTbaa@37F0@  is the friction coefficient between tires and road, and g is the gravitational acceleration.   Find a formula for the shortest possible time to reach B.

 

We can use the formula: in normal-tangential coordinates

a= dV dt t+ V 2 R n= a t t+ V 2 R n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahggacqGH9a qpdaWcaaqaaiaadsgacaWGwbaabaGaamizaiaadshaaaGaaCiDaiab gUcaRmaalaaabaGaamOvamaaCaaaleqabaGaaGOmaaaaaOqaaiaadk faaaGaaCOBaiabg2da9iaadggadaWgaaWcbaGaamiDaaqabaGccaWH 0bGaey4kaSYaaSaaaeaacaWGwbWaaWbaaSqabeaacaaIYaaaaaGcba GaamOuaaaacaWHUbaaaa@4A11@

Since we are told a t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadggadaWgaa WcbaGaamiDaaqabaaaaa@3845@  is constant and the car is at rest at t=0 we can use the constant acceleration formula to find V

dV dt = a t V= a t t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam izaiaadAfaaeaacaWGKbGaamiDaaaacqGH9aqpcaWGHbWaaSbaaSqa aiaadshaaeqaaOGaeyO0H4TaamOvaiabg2da9iaadggadaWgaaWcba GaamiDaaqabaGccaWG0baaaa@4457@

Furthermore

ds dt = a t ts= 1 2 a t t 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam izaiaadohaaeaacaWGKbGaamiDaaaacqGH9aqpcaWGHbWaaSbaaSqa aiaadshaaeqaaOGaamiDaiabgkDiElaadohacqGH9aqpdaWcaaqaai aaigdaaeaacaaIYaaaaiaadggadaWgaaWcbaGaamiDaaqabaGccaWG 0bWaaWbaaSqabeaacaaIYaaaaaaa@47FA@

The arc-length from A to B is πR MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabec8aWjaadk faaaa@38CE@  so the time to travel from A to B follows as

T= 2πR/ a t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsfacqGH9a qpdaGcaaqaaiaaikdacqaHapaCcaWGsbGaai4laiaadggadaWgaaWc baGaamiDaaqabaaabeaaaaa@3E37@

We can find a t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadggadaWgaa WcbaGaamiDaaqabaaaaa@3845@  from the condition that | a |μg MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaemaabaGaaC yyaaGaay5bSlaawIa7aiabgsMiJkabeY7aTjaadEgaaaa@3E9D@  

| a |=| a t t+ V 2 R n |= a t 2 + V 4 R 2 = a t 2 + ( a t t ) 4 R 2 μg MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaemaabaGaaC yyaaGaay5bSlaawIa7aiabg2da9maaemaabaGaamyyamaaBaaaleaa caWG0baabeaakiaahshacqGHRaWkdaWcaaqaaiaadAfadaahaaWcbe qaaiaaikdaaaaakeaacaWGsbaaaiaah6gaaiaawEa7caGLiWoacqGH 9aqpdaGcaaqaaiaadggadaqhaaWcbaGaamiDaaqaaiaaikdaaaGccq GHRaWkdaWcaaqaaiaadAfadaahaaWcbeqaaiaaisdaaaaakeaacaWG sbWaaWbaaSqabeaacaaIYaaaaaaaaeqaaOGaeyypa0ZaaOaaaeaaca WGHbWaa0baaSqaaiaadshaaeaacaaIYaaaaOGaey4kaSYaaSaaaeaa daqadaqaaiaadggadaWgaaWcbaGaamiDaaqabaGccaWG0baacaGLOa GaayzkaaWaaWbaaSqabeaacaaI0aaaaaGcbaGaamOuamaaCaaaleqa baGaaGOmaaaaaaaabeaakiabgsMiJkabeY7aTjaadEgaaaa@5F09@

(notice that we just used the usual Cartesian formula to find the magnitude of the vector). The maximum acceleration occurs at B, so we can substitute for t and solve for a t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGHbWaaSbaaS qaaiaadshaaeqaaaaa@3781@  

| a |= a t 2 +4 π 2 a t 2 μg a t μg/ 1+4 π 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaWaaqWaae aacaWHHbaacaGLhWUaayjcSdGaeyypa0ZaaOaaaeaacaWGHbWaa0ba aSqaaiaadshaaeaacaaIYaaaaOGaey4kaSIaaGinaiabec8aWnaaCa aaleqabaGaaGOmaaaakiaadggadaqhaaWcbaGaamiDaaqaaiaaikda aaaabeaakiabgsMiJkabeY7aTjaadEgaaeaacqGHshI3caWGHbWaaS baaSqaaiaadshaaeqaaOGaeyizImQaeqiVd0Maam4zaiaac+cadaGc aaqaaiaaigdacqGHRaWkcaaI0aGaeqiWda3aaWbaaSqabeaacaaIYa aaaaqabaaaaaa@583B@

Finally remember T= 2πR/ a t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsfacqGH9a qpdaGcaaqaaiaaikdacqaHapaCcaWGsbGaai4laiaadggadaWgaaWc baGaamiDaaqabaaabeaaaaa@3E37@  so

T 2πR μg ( 1+4 π 2 ) 1/4 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsfacqGHLj YSdaGcaaqaamaalaaabaGaaGOmaiabec8aWjaadkfaaeaacqaH8oqB caWGNbaaaaWcbeaakmaabmaabaGaaGymaiabgUcaRiaaisdacqaHap aCdaahaaWcbeqaaiaaikdaaaaakiaawIcacaGLPaaadaahaaWcbeqa aiaaigdacaGGVaGaaGinaaaaaaa@47ED@

 

 

Example: Design speed limit for a curvy road:  As a consulting firm specializing in highway design, we have been asked to develop a design formula that can be used to calculate the speed limit for cars that travel along a curvy road.

 

The following procedure will be used:

·         The curvy road will be approximated as a sine wave y=Asin(2πx/L) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 da9iaadgeaciGGZbGaaiyAaiaac6gacaGGOaGaaGOmaiabec8aWjaa dIhacaGGVaGaamitaiaacMcaaaa@41EB@  as shown in the figure MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  for a given road, engineers will measure values of A and L that fit the path.

·         Vehicles will be assumed to travel at constant speed V around the path MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  your mission is to calculate the maximum allowable value of V

·         For safety, the magnitude of the acceleration of the car at any point along the path must be less than 0.2g, where g is the gravitational acceleration. (Again, note that constant speed does not mean constant acceleration, because the car’s direction is changing with time).

 

Our goal, then, is to calculate a formula for the magnitude of the acceleration in terms of V, A and L.  The result can be used to deduce a formula for the speed limit.

 

 

 

Calculation:

 

We can solve this problem quickly using normal-tangential coordinates.  Since the speed is constant, the acceleration vector is

a= V 2 R n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabg2 da9maalaaabaGaamOvamaaCaaaleqabaGaaGOmaaaaaOqaaiaadkfa aaGaaCOBaaaa@3C9A@

Our only problem is that we don’t know R MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqacKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E8@  but we can use what we know about vectors and normal-tangential coordinates to figure it out.

 

The position vector is

r=xi+Asin(2πx/L)j MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadIhacaWHPbGaey4kaSIaamyqaiGacohacaGGPbGaaiOBaiaa cIcacaaIYaGaeqiWdaNaamiEaiaac+cacaWGmbGaaiykaiaahQgaaa a@46B4@ ,

where x is some unknown function of distance s traveled along the path.  We can calculate the tangent from the formula for the tangent (and the chain rule)

t= dr ds = dr dx dx ds =( i+ 2π L Acos( 2πx/L )j ) dx ds MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahshacqGH9a qpdaWcaaqaaiaadsgacaWHYbaabaGaamizaiaadohaaaGaeyypa0Za aSaaaeaacaWGKbGaaCOCaaqaaiaadsgacaWG4baaamaalaaabaGaam izaiaadIhaaeaacaWGKbGaam4CaaaacqGH9aqpdaqadaqaaiaahMga cqGHRaWkdaWcaaqaaiaaikdacqaHapaCaeaacaWGmbaaaiaadgeaci GGJbGaai4BaiaacohadaqadaqaaiaaikdacqaHapaCcaWG4bGaai4l aiaadYeaaiaawIcacaGLPaaacaWHQbaacaGLOaGaayzkaaWaaSaaae aacaWGKbGaamiEaaqaaiaadsgacaWGZbaaaaaa@5B6C@

We know that t must be a unit vector, therefore

tt=1( 1+ 4 π 2 L 2 A 2 cos 2 ( 2πx/L ) ) ( dx ds ) 2 dx ds = ( 1+ 4 π 2 L 2 A 2 cos 2 ( 2πx/L ) ) 1/2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaaCiDai abgwSixlaahshacqGH9aqpcaaIXaGaeyO0H49aaeWaaeaacaaIXaGa ey4kaSYaaSaaaeaacaaI0aGaeqiWda3aaWbaaSqabeaacaaIYaaaaa GcbaGaamitamaaCaaaleqabaGaaGOmaaaaaaGccaWGbbWaaWbaaSqa beaacaaIYaaaaOGaci4yaiaac+gacaGGZbWaaWbaaSqabeaacaaIYa aaaOWaaeWaaeaacaaIYaGaeqiWdaNaamiEaiaac+cacaWGmbaacaGL OaGaayzkaaaacaGLOaGaayzkaaWaaeWaaeaadaWcaaqaaiaadsgaca WG4baabaGaamizaiaadohaaaaacaGLOaGaayzkaaWaaWbaaSqabeaa caaIYaaaaaGcbaGaeyO0H49aaSaaaeaacaWGKbGaamiEaaqaaiaads gacaWGZbaaaiabg2da9maabmaabaGaaGymaiabgUcaRmaalaaabaGa aGinaiabec8aWnaaCaaaleqabaGaaGOmaaaaaOqaaiaadYeadaahaa WcbeqaaiaaikdaaaaaaOGaamyqamaaCaaaleqabaGaaGOmaaaakiGa cogacaGGVbGaai4CamaaCaaaleqabaGaaGOmaaaakmaabmaabaGaaG Omaiabec8aWjaadIhacaGGVaGaamitaaGaayjkaiaawMcaaaGaayjk aiaawMcaamaaCaaaleqabaGaeyOeI0IaaGymaiaac+cacaaIYaaaaa aaaa@7848@

We could separate variables and integrate this to calculate x(s) if we need it MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  but in practice we don’t need to bother.

 

So now we know that

t= ( i+ 2π L Acos( 2πx/L )j ) ( 1+ 4 π 2 L 2 A 2 cos 2 ( 2πx/L ) ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahshacqGH9a qpdaWcaaqaamaabmaabaGaaCyAaiabgUcaRmaalaaabaGaaGOmaiab ec8aWbqaaiaadYeaaaGaamyqaiGacogacaGGVbGaai4Camaabmaaba GaaGOmaiabec8aWjaadIhacaGGVaGaamitaaGaayjkaiaawMcaaiaa hQgaaiaawIcacaGLPaaaaeaadaGcaaqaamaabmaabaGaaGymaiabgU caRmaalaaabaGaaGinaiabec8aWnaaCaaaleqabaGaaGOmaaaaaOqa aiaadYeadaahaaWcbeqaaiaaikdaaaaaaOGaamyqamaaCaaaleqaba GaaGOmaaaakiGacogacaGGVbGaai4CamaaCaaaleqabaGaaGOmaaaa kmaabmaabaGaaGOmaiabec8aWjaadIhacaGGVaGaamitaaGaayjkai aawMcaaaGaayjkaiaawMcaaaWcbeaaaaaaaa@5E98@

The normal vector n must be perpendicular to both t and k, so we can create it using

n=±k×t=±k× ( i+ 2π L Acos( 2πx/L )j ) ( 1+ 4 π 2 L 2 A 2 cos 2 ( 2πx/L ) ) =± ( 2π L Acos( 2πx/L )i+j ) ( 1+ 4 π 2 L 2 A 2 cos 2 ( 2πx/L ) ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaah6gacqGH9a qpcqGHXcqScaWHRbGaey41aqRaaCiDaiabg2da9iabgglaXkaahUga cqGHxdaTdaWcaaqaamaabmaabaGaaCyAaiabgUcaRmaalaaabaGaaG Omaiabec8aWbqaaiaadYeaaaGaamyqaiGacogacaGGVbGaai4Camaa bmaabaGaaGOmaiabec8aWjaadIhacaGGVaGaamitaaGaayjkaiaawM caaiaahQgaaiaawIcacaGLPaaaaeaadaGcaaqaamaabmaabaGaaGym aiabgUcaRmaalaaabaGaaGinaiabec8aWnaaCaaaleqabaGaaGOmaa aaaOqaaiaadYeadaahaaWcbeqaaiaaikdaaaaaaOGaamyqamaaCaaa leqabaGaaGOmaaaakiGacogacaGGVbGaai4CamaaCaaaleqabaGaaG OmaaaakmaabmaabaGaaGOmaiabec8aWjaadIhacaGGVaGaamitaaGa ayjkaiaawMcaaaGaayjkaiaawMcaaaWcbeaaaaGccqGH9aqpcqGHXc qSdaWcaaqaamaabmaabaGaeyOeI0YaaSaaaeaacaaIYaGaeqiWdaha baGaamitaaaacaWGbbGaci4yaiaac+gacaGGZbWaaeWaaeaacaaIYa GaeqiWdaNaamiEaiaac+cacaWGmbaacaGLOaGaayzkaaGaaCyAaiab gUcaRiaahQgaaiaawIcacaGLPaaaaeaadaGcaaqaamaabmaabaGaaG ymaiabgUcaRmaalaaabaGaaGinaiabec8aWnaaCaaaleqabaGaaGOm aaaaaOqaaiaadYeadaahaaWcbeqaaiaaikdaaaaaaOGaamyqamaaCa aaleqabaGaaGOmaaaakiGacogacaGGVbGaai4CamaaCaaaleqabaGa aGOmaaaakmaabmaabaGaaGOmaiabec8aWjaadIhacaGGVaGaamitaa GaayjkaiaawMcaaaGaayjkaiaawMcaaaWcbeaaaaaaaa@94CD@

 

(the ± MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgglaXcaa@3828@  is because there are two vectors normal to both t and k). Finally we know that

n=R dt ds = ( 4 π 2 L 2 Asin( 2πx/L )j ) ( 1+ 4 π 2 L 2 A 2 cos 2 ( 2πx/L ) ) dx ds +( i+ 2π L Acos( 2πx/L )j ) d ds { 1 ( 1+ 4 π 2 L 2 A 2 cos 2 ( 2πx/L ) ) } MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaah6gacqGH9a qpcaWGsbWaaSaaaeaacaWGKbGaaCiDaaqaaiaadsgacaWGZbaaaiab g2da9maalaaabaWaaeWaaeaacqGHsisldaWcaaqaaiaaisdacqaHap aCdaahaaWcbeqaaiaaikdaaaaakeaacaWGmbWaaWbaaSqabeaacaaI YaaaaaaakiaadgeaciGGZbGaaiyAaiaac6gadaqadaqaaiaaikdacq aHapaCcaWG4bGaai4laiaadYeaaiaawIcacaGLPaaacaWHQbaacaGL OaGaayzkaaaabaWaaOaaaeaadaqadaqaaiaaigdacqGHRaWkdaWcaa qaaiaaisdacqaHapaCdaahaaWcbeqaaiaaikdaaaaakeaacaWGmbWa aWbaaSqabeaacaaIYaaaaaaakiaadgeadaahaaWcbeqaaiaaikdaaa GcciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaikdaaaGcdaqadaqa aiaaikdacqaHapaCcaWG4bGaai4laiaadYeaaiaawIcacaGLPaaaai aawIcacaGLPaaaaSqabaaaaOWaaSaaaeaacaWGKbGaamiEaaqaaiaa dsgacaWGZbaaaiabgUcaRmaabmaabaGaaCyAaiabgUcaRmaalaaaba GaaGOmaiabec8aWbqaaiaadYeaaaGaamyqaiGacogacaGGVbGaai4C amaabmaabaGaaGOmaiabec8aWjaadIhacaGGVaGaamitaaGaayjkai aawMcaaiaahQgaaiaawIcacaGLPaaadaWcaaqaaiaadsgaaeaacaWG KbGaam4CaaaadaGadaqaamaalaaabaGaaGymaaqaamaakaaabaWaae WaaeaacaaIXaGaey4kaSYaaSaaaeaacaaI0aGaeqiWda3aaWbaaSqa beaacaaIYaaaaaGcbaGaamitamaaCaaaleqabaGaaGOmaaaaaaGcca WGbbWaaWbaaSqabeaacaaIYaaaaOGaci4yaiaac+gacaGGZbWaaWba aSqabeaacaaIYaaaaOWaaeWaaeaacaaIYaGaeqiWdaNaamiEaiaac+ cacaWGmbaacaGLOaGaayzkaaaacaGLOaGaayzkaaaaleqaaaaaaOGa ay5Eaiaaw2haaaaa@963A@

Since n is a unit vector, we can take the dot product of both sides of this expression with n (from above)

nn=1=±R ( 4 π 2 L 2 Asin( 2πx/L )j ) ( 1+ 4 π 2 L 2 A 2 cos 2 ( 2πx/L ) ) dx ds MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaah6gacqGHfl Y1caWHUbGaeyypa0JaaGymaiabg2da9iabgglaXkaadkfadaWcaaqa amaabmaabaGaeyOeI0YaaSaaaeaacaaI0aGaeqiWda3aaWbaaSqabe aacaaIYaaaaaGcbaGaamitamaaCaaaleqabaGaaGOmaaaaaaGccaWG bbGaci4CaiaacMgacaGGUbWaaeWaaeaacaaIYaGaeqiWdaNaamiEai aac+cacaWGmbaacaGLOaGaayzkaaGaaCOAaaGaayjkaiaawMcaaaqa amaabmaabaGaaGymaiabgUcaRmaalaaabaGaaGinaiabec8aWnaaCa aaleqabaGaaGOmaaaaaOqaaiaadYeadaahaaWcbeqaaiaaikdaaaaa aOGaamyqamaaCaaaleqabaGaaGOmaaaakiGacogacaGGVbGaai4Cam aaCaaaleqabaGaaGOmaaaakmaabmaabaGaaGOmaiabec8aWjaadIha caGGVaGaamitaaGaayjkaiaawMcaaaGaayjkaiaawMcaaaaadaWcaa qaaiaadsgacaWG4baabaGaamizaiaadohaaaaaaa@6B1B@

Therefore (using our earlier expression for dx/ds MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsgacaWG4b Gaai4laiaadsgacaWGZbaaaa@3AB4@ , and noting that R is positive by definition)

1 R = ( 4 π 2 L 2 Asin( 2πx/L )j ) ( 1+ 4 π 2 L 2 A 2 cos 2 ( 2πx/L ) ) 3/2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG ymaaqaaiaadkfaaaGaeyypa0ZaaSaaaeaadaqadaqaamaalaaabaGa aGinaiabec8aWnaaCaaaleqabaGaaGOmaaaaaOqaaiaadYeadaahaa WcbeqaaiaaikdaaaaaaOGaamyqaiGacohacaGGPbGaaiOBamaabmaa baGaaGOmaiabec8aWjaadIhacaGGVaGaamitaaGaayjkaiaawMcaai aahQgaaiaawIcacaGLPaaaaeaadaqadaqaaiaaigdacqGHRaWkdaWc aaqaaiaaisdacqaHapaCdaahaaWcbeqaaiaaikdaaaaakeaacaWGmb WaaWbaaSqabeaacaaIYaaaaaaakiaadgeadaahaaWcbeqaaiaaikda aaGcciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaikdaaaGcdaqada qaaiaaikdacqaHapaCcaWG4bGaai4laiaadYeaaiaawIcacaGLPaaa aiaawIcacaGLPaaadaahaaWcbeqaaiaaiodacaGGVaGaaGOmaaaaaa aaaa@6194@

So now we know the acceleration vector is

 

a= A (2πV/L) 2 sin(2πx/L) ( 1+ 4 π 2 L 2 A 2 cos 2 ( 2πx/L ) ) 3/2 n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabg2 da9maalaaabaGaamyqaiaacIcacaaIYaGaeqiWdaNaamOvaiaac+ca caWGmbGaaiykamaaCaaaleqabaGaaGOmaaaakiGacohacaGGPbGaai OBaiaacIcacaaIYaGaeqiWdaNaamiEaiaac+cacaWGmbGaaiykaaqa amaabmaabaGaaGymaiabgUcaRmaalaaabaGaaGinaiabec8aWnaaCa aaleqabaGaaGOmaaaaaOqaaiaadYeadaahaaWcbeqaaiaaikdaaaaa aOGaamyqamaaCaaaleqabaGaaGOmaaaakiGacogacaGGVbGaai4Cam aaCaaaleqabaGaaGOmaaaakmaabmaabaGaaGOmaiabec8aWjaadIha caGGVaGaamitaaGaayjkaiaawMcaaaGaayjkaiaawMcaamaaCaaale qabaGaaG4maiaac+cacaaIYaaaaaaakiaah6gaaaa@60CF@

We are interested in the magnitude of the acceleration… 

| a |= A (2πV/L) 2 sin(2πx/L) ( 1+ 4 π 2 L 2 A 2 cos 2 ( 2πx/L ) ) 3/2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca WHHbaacaGLhWUaayjcSdGaeyypa0ZaaSaaaeaacaWGbbGaaiikaiaa ikdacqaHapaCcaWGwbGaai4laiaadYeacaGGPaWaaWbaaSqabeaaca aIYaaaaOGaci4CaiaacMgacaGGUbGaaiikaiaaikdacqaHapaCcaWG 4bGaai4laiaadYeacaGGPaaabaWaaeWaaeaacaaIXaGaey4kaSYaaS aaaeaacaaI0aGaeqiWda3aaWbaaSqabeaacaaIYaaaaaGcbaGaamit amaaCaaaleqabaGaaGOmaaaaaaGccaWGbbWaaWbaaSqabeaacaaIYa aaaOGaci4yaiaac+gacaGGZbWaaWbaaSqabeaacaaIYaaaaOWaaeWa aeaacaaIYaGaeqiWdaNaamiEaiaac+cacaWGmbaacaGLOaGaayzkaa aacaGLOaGaayzkaaWaaWbaaSqabeaacaaIZaGaai4laiaaikdaaaaa aaaa@62F0@

We see from this that the car has the biggest acceleration when x=L/2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 da9iaadYeacaGGVaGaaGOmaaaa@3A39@ . The maximum acceleration follows as

a max =A (2πV/L) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyamaaBa aaleaaciGGTbGaaiyyaiaacIhaaeqaaOGaeyypa0JaamyqaiaacIca caaIYaGaeqiWdaNaamOvaiaac+cacaWGmbGaaiykamaaCaaaleqaba GaaGOmaaaaaaa@42CC@

The formula for the speed limit is therefore V<(L/2π) 0.2g/A MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabgY da8iaacIcacaWGmbGaai4laiaaikdacqaHapaCcaGGPaWaaOaaaeaa caaIWaGaaiOlaiaaikdacaWGNbGaai4laiaadgeaaSqabaaaaa@41D3@

 

Now we send in a bill for a big consulting fee…

 

 

3.1.6 Position, Velocity and Acceleration in cylindrical-polar coordinates.

 

When solving problems involving central forces (forces that attract particles towards a fixed point) it is often convenient to describe motion using polar coordinates.

 

Polar coordinate formulas

 

Polar coordinates are related to x,y coordinates through

r= x 2 + y 2 θ= tan 1 (y/x) x=rcosθy=rsinθ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGYb Gaeyypa0ZaaOaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4k aSIaamyEamaaCaaaleqabaGaaGOmaaaaaeqaaOGaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaeqiUdeNaeyypa0JaciiDaiaacggacaGGUbWaaWbaaS qabeaacqGHsislcaaIXaaaaOGaaiikaiaadMhacaGGVaGaamiEaiaa cMcaaeaacaWG4bGaeyypa0JaamOCaiGacogacaGGVbGaai4CaiabeI 7aXjaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaamyEaiabg2da9iaadkhaciGGZbGaaiyAaiaac6gacqaH4oqCaa aa@7730@

We can also specify height out of the plane of the picture using the usual z coordinate.

 

Suppose that the position of a particle is specified by its ‘polar coordinates’ (r,θ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaacIcacaWGYb GaaiilaiabeI7aXjaacMcaaaa@3AF0@  relative to a fixed origin, as shown in the figure. Let e r MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahwgadaWgaa WcbaGaamOCaaqabaaaaa@384B@  be a unit vector pointing in the radial direction, and let e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahwgadaWgaa WcbaGaeqiUdehabeaaaaa@390A@  be a unit vector pointing in the tangential direction, i.e

e r =cosθi+sinθj e θ =sinθi+cosθj MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaaCyzam aaBaaaleaacaWGYbaabeaakiabg2da9iGacogacaGGVbGaai4Caiab eI7aXjaahMgacqGHRaWkciGGZbGaaiyAaiaac6gacqaH4oqCcaWHQb aabaGaaCyzamaaBaaaleaacqaH4oqCaeqaaOGaeyypa0JaeyOeI0Ia ci4CaiaacMgacaGGUbGaeqiUdeNaaCyAaiabgUcaRiGacogacaGGVb Gaai4CaiabeI7aXjaahQgaaaaa@55EB@

The position, velocity and acceleration of the particle can then be expressed as

r=r e r +zk v= dr dt e r +r dθ dt e θ + dz dt k a=( d 2 r d t 2 r ( dθ dt ) 2 ) e r +( r d 2 θ d t 2 +2 dr dt dθ dt ) e θ + d 2 z d t 2 k MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaaCOCai abg2da9iaadkhacaWHLbWaaSbaaSqaaiaadkhaaeqaaOGaey4kaSIa amOEaiaahUgaaeaacaWH2bGaeyypa0ZaaSaaaeaacaWGKbGaamOCaa qaaiaadsgacaWG0baaaiaahwgadaWgaaWcbaGaamOCaaqabaGccqGH RaWkcaWGYbWaaSaaaeaacaWGKbGaeqiUdehabaGaamizaiaadshaaa GaaCyzamaaBaaaleaacqaH4oqCaeqaaOGaey4kaSYaaSaaaeaacaWG KbGaamOEaaqaaiaadsgacaWG0baaaiaahUgaaeaacaWHHbGaeyypa0 ZaaeWaaeaadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG YbaabaGaamizaiaadshadaahaaWcbeqaaiaaikdaaaaaaOGaeyOeI0 IaamOCamaabmaabaWaaSaaaeaacaWGKbGaeqiUdehabaGaamizaiaa dshaaaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGccaGLOa GaayzkaaGaaCyzamaaBaaaleaacaWGYbaabeaakiabgUcaRmaabmaa baGaamOCamaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiabeI 7aXbqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaakiabgUca RiaaikdadaWcaaqaaiaadsgacaWGYbaabaGaamizaiaadshaaaWaaS aaaeaacaWGKbGaeqiUdehabaGaamizaiaadshaaaaacaGLOaGaayzk aaGaaCyzamaaBaaaleaacqaH4oqCaeqaaOGaey4kaSYaaSaaaeaaca WGKbWaaWbaaSqabeaacaaIYaaaaOGaamOEaaqaaiaadsgacaWG0bWa aWbaaSqabeaacaaIYaaaaaaakiaahUgaaaaa@8797@

In most problems we solve, we just substitute known information about (r,θ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaacIcacaWGYb GaaiilaiabeI7aXjaacMcaaaa@3AF0@  into these formulas.

 

Deriving the polar coordinate formulas

 

The formulas for polar coordinate can be derived using the same ideas we used to set up normal-tangential coordinates.   The general procedure to set up any new coordinate system in Newtonian physics is:

(1)   Choose an inertial frame MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  this defines a stationary Cartesian {i,j,k} MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaGG7bGaaCyAai aacYcacaWHQbGaaiilaiaahUgacaGG9baaaa@3BAF@  basis

(2)   Choose some new coordinates MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  here we use (r,θ,z) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaGGOaGaamOCai aacYcacqaH4oqCcaGGSaGaamOEaiaacMcaaaa@3BDB@  and write down position vector in the inertial frame in terms of these new coordinates.   For the cylindrical-polar system

r=rcosθi+rsinθj+zk MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHYbGaeyypa0 JaamOCaiGacogacaGGVbGaai4CaiabeI7aXjaahMgacqGHRaWkcaWG YbGaci4CaiaacMgacaGGUbGaeqiUdeNaaCOAaiabgUcaRiaadQhaca WHRbaaaa@4818@

(3)   We now think about making very small changes to each of r,θ,z MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGYbGaaiilai abeI7aXjaacYcacaWG6baaaa@3A82@  .  As we do so, r will change.   We define unit vectors that point in each of the direction associated with making changes to r,θ,z MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGYbGaaiilai abeI7aXjaacYcacaWG6baaaa@3A82@ : mathematically this operation is

e r = 1 | dr/dr | dr dr =cosθi+sinθj e θ = 1 | dr/dθ | dr dθ = rsinθi+rcosθj r =sinθi+cosθj e z = 1 | dr/dz | dr dz =k MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaahwgada WgaaWcbaGaamOCaaqabaGccqGH9aqpdaWcaaqaaiaaigdaaeaadaab daqaaiaadsgacaWHYbGaai4laiaadsgacaWGYbaacaGLhWUaayjcSd aaamaalaaabaGaamizaiaahkhaaeaacaWGKbGaamOCaaaacqGH9aqp ciGGJbGaai4BaiaacohacqaH4oqCcaWHPbGaey4kaSIaci4CaiaacM gacaGGUbGaeqiUdeNaaCOAaaqaaiaahwgadaWgaaWcbaGaeqiUdeha beaakiabg2da9maalaaabaGaaGymaaqaamaaemaabaGaamizaiaahk hacaGGVaGaamizaiabeI7aXbGaay5bSlaawIa7aaaadaWcaaqaaiaa dsgacaWHYbaabaGaamizaiabeI7aXbaacqGH9aqpdaWcaaqaaiabgk HiTiaadkhaciGGZbGaaiyAaiaac6gacqaH4oqCcaWHPbGaey4kaSIa amOCaiGacogacaGGVbGaai4CaiabeI7aXjaahQgaaeaacaWGYbaaai abg2da9iabgkHiTiGacohacaGGPbGaaiOBaiabeI7aXjaahMgacqGH RaWkciGGJbGaai4BaiaacohacqaH4oqCcaWHQbaabaGaaCyzamaaBa aaleaacaWG6baabeaakiabg2da9maalaaabaGaaGymaaqaamaaemaa baGaamizaiaahkhacaGGVaGaamizaiaadQhaaiaawEa7caGLiWoaaa WaaSaaaeaacaWGKbGaaCOCaaqaaiaadsgacaWG6baaaiabg2da9iaa hUgaaaaa@9353@

For the polar coordinate system it turns out that { e r , e θ ,k} MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaGG7bGaaCyzam aaBaaaleaacaWGYbaabeaakiaacYcacaWHLbWaaSbaaSqaaiabeI7a XbqabaGccaGGSaGaaC4Aaiaac2haaaa@3EBF@  are mutually perpendicular and so are a Cartesian basis.   But note that the directions { e r , e θ ,k} MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaGG7bGaaCyzam aaBaaaleaacaWGYbaabeaakiaacYcacaWHLbWaaSbaaSqaaiabeI7a XbqabaGccaGGSaGaaC4Aaiaac2haaaa@3EBF@  are functions of r,θ,z MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGYbGaaiilai abeI7aXjaacYcacaWG6baaaa@3A82@ .   Regardless, we can express any vectors as components in our new basis using the usual ideas MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  the vector will be made up of contributions parallel to each of the new basis vectors.

 

The polar coordinate formulas now follow by simple calculus.  Since we have chosen to work with r,θ,z MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGYbGaaiilai abeI7aXjaacYcacaWG6baaaa@3A82@  instead of x,y,z , and { e r , e θ ,k} MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaGG7bGaaCyzam aaBaaaleaacaWGYbaabeaakiaacYcacaWHLbWaaSbaaSqaaiabeI7a XbqabaGccaGGSaGaaC4Aaiaac2haaaa@3EBF@  instead of  {i,j,k} MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaGG7bGaaCyAai aacYcacaWHQbGaaiilaiaahUgacaGG9baaaa@3BAF@ , we would like to write down position, velocity and acceleration in the { e r , e θ ,k} MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaGG7bGaaCyzam aaBaaaleaacaWGYbaabeaakiaacYcacaWHLbWaaSbaaSqaaiabeI7a XbqabaGccaGGSaGaaC4Aaiaac2haaaa@3EBF@  basis, terms of time derivatives of  r,θ,z MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGYbGaaiilai abeI7aXjaacYcacaWG6baaaa@3A82@ .  Before we work through the details we have to do some busy-work.   Since { e r , e θ ,k} MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaGG7bGaaCyzam aaBaaaleaacaWGYbaabeaakiaacYcacaWHLbWaaSbaaSqaaiabeI7a XbqabaGccaGGSaGaaC4Aaiaac2haaaa@3EBF@  are functions of r,θ,z MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGYbGaaiilai abeI7aXjaacYcacaWG6baaaa@3A82@  we will need to know how to differentiate them with respect to r,θ,z MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGYbGaaiilai abeI7aXjaacYcacaWG6baaaa@3A82@ To do this we go back to their original definitions:

e r =cosθi+sinθj d e r dr = d e r dz =0 d e r dθ =sinθi+cosθj= e θ e θ =sinθi+cosθj d e θ dr = d e θ dz =0 d e θ dθ =cosθisinθj= e r e z =k d e z dr = d e z dθ = d e z dz =0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaahwgada WgaaWcbaGaamOCaaqabaGccqGH9aqpciGGJbGaai4BaiaacohacqaH 4oqCcaWHPbGaey4kaSIaci4CaiaacMgacaGGUbGaeqiUdeNaaCOAai abgkDiEpaalaaabaGaamizaiaahwgadaWgaaWcbaGaamOCaaqabaaa keaacaWGKbGaamOCaaaacqGH9aqpdaWcaaqaaiaadsgacaWHLbWaaS baaSqaaiaadkhaaeqaaaGcbaGaamizaiaadQhaaaGaeyypa0JaaCim aiaaykW7caaMc8UaaGPaVlaaykW7daWcaaqaaiaadsgacaWHLbWaaS baaSqaaiaadkhaaeqaaaGcbaGaamizaiabeI7aXbaacqGH9aqpcqGH sislciGGZbGaaiyAaiaac6gacqaH4oqCcaWHPbGaey4kaSIaci4yai aac+gacaGGZbGaeqiUdeNaaCOAaiabg2da9iaahwgadaWgaaWcbaGa eqiUdehabeaaaOqaaiaahwgadaWgaaWcbaGaeqiUdehabeaakiabg2 da9iabgkHiTiGacohacaGGPbGaaiOBaiabeI7aXjaahMgacqGHRaWk ciGGJbGaai4BaiaacohacqaH4oqCcaWHQbGaeyO0H49aaSaaaeaaca WGKbGaaCyzamaaBaaaleaacqaH4oqCaeqaaaGcbaGaamizaiaadkha aaGaeyypa0ZaaSaaaeaacaWGKbGaaCyzamaaBaaaleaacqaH4oqCae qaaaGcbaGaamizaiaadQhaaaGaeyypa0JaaCimaiaaykW7caaMc8Ua aGPaVlaaykW7daWcaaqaaiaadsgacaWHLbWaaSbaaSqaaiabeI7aXb qabaaakeaacaWGKbGaeqiUdehaaiabg2da9iabgkHiTiGacogacaGG VbGaai4CaiabeI7aXjaahMgacqGHsislciGGZbGaaiyAaiaac6gacq aH4oqCcaWHQbGaeyypa0JaeyOeI0IaaCyzamaaBaaaleaacaWGYbaa beaaaOqaaiaahwgadaWgaaWcbaGaamOEaaqabaGccqGH9aqpcaWHRb GaeyO0H49aaSaaaeaacaWGKbGaaCyzamaaBaaaleaacaWG6baabeaa aOqaaiaadsgacaWGYbaaaiabg2da9maalaaabaGaamizaiaahwgada WgaaWcbaGaamOEaaqabaaakeaacaWGKbGaeqiUdehaaiabg2da9maa laaabaGaamizaiaahwgadaWgaaWcbaGaamOEaaqabaaakeaacaWGKb GaamOEaaaacqGH9aqpcaWHWaaaaaa@CAE2@

 

The rest is just a tedious exercise in using the chain rule.  The position vector can be written down by inspection as r=r e r +zk MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHYbGaeyypa0 JaamOCaiaahwgadaWgaaWcbaGaamOCaaqabaGccqGHRaWkcaWG6bGa aC4Aaaaa@3D5E@  .  Then

r=r e r +zk v= dr dt = dr dt e r +r d e r dθ dθ dt + dz dt k= dr dt e r +r dθ dt e θ + dz dt k a= dv dt = d 2 r d t 2 e r + dr dt d e r dθ dθ dt + dr dt dθ dt e θ +r d 2 θ d t 2 e θ + dr dt dθ dt d e θ dθ dθ dt + d 2 z d t 2 k =( d 2 r d t 2 r ( dθ dt ) 2 ) e r +( r d 2 θ d t 2 +2 dr dt dθ dt ) e θ + d 2 z d t 2 k MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaahkhacq GH9aqpcaWGYbGaaCyzamaaBaaaleaacaWGYbaabeaakiabgUcaRiaa dQhacaWHRbaabaGaaCODaiabg2da9maalaaabaGaamizaiaahkhaae aacaWGKbGaamiDaaaacqGH9aqpdaWcaaqaaiaadsgacaWGYbaabaGa amizaiaadshaaaGaaCyzamaaBaaaleaacaWGYbaabeaakiabgUcaRi aadkhadaWcaaqaaiaadsgacaWHLbWaaSbaaSqaaiaadkhaaeqaaaGc baGaamizaiabeI7aXbaadaWcaaqaaiaadsgacqaH4oqCaeaacaWGKb GaamiDaaaacqGHRaWkdaWcaaqaaiaadsgacaWG6baabaGaamizaiaa dshaaaGaaC4Aaiabg2da9maalaaabaGaamizaiaadkhaaeaacaWGKb GaamiDaaaacaWHLbWaaSbaaSqaaiaadkhaaeqaaOGaey4kaSIaamOC amaalaaabaGaamizaiabeI7aXbqaaiaadsgacaWG0baaaiaahwgada WgaaWcbaGaeqiUdehabeaakiabgUcaRmaalaaabaGaamizaiaadQha aeaacaWGKbGaamiDaaaacaWHRbaabaGaaCyyaiabg2da9maalaaaba GaamizaiaahAhaaeaacaWGKbGaamiDaaaacqGH9aqpdaWcaaqaaiaa dsgadaahaaWcbeqaaiaaikdaaaGccaWGYbaabaGaamizaiaadshada ahaaWcbeqaaiaaikdaaaaaaOGaaCyzamaaBaaaleaacaWGYbaabeaa kiabgUcaRmaalaaabaGaamizaiaadkhaaeaacaWGKbGaamiDaaaada WcaaqaaiaadsgacaWHLbWaaSbaaSqaaiaadkhaaeqaaaGcbaGaamiz aiabeI7aXbaadaWcaaqaaiaadsgacqaH4oqCaeaacaWGKbGaamiDaa aacqGHRaWkdaWcaaqaaiaadsgacaWGYbaabaGaamizaiaadshaaaWa aSaaaeaacaWGKbGaeqiUdehabaGaamizaiaadshaaaGaaCyzamaaBa aaleaacqaH4oqCaeqaaOGaey4kaSIaamOCamaalaaabaGaamizamaa CaaaleqabaGaaGOmaaaakiabeI7aXbqaaiaadsgacaWG0bWaaWbaaS qabeaacaaIYaaaaaaakiaahwgadaWgaaWcbaGaeqiUdehabeaakiab gUcaRmaalaaabaGaamizaiaadkhaaeaacaWGKbGaamiDaaaadaWcaa qaaiaadsgacqaH4oqCaeaacaWGKbGaamiDaaaadaWcaaqaaiaadsga caWHLbWaaSbaaSqaaiabeI7aXbqabaaakeaacaWGKbGaeqiUdehaam aalaaabaGaamizaiabeI7aXbqaaiaadsgacaWG0baaaiabgUcaRmaa laaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadQhaaeaacaWGKb GaamiDamaaCaaaleqabaGaaGOmaaaaaaGccaWHRbaabaGaaGPaVlaa ykW7caaMc8UaaGPaVlabg2da9maabmaabaWaaSaaaeaacaWGKbWaaW baaSqabeaacaaIYaaaaOGaamOCaaqaaiaadsgacaWG0bWaaWbaaSqa beaacaaIYaaaaaaakiabgkHiTiaadkhadaqadaqaamaalaaabaGaam izaiabeI7aXbqaaiaadsgacaWG0baaaaGaayjkaiaawMcaamaaCaaa leqabaGaaGOmaaaaaOGaayjkaiaawMcaaiaahwgadaWgaaWcbaGaam OCaaqabaGccqGHRaWkdaqadaqaaiaadkhadaWcaaqaaiaadsgadaah aaWcbeqaaiaaikdaaaGccqaH4oqCaeaacaWGKbGaamiDamaaCaaale qabaGaaGOmaaaaaaGccqGHRaWkcaaIYaWaaSaaaeaacaWGKbGaamOC aaqaaiaadsgacaWG0baaamaalaaabaGaamizaiabeI7aXbqaaiaads gacaWG0baaaaGaayjkaiaawMcaaiaahwgadaWgaaWcbaGaeqiUdeha beaakiabgUcaRmaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaaki aadQhaaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGccaWH Rbaaaaa@FA74@

 

 

 

Examples using polar coordinates

 

Example The robotic manipulator shown in the figure rotates with constant angular speed ω MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyYdChaaa@38CB@  about the k axis.   Find a formula for the maximum allowable (constant) rate of extension dL/dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadY eacaGGVaGaamizaiaadshaaaa@3B4D@  if the acceleration of the gripper may not exceed g.

 

We can simply write down the acceleration vector, using polar coordinates.  We identify ω=dθ/dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyYdCNaey ypa0JaamizaiabeI7aXjaac+cacaWGKbGaamiDaaaa@3F05@  and r=L, so that

a=( L ω 2 ) e r +( 2 dL dt ω ) e θ | a | 2 = L 2 ω 4 +4 ( dL dt ω ) 2 < g 2 dL dt < 1 4 g 2 / ω 2 L 2 ω 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahggacqGH9a qpdaqadaqaaiabgkHiTiaadYeacqaHjpWDdaahaaWcbeqaaiaaikda aaaakiaawIcacaGLPaaacaWHLbWaaSbaaSqaaiaadkhaaeqaaOGaey 4kaSYaaeWaaeaacaaIYaWaaSaaaeaacaWGKbGaamitaaqaaiaadsga caWG0baaaiabeM8a3bGaayjkaiaawMcaaiaahwgadaWgaaWcbaGaeq iUdehabeaakiabgkDiEpaaemaabaGaaCyyaaGaay5bSlaawIa7amaa CaaaleqabaGaaGOmaaaakiabg2da9iaadYeadaahaaWcbeqaaiaaik daaaGccqaHjpWDdaahaaWcbeqaaiaaisdaaaGccqGHRaWkcaaI0aWa aeWaaeaadaWcaaqaaiaadsgacaWGmbaabaGaamizaiaadshaaaGaeq yYdChacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaeyipaWJa am4zamaaCaaaleqabaGaaGOmaaaakiabgkDiEpaalaaabaGaamizai aadYeaaeaacaWGKbGaamiDaaaacqGH8aapdaWcaaqaaiaaigdaaeaa caaI0aaaamaakaaabaGaam4zamaaCaaaleqabaGaaGOmaaaakiaac+ cacqaHjpWDdaahaaWcbeqaaiaaikdaaaGccqGHsislcaWGmbWaaWba aSqabeaacaaIYaaaaOGaeqyYdC3aaWbaaSqabeaacaaIYaaaaaqaba aaaa@7885@

 

 

Example: The position of a particle in polar coordinates is given by θ= t 2 r=t/ π MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH4oqCcqGH9a qpcaWG0bWaaWbaaSqabeaacaaIYaaaaOGaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWGYbGaeyypa0JaamiDai aac+cadaGcaaqaaiabec8aWbWcbeaaaaa@4BF7@  (meters).  At the instant when θ=π MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH4oqCcqGH9a qpcqaHapaCaaa@39EF@  , calculate the following quantities:

 

·       The position vector in i,j MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHPbGaaiilai aahQgaaaa@380B@  components and in e r , e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHLbWaaSbaaS qaaiaadkhaaeqaaOGaaiilaiaahwgadaWgaaWcbaGaeqiUdehabeaa aaa@3B11@  components

 

When θ=π,r=1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacqaH4oqCcqGH9a qpcqaHapaCcaGGSaGaamOCaiabg2da9iaaigdaaaa@3D57@  so

r=i(meters) r= e r (meters) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakqaabeqaaiaahkhacq GH9aqpcqGHsislcaWHPbGaaGPaVlaaykW7caaMc8Uaaiikaiaad2ga caWGLbGaamiDaiaadwgacaWGYbGaam4CaiaacMcaaeaacaWHYbGaey ypa0JaaCyzamaaBaaaleaacaWGYbaabeaakiaaykW7caaMc8Uaaiik aiaad2gacaWGLbGaamiDaiaadwgacaWGYbGaam4CaiaacMcaaaaa@533E@

 

·       The velocity vector in e r , e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHLbWaaSbaaS qaaiaadkhaaeqaaOGaaiilaiaahwgadaWgaaWcbaGaeqiUdehabeaa aaa@3B11@  and i,j MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHPbGaaiilai aahQgaaaa@380B@

 

The polar coordinate formula is v= dr dt e r +r dθ dt e θ = 1 π e r +2t e θ = 1 π e r +2 π e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAhacqGH9a qpdaWcaaqaaiaadsgacaWGYbaabaGaamizaiaadshaaaGaaCyzamaa BaaaleaacaWGYbaabeaakiabgUcaRiaadkhadaWcaaqaaiaadsgacq aH4oqCaeaacaWGKbGaamiDaaaacaWHLbWaaSbaaSqaaiabeI7aXbqa baGccqGH9aqpdaWcaaqaaiaaigdaaeaadaGcaaqaaiabec8aWbWcbe aaaaGccaWHLbWaaSbaaSqaaiaadkhaaeqaaOGaey4kaSIaaGOmaiaa dshacaWHLbWaaSbaaSqaaiabeI7aXbqabaGccqGH9aqpdaWcaaqaai aaigdaaeaadaGcaaqaaiabec8aWbWcbeaaaaGccaWHLbWaaSbaaSqa aiaadkhaaeqaaOGaey4kaSIaaGOmamaakaaabaGaeqiWdahaleqaaO GaaCyzamaaBaaaleaacqaH4oqCaeqaaaaa@5ECD@  m/s

By inspection we see that i= e r ,j= e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHPbGaeyypa0 JaeyOeI0IaaCyzamaaBaaaleaacaWGYbaabeaakiaacYcacaWHQbGa eyypa0JaeyOeI0IaaCyzamaaBaaaleaacqaH4oqCaeqaaaaa@40DC@   at the instant of interest, so v= 1 π i2 π j MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAhacqGH9a qpcqGHsisldaWcaaqaaiaaigdaaeaadaGcaaqaaiabec8aWbWcbeaa aaGccaWHPbGaeyOeI0IaaGOmamaakaaabaGaeqiWdahaleqaaOGaaC OAaaaa@4149@

 

 

·       The acceleration vector in e r , e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHLbWaaSbaaS qaaiaadkhaaeqaaOGaaiilaiaahwgadaWgaaWcbaGaeqiUdehabeaa aaa@3B11@  components

 

a=( d 2 r d t 2 r ( dθ dt ) 2 ) e r +( r d 2 θ d t 2 +2 dr dt dθ dt ) e θ =4 t 2 e r +(2+4 1 π t) e θ =4π e r +6 e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaaCyyai abg2da9maabmaabaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaa aOGaamOCaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaaki abgkHiTiaadkhadaqadaqaamaalaaabaGaamizaiabeI7aXbqaaiaa dsgacaWG0baaaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaO GaayjkaiaawMcaaiaahwgadaWgaaWcbaGaamOCaaqabaGccqGHRaWk daqadaqaaiaadkhadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaa GccqaH4oqCaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGc cqGHRaWkcaaIYaWaaSaaaeaacaWGKbGaamOCaaqaaiaadsgacaWG0b aaamaalaaabaGaamizaiabeI7aXbqaaiaadsgacaWG0baaaaGaayjk aiaawMcaaiaahwgadaWgaaWcbaGaeqiUdehabeaaaOqaaiabg2da9i abgkHiTiaaisdacaWG0bWaaWbaaSqabeaacaaIYaaaaOGaaCyzamaa BaaaleaacaWGYbaabeaakiabgUcaRiaacIcacaaIYaGaey4kaSIaaG inamaalaaabaGaaGymaaqaamaakaaabaGaeqiWdahaleqaaaaakiaa dshacaGGPaGaaCyzamaaBaaaleaacqaH4oqCaeqaaOGaeyypa0Jaey OeI0IaaGinaiabec8aWjaahwgadaWgaaWcbaGaamOCaaqabaGccqGH RaWkcaaI2aGaaCyzamaaBaaaleaacqaH4oqCaeqaaaaaaa@7DFC@  m/s2

 

·       Unit vectors t,n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWH0bGaaiilai aah6gaaaa@381A@  tangent and normal to the path, in e r , e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHLbWaaSbaaS qaaiaadkhaaeqaaOGaaiilaiaahwgadaWgaaWcbaGaeqiUdehabeaa aaa@3B11@ .   (Choose n to point towards the center of curvature)

 

We know t is parallel to v so

t= v | v | = 1 4π+1/π ( 1 π e r +2 π e θ )= 1 4 π 2 +1 ( e r +2π e θ ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahshacqGH9a qpdaWcaaqaaiaahAhaaeaadaabdaqaaiaahAhaaiaawEa7caGLiWoa aaGaeyypa0ZaaSaaaeaacaaIXaaabaWaaOaaaeaacaaI0aGaeqiWda Naey4kaSIaaGymaiaac+cacqaHapaCaSqabaaaaOWaaeWaaeaadaWc aaqaaiaaigdaaeaadaGcaaqaaiabec8aWbWcbeaaaaGccaWHLbWaaS baaSqaaiaadkhaaeqaaOGaey4kaSIaaGOmamaakaaabaGaeqiWdaha leqaaOGaaCyzamaaBaaaleaacqaH4oqCaeqaaaGccaGLOaGaayzkaa Gaeyypa0ZaaSaaaeaacaaIXaaabaWaaOaaaeaacaaI0aGaeqiWda3a aWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGymaaWcbeaaaaGcdaqada qaaiaahwgadaWgaaWcbaGaamOCaaqabaGccqGHRaWkcaaIYaGaeqiW daNaaCyzamaaBaaaleaacqaH4oqCaeqaaaGccaGLOaGaayzkaaaaaa@6370@

 

We can find n in three ways: first, we know n must be perpendicular to t and must lie in the i,j plane (and hence is perpendicular to k).   Remember that you can create a vector perpendicular to two others using a cross product, so n=±k×t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHUbGaeyypa0 JaeyySaeRaaC4AaiabgEna0kaahshaaaa@3D69@  .   The positive choice points towards the center of curvature (by inspection), and note k× e r = e θ k× e θ = e r MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHRbGaey41aq RaaCyzamaaBaaaleaacaWGYbaabeaakiabg2da9iaahwgadaWgaaWc baGaeqiUdehabeaakiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaC4AaiabgEna0kaahwgadaWgaaWcbaGaeqiU dehabeaakiabg2da9iabgkHiTiaahwgadaWgaaWcbaGaamOCaaqaba aaaa@54BD@   so  n= 1 4 π 2 +1 ( 2π e r + e θ ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaah6gacqGH9a qpdaWcaaqaaiaaigdaaeaadaGcaaqaaiaaisdacqaHapaCdaahaaWc beqaaiaaikdaaaGccqGHRaWkcaaIXaaaleqaaaaakmaabmaabaGaey OeI0IaaGOmaiabec8aWjaahwgadaWgaaWcbaGaamOCaaqabaGccqGH RaWkcaWHLbWaaSbaaSqaaiabeI7aXbqabaaakiaawIcacaGLPaaaaa a@48F8@

 

You can also use the condition tn=0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWH0bGaeyyXIC TaaCOBaiabg2da9iaaicdaaaa@3B74@   - if we assume n= n r e r + n θ e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHUbGaeyypa0 JaamOBamaaBaaaleaacaWGYbaabeaakiaahwgadaWgaaWcbaGaamOC aaqabaGccqGHRaWkcaWGUbWaaSbaaSqaaiabeI7aXbqabaGccaWHLb WaaSbaaSqaaiabeI7aXbqabaaaaa@423F@   then

  tn=0 1 4 π 2 +1 ( e r +2π e θ )( n r e r + n θ e θ )=0 n r +2π n θ =0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaaCiDai abgwSixlaah6gacqGH9aqpcaaIWaGaeyO0H49aaSaaaeaacaaIXaaa baWaaOaaaeaacaaI0aGaeqiWda3aaWbaaSqabeaacaaIYaaaaOGaey 4kaSIaaGymaaWcbeaaaaGcdaqadaqaaiaahwgadaWgaaWcbaGaamOC aaqabaGccqGHRaWkcaaIYaGaeqiWdaNaaCyzamaaBaaaleaacqaH4o qCaeqaaaGccaGLOaGaayzkaaGaeyyXIC9aaeWaaeaacaWGUbWaaSba aSqaaiaadkhaaeqaaOGaaCyzamaaBaaaleaacaWGYbaabeaakiabgU caRiaad6gadaWgaaWcbaGaeqiUdehabeaakiaahwgadaWgaaWcbaGa eqiUdehabeaaaOGaayjkaiaawMcaaiabg2da9iaaicdaaeaacqGHsh I3caWGUbWaaSbaaSqaaiaadkhaaeqaaOGaey4kaSIaaGOmaiabec8a Wjaad6gadaWgaaWcbaGaeqiUdehabeaakiabg2da9iaaicdaaaaa@6B50@

Any n r , n θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGUbWaaSbaaS qaaiaadkhaaeqaaOGaaiilaiaad6gadaWgaaWcbaGaeqiUdehabeaa aaa@3B1B@  that satisfies this (eg n r =2π, n θ =1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGUbWaaSbaaS qaaiaadkhaaeqaaOGaeyypa0JaeyOeI0IaaGOmaiabec8aWjaacYca caWGUbWaaSbaaSqaaiabeI7aXbqabaGccqGH9aqpcaaIXaaaaa@4152@  ) is perpendicular to t.  But n must be a unit vector, and we know we want the vector to point towards the origin (because the center of curvature of the path is inside the turn).   So we have to choose

n= 1 4 π 2 +1 ( 2π e r + e θ ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaah6gacqGH9a qpdaWcaaqaaiaaigdaaeaadaGcaaqaaiaaisdacqaHapaCdaahaaWc beqaaiaaikdaaaGccqGHRaWkcaaIXaaaleqaaaaakmaabmaabaGaey OeI0IaaGOmaiabec8aWjaahwgadaWgaaWcbaGaamOCaaqabaGccqGH RaWkcaWHLbWaaSbaaSqaaiabeI7aXbqabaaakiaawIcacaGLPaaaaa a@48F8@

 

The last (cumbersome, but general) way to do the calculation is to note that a(at)t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHHbGaeyOeI0 IaaiikaiaahggacqGHflY1caWH0bGaaiykaiaahshaaaa@3DD4@  must be parallel to n

4π e r +6 e θ [ ( 4π e r +6 e θ ) 1 4 π 2 +1 ( e r +2π e θ ) ] 1 4 π 2 +1 ( e r +2π e θ ) =4π e r +6 e θ 8π 4 π 2 +1 ( e r +2π e θ )= 4π(3+4 π 2 ) 4 π 2 +1 e r + 2(3+4 π 2 ) 4 π 2 +1 e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaeyOeI0 IaaGinaiabec8aWjaahwgadaWgaaWcbaGaamOCaaqabaGccqGHRaWk caaI2aGaaCyzamaaBaaaleaacqaH4oqCaeqaaOGaeyOeI0YaamWaae aadaqadaqaaiabgkHiTiaaisdacqaHapaCcaWHLbWaaSbaaSqaaiaa dkhaaeqaaOGaey4kaSIaaGOnaiaahwgadaWgaaWcbaGaeqiUdehabe aaaOGaayjkaiaawMcaaiabgwSixpaalaaabaGaaGymaaqaamaakaaa baGaaGinaiabec8aWnaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaig daaSqabaaaaOWaaeWaaeaacaWHLbWaaSbaaSqaaiaadkhaaeqaaOGa ey4kaSIaaGOmaiabec8aWjaahwgadaWgaaWcbaGaeqiUdehabeaaaO GaayjkaiaawMcaaaGaay5waiaaw2faamaalaaabaGaaGymaaqaamaa kaaabaGaaGinaiabec8aWnaaCaaaleqabaGaaGOmaaaakiabgUcaRi aaigdaaSqabaaaaOWaaeWaaeaacaWHLbWaaSbaaSqaaiaadkhaaeqa aOGaey4kaSIaaGOmaiabec8aWjaahwgadaWgaaWcbaGaeqiUdehabe aaaOGaayjkaiaawMcaaaqaaiabg2da9iabgkHiTiaaisdacqaHapaC caWHLbWaaSbaaSqaaiaadkhaaeqaaOGaey4kaSIaaGOnaiaahwgada WgaaWcbaGaeqiUdehabeaakiabgkHiTmaalaaabaGaaGioaiabec8a WbqaaiaaisdacqaHapaCdaahaaWcbeqaaiaaikdaaaGccqGHRaWkca aIXaaaamaabmaabaGaaCyzamaaBaaaleaacaWGYbaabeaakiabgUca RiaaikdacqaHapaCcaWHLbWaaSbaaSqaaiabeI7aXbqabaaakiaawI cacaGLPaaacqGH9aqpcqGHsisldaWcaaqaaiaaisdacqaHapaCcaGG OaGaaG4maiabgUcaRiaaisdacqaHapaCdaahaaWcbeqaaiaaikdaaa GccaGGPaaabaGaaGinaiabec8aWnaaCaaaleqabaGaaGOmaaaakiab gUcaRiaaigdaaaGaaCyzamaaBaaaleaacaWGYbaabeaakiabgUcaRm aalaaabaGaaGOmaiaacIcacaaIZaGaey4kaSIaaGinaiabec8aWnaa CaaaleqabaGaaGOmaaaakiaacMcaaeaacaaI0aGaeqiWda3aaWbaaS qabeaacaaIYaaaaOGaey4kaSIaaGymaaaacaWHLbWaaSbaaSqaaiab eI7aXbqabaaaaaa@AFF7@

Dividing by the magnitude of this vector (to create a unit vector) gives the same answer as before.

 

·       Tangential and normal components of acceleration a t , a n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGHbWaaSbaaS qaaiaadshaaeqaaOGaaiilaiaadggadaWgaaWcbaGaamOBaaqabaaa aa@3A40@  : we know (from ENGN30) that we can find the component of a vector in a basis by dotting it with the basis vectors, so

a t =at= 8π 1+4 π 2 a n =an= 8 π 2 +6 1+4 π 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGHbWaaSbaaS qaaiaadshaaeqaaOGaeyypa0JaaCyyaiabgwSixlaahshacqGH9aqp daWcaaqaaiaaiIdacqaHapaCaeaadaGcaaqaaiaaigdacqGHRaWkca aI0aGaeqiWda3aaWbaaSqabeaacaaIYaaaaaqabaaaaOGaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaadggadaWgaaWcbaGaamOBaaqabaGc cqGH9aqpcaWHHbGaeyyXICTaaCOBaiabg2da9maalaaabaGaaGioai abec8aWnaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiAdaaeaadaGc aaqaaiaaigdacqGHRaWkcaaI0aGaeqiWda3aaWbaaSqabeaacaaIYa aaaaqabaaaaaaa@6BFC@  m/s2

You can also do this problem using the formula

a= dV dt t+ V 2 R n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHHbGaeyypa0 ZaaSaaaeaacaWGKbGaamOvaaqaaiaadsgacaWG0baaaiaahshacqGH RaWkdaWcaaqaaiaadAfadaahaaWcbeqaaiaaikdaaaaakeaacaWGsb aaaiaah6gaaaa@40A7@  .   This shows a t = dV dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGHbWaaSbaaS qaaiaadshaaeqaaOGaeyypa0ZaaSaaaeaacaWGKbGaamOvaaqaaiaa dsgacaWG0baaaaaa@3C47@  , which is not too hard to calculate (but is a pain so I can’t be bothered).   To find a n = V 2 /R MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGHbWaaSbaaS qaaiaad6gaaeqaaOGaeyypa0JaamOvamaaCaaaleqabaGaaGOmaaaa kiaac+cacaWGsbaaaa@3BE3@   we would either have to use our MA0200 super-powers to find the radius of curvature of the path, or else use a dV dt t=+ V 2 R n= a n n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHHbGaeyOeI0 YaaSaaaeaacaWGKbGaamOvaaqaaiaadsgacaWG0baaaiaahshacqGH 9aqpcqGHRaWkdaWcaaqaaiaadAfadaahaaWcbeqaaiaaikdaaaaake aacaWGsbaaaiaah6gacqGH9aqpcaWGHbWaaSbaaSqaaiaad6gaaeqa aOGaaCOBaaaa@45A0@  and then take the magnitude of the vector on the left to get a n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGHbWaaSbaaS qaaiaad6gaaeqaaaaa@377B@  .  

 

 

3.1.7 Measuring position, velocity and acceleration

 

If you are designing a control system, you will need some way to detect the motion of the system you are trying to control.  A vast array of different sensors is available for you to choose from: see for example the list at http://www.sensorland.com/HowPage001.html .  A very short list of common sensors is given below

1.      GPS MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  determines position on the earth’s surface by measuring the time for electromagnetic waves to travel from satellites in known positions in space to the sensor.   Can be accurate down to cm distances, but the sensor needs to be left in position for a long time for this kind of accuracy.  A few m is more common.

2.      Optical or radio frequency position sensing MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  measure position by (a) monitoring deflection of laser beams off a target; or measuring the time for signals to travel from a set of radio emitters with known positions to the sensor.  Precision can vary from cm accuracy down to light wavelengths.

3.      Capacitative displacement sensing MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  determine position by measuring the capacitance between two parallel plates.  The device needs to be physically connected to the object you are tracking and a reference point. Can only measure distances of mm or less, but precision can be down to micron accuracy.

4.      Electromagnetic displacement sensing MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  measures position by detecting electromagnetic fields between conducting coils, or coil/magnet combinations within the sensor.  Needs to be physically connected to the object you are tracking and a reference point.  Measures displacements of order cm down to microns.

5.      Radar velocity sensing MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  measures velocity by detecting the change in frequency of electromagnetic waves reflected off the traveling object.

6.      Inertial accelerometers: measure accelerations by detecting the deflection of a spring acting on a mass.

 

Accelerometers are also often used to construct an ‘inertial platform,’ which uses gyroscopes to maintain a fixed orientation in space, and has three accelerometers that can detect motion in three mutually perpendicular directions.  These accelerations can then be integrated to determine the position.  They are used in aircraft, marine applications, and space vehicles where GPS cannot be used or where a backup is needed for GPS.   They are often combined with GPS receivers as well: accelerometers are very good for measuring changes in velocity and position over a short time interval, and GPS is very good over long time intervals, so you can use ‘sensor fusion’ to make a sensor that uses both signals to get the best possible measurement.

 

 

 

 

3.2 Calculating forces required to cause prescribed motion of a particle

 

3.2.1 The Newtonian Inertial Frame.

 

Newton’s laws are very familiar, and it is easy to write them down without much thought.  They do have a flaw, however.

 

When we use Newton’s laws, we assume that:

(1)   We can identify some point in the universe that is not accelerating

(2)   We can identify three mutually perpendicular directions that are ‘fixed’ in the sense that they do not rotate.  

Together, these define a so-called ‘inertial frame’ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  a Cartesian coordinate system in which motion obeys Newton’s laws.

 

For engineering calculations, identifying a suitable origin and fixed directions usually poses no difficulty.  If we are solving problems involving terrestrial motion over short distances compared with the earth’s radius, we simply take a point on the earth’s surface as fixed, and take three directions relative to the earth’s surface to be fixed.  If we are solving problems involving motion in space near the earth, or modeling weather, we take the center of the earth as a fixed point, (or for more complex calculations the center of the sun); and choose axes to have a fixed direction relative to nearby stars.    Experiments show that Newton’s laws predict motion sufficiently accurately for our needs.   But there will always be some very small error.  

 

In reality, an unambiguous inertial frame does not exist.  We can only describe the relative motion of the mass in the universe, not its absolute motion.  The general theory of relativity gives us a framework that avoids having to choose an inertial frame.  More elaborate calculations show that Newton’s laws are rigorous approximations to the general equations (in the sense of a Taylor expansion of the more general equations for low particle speeds compared with the speed of light), and would also (in principle) tell us the best choice of directions to set up a Newtonian frame at any point in space.

 

3.2.2 Newton’s laws of motion for a particle

 

If we ignore these conceptual difficulties, Newton’s laws for a particle are very simple.  Let

1.      m denote the mass of the particle

2.      F denote the resultant force acting on the particle (as a vector, in the inertial frame)

3.      a denote the acceleration of the particle (again, as a vector in the inertial frame).  Then

 

F=ma MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbaaaa@39A7@

 

Occasionally, we use a particle idealization to model systems which, strictly speaking, are not particles.  These are:

1.      A large mass, which moves without rotation (e.g. a car moving along a straight line)

2.      A single particle which is attached to a rigid frame with negligible mass (e.g. a person on a bicycle)

 

In these cases it may be necessary to consider the moments acting on the mass (or frame) in order to calculate unknown reaction forces. 

1.      For a large mass which moves without rotation, the resultant moment of external forces about the center of mass must vanish.

2.      For a particle attached to a massless frame, the resultant moment of external forces acting on the frame about the particle must vanish.

M C =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytamaaBa aaleaacaWGdbaabeaakiabg2da9iaahcdaaaa@3989@

We will see where this equation comes from when we analyze rigid body dynamics, and we’ll also understand when it is no longer correct.

 

It is very important to take moments about the correct point in dynamics problems! Forgetting this is the most common reason to screw up a dynamics problem…

 

If you need to solve a problem where more than one particle is attached to a massless frame, you have to draw a separate free body diagram for each particle, and for the frame.   The particles must obey Newton’s laws F=ma MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbaaaa@39A7@ .   The forces acting on the frame must obey F=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaahcdaaaa@3884@  and M C =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytamaaBa aaleaacaWGdbaabeaakiabg2da9iaahcdaaaa@3989@ , (because the frame has no mass). 

 

Newton’s laws of motion can be used to calculate the forces required to make a particle move in a particular way.  

 

We use the following general procedure to solve problems like this

(1) Decide how to idealize the system (what are the particles?)

(2) Draw a free body diagram showing the forces acting on each particle

(3) Consider the kinematics of the problem. The goal is to calculate the acceleration of each particle in the system MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  you may be able to start by writing down the position vector and differentiating it, or you may be able to relate the accelerations of two particles (eg if two particles move together, their accelerations must be equal).

(4) Write down F=ma for each particle.

(5) If you are solving a problem involving a massless frames (see, e.g. Example 3, involving a bicycle with negligible mass) you also need to write down M C =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytamaaBa aaleaacaWGdbaabeaakiabg2da9iaahcdaaaa@3989@  about the particle.

(5) Solve the resulting equations for any unknown components of force or acceleration (this is just like a statics problem, except the right hand side is not zero).

 

It is best to show how this is done by means of examples. 

 

 

Example 1: Estimate the minimum thrust that must be produced by the engines of an aircraft in order to take off from the deck of an aircraft carrier (the figure is from www.lakehurst.navy.mil/NLWeb/media-library.asp)

 

We will estimate the acceleration required to reach takeoff speed, assuming the aircraft accelerates from zero speed to takeoff speed along the deck of the carrier, and then use Newton’s laws to deduce the force.

 

Data/ Assumptions:

1.      The flight deck of a Nimitz class aircraft carrier is about 300m long (http://www.naval-technology.com/projects/nimitz/) but only a fraction of this is used for takeoff (the angled runway is used for landing).   We will take the length of the runway to be d=200m

2.      We will assume that the acceleration during takeoff roll is constant.

3.      We will assume that the aircraft carrier is not moving (this is wrong MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  actually the aircraft carrier always moves at high speed during takeoff.  We neglect motion to make the calculation simpler)

4.      The FA18 Super Hornet is a typical aircraft used on a carrier MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  it has max catapult weight of m=15000kg  http://www.boeing.com/defense-space/military/fa18ef/docs/EF_overview.pdf

5.      The manufacturers are somewhat reticent about performance specifications for the Hornet but v t = MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaWG0baabeaakiabg2da9aaa@3926@  150 knots (77 m/s) is a reasonable guess for a minimum controllable airspeed for this aircraft.

 

Calculations:

1.      Idealization: We will idealize the aircraft as a particle.  We can do this because the aircraft is not rotating during takeoff.

 

2.      FBD: The figure shows a free body diagram.  F T MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGubaabeaaaaa@37C6@  represents the (unknown) force exerted on the aircraft due to its engines.

 

3.      Kinematics: We must calculate the acceleration required to reach takeoff speed.  We are given (i) the distance to takeoff d, (ii) the takeoff speed v t MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaWG0baabeaaaaa@3816@  and (iii) the aircraft is at rest at the start of the takeoff roll. We can therefore write down the position vector r and velocity v of the aircraft at takeoff, and use the straight line motion formulas for r and v to calculate the time t to reach takeoff speed and the acceleration a.  Taking the origin at the initial position of the aircraft, we have that, at the instant of takeoff

r=di= 1 2 a t 2 iv= v t i=ati MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadsgacaWHPbGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaa caWGHbGaamiDamaaCaaaleqabaGaaGOmaaaakiaahMgacaaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caWH2bGaeyypa0JaamODamaaBaaaleaacaWG0b aabeaakiaahMgacqGH9aqpcaWGHbGaamiDaiaahMgaaaa@5CA6@

This gives two scalar equations which can be solved for a and t

d= 1 2 a t 2 v t =ata= v t 2 2d t= 2d v t MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiabg2 da9maalaaabaGaaGymaaqaaiaaikdaaaGaamyyaiaadshadaahaaWc beqaaiaaikdaaaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaamODamaaBaaaleaacaWG0baabeaakiabg2da9iaaykW7 caWGHbGaamiDaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaeyO0H4TaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caWGHbGaeyypa0ZaaSaaaeaacaWG2bWaa0 baaSqaaiaadshaaeaacaaIYaaaaaGcbaGaaGOmaiaadsgaaaGaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamiDaiabg2da9maala aabaGaaGOmaiaadsgaaeaacaWG2bWaaSbaaSqaaiaadshaaeqaaaaa aaa@7F57@

 

4.      EOM: The vector equation of motion for this problem is

F T i=mai=m v t 2 2d i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGubaabeaakiaahMgacqGH9aqpcaWGTbGaamyyaiaahMga cqGH9aqpcaWGTbWaaSaaaeaacaWG2bWaa0baaSqaaiaadshaaeaaca aIYaaaaaGcbaGaaGOmaiaadsgaaaGaaCyAaaaa@451F@

 

5.      Solution: The i component of the equation of motion gives an equation for the unknown force in terms of known quantities

F T =m v t 2 2d MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGubaabeaakiabg2da9iaad2gadaWcaaqaaiaadAhadaqh aaWcbaGaamiDaaqaaiaaikdaaaaakeaacaaIYaGaamizaaaaaaa@3F6B@

Substituting numbers gives the magnitude of the force as F=222 kN.  This is very close, but slightly greater than, the 200kN (44000lb) thrust quoted on the spec sheet for the Hornet.  Using a catapult to accelerate the aircraft, speeding up the aircraft carrier, and increasing thrust using an afterburner buys a margin of safety. 

 

 

 

Example 2: Mechanics of Magic! You have no doubt seen the simple `tablecloth trick’ in which a tablecloth is whipped out from underneath a fully set table (if not, you can watch it at http://wm.kusa.gannett.edgestreams.net/news/1132187192333-11-16-05-spangler-2p.wmv)

 

In this problem we shall estimate the critical acceleration that must be imposed on the tablecloth to pull it from underneath the objects placed upon it.

 

We wish to determine conditions for the tablecloth to slip out from under the glass. We can do this by calculating the reaction forces acting between the glass and the tablecloth, and see whether or not slip will occur.   It is best to calculate the forces required to make the glass move with the tablecloth (i.e. to prevent slip), and see if these forces are big enough to cause slip.

 

1.      Idealization: We will assume that the glass behaves like a particle (again, we can do this because the glass does not rotate)

 

2.      FBD. The figure shows a free body diagram for the glass.  The forces include (i) the weight; and (ii) the normal and tangential components of reaction at the contact between the tablecloth and the glass.   The normal and tangential forces must act somewhere inside the contact area, but their position is unknown.   For a more detailed discussion of contact forces see Sects 2.4 and 2.5.

 

3.      Kinematics We are assuming that the glass has the same acceleration as the tablecloth. The table cloth is moving in the i direction, and has magnitude a. The acceleration vector is therefore a=ai MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabg2 da9iaadggacaWHPbaaaa@39BE@ .

4.      EOM. Newton’s laws of motion yield

F=maTi+(Nmg)j=mai MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbGaeyO0H4TaamivaiaahMgacqGHRaWkcaGGOaGa amOtaiabgkHiTiaad2gacaWGNbGaaiykaiaahQgacqGH9aqpcaWGTb GaamyyaiaahMgaaaa@486B@

5.      Solution: The i and j components of the vector equation must each be satisfied (just as when you solve a statics problem), so that

T=maNmg=0N=mg MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 da9iaad2gacaWGHbGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caWGobGaeyOeI0IaamyBaiaadEgacqGH 9aqpcaaIWaGaeyO0H4TaamOtaiabg2da9iaad2gacaWGNbaaaa@5F5A@

Finally, we must use the friction law to decide whether or not the tablecloth will slip from under the glass.   Recall that, for no slip, the friction force must satisfy

| T |<μN MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca WGubaacaGLhWUaayjcSdGaeyipaWJaeqiVd0MaamOtaaaa@3D7E@

where μ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiVd0gaaa@37AC@  is the friction coefficient.  Substituting for T and N from (5) shows that for no slip

| a |<μg MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca WGHbaacaGLhWUaayjcSdGaeyipaWJaeqiVd0Maam4zaaaa@3DA4@

To do the trick, therefore, the acceleration must exceed μg MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiVd0Maam 4zaaaa@3898@ .  For a friction coefficient of order 0.1, this gives an acceleration of order 1m/ s 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiaad2 gacaGGVaGaam4CamaaCaaaleqabaGaaGOmaaaaaaa@3A37@ .   There is a special trick to pulling the tablecloth with a large acceleration MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  but that’s a secret.

 

 

Example 3: Bicycle Safety.  If a bike rider brakes too hard on the front wheel, his or her bike will tip over (the figure is from http://www.thosefunnypictures.com/picture/7658/bike-flip.html).  In this example we investigate the conditions that will lead the bike to capsize, and identify design variables that can influence these conditions.

 

If the bike tips over, the rear wheel leaves the ground.  If this happens, the reaction force acting on the wheel must be zero MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  so we can detect the point where the bike is just on the verge of tipping over by calculating the reaction forces, and finding the conditions where the reaction force on the rear wheel is zero.

 

1.      Idealization:

a.       We will idealize the rider as a particle (apologies to bike racers MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  but that’s how we think of you…). The particle is located at the center of mass of the rider.  The figure shows the most important design parameters- these are the height of the rider’s COM, the wheelbase L and the distance of the COM from the rear wheel.

b.      We assume that the bike is a massless frame.   The wheels are also assumed to have no mass.  This means that the forces acting on the wheels must satisfy F=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaahcdaaaa@3884@  and M=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytaiabg2 da9iaahcdaaaa@388B@  - and can be analyzed using methods of statics.  If you’ve forgotten how to think about statics of wheels, you should re-read the notes on this topic MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  in particular, make sure you understand the nature of the forces acting on a freely rotating wheel (Section 2.4.6 of the reference notes).

c.       We assume that the rider brakes so hard that the front wheel is prevented from rotating.  It must therefore skid over the ground.  Friction will resist this sliding. We denote the friction coefficient at the contact point B by μ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiVd0gaaa@37AC@ .

d.      The rear wheel is assumed to rotate freely.

e.       We neglect air resistance.

2.      FBD. The figure shows a free body diagram for the rider and for the bike together.  Note that

a.       A normal and tangential force acts at the contact point on the front wheel (in general, both normal and tangential forces always act at contact points, unless the contact happens to be frictionless).  Because the contact is slipping it is essential to draw the friction force in the correct direction MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  the force must resist the motion of the bike;

b.      Only a normal force acts at the contact point on the rear wheel because it is freely rotating, and behaves like a 2-force member.

 

3.      Kinematics The bike is moving in the i direction. As a vector, its acceleration is therefore a=ai MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabg2 da9iaadggacaWHPbaaaa@39BE@ , where a is unknown. 

 

4.      EOM: Because this problem includes a massless frame, we must use two equations of motion ( F=ma MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbaaaa@39A7@  and M C =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytamaaBa aaleaacaWGdbaabeaakiabg2da9iaahcdaaaa@3989@  ).  It is essential to take moments about the particle (i.e. the rider’s COM).

 

F=ma MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbaaaa@39A7@  gives T B i+( N A + N B W)j=mai MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Iaam ivamaaBaaaleaacaWGcbaabeaakiaahMgacqGHRaWkcaGGOaGaamOt amaaBaaaleaacaWGbbaabeaakiabgUcaRiaad6eadaWgaaWcbaGaam OqaaqabaGccqGHsislcaWGxbGaaiykaiaahQgacqGH9aqpcaWGTbGa amyyaiaahMgaaaa@46F3@

M C =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytamaaBa aaleaacaWGdbaabeaakiabg2da9iaahcdaaaa@3989@  gives N B (Ld)k N A dk T B hk=0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad6eadaWgaa WcbaGaamOqaaqabaGccaGGOaGaamitaiabgkHiTiaadsgacaGGPaGa aC4AaiabgkHiTiaad6eadaWgaaWcbaGaamyqaaqabaGccaWGKbGaaC 4AaiabgkHiTiaadsfadaWgaaWcbaGaamOqaaqabaGccaWGObGaaC4A aiabg2da9iaahcdaaaa@47FA@  

 

The two nonzero components of F=ma MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbaaaa@39A7@  and the one nonzero component of M C =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytamaaBa aaleaacaWGdbaabeaakiabg2da9iaahcdaaaa@3989@  give us three scalar equations

T B =ma ( N A + N B W)=0 N B (Ld) N A d T B h=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsi slcaWGubWaaSbaaSqaaiaadkeaaeqaaOGaeyypa0JaamyBaiaadgga aeaacaGGOaGaamOtamaaBaaaleaacaWGbbaabeaakiabgUcaRiaad6 eadaWgaaWcbaGaamOqaaqabaGccqGHsislcaWGxbGaaiykaiabg2da 9iaaicdaaeaacaWGobWaaSbaaSqaaiaadkeaaeqaaOGaaiikaiaadY eacqGHsislcaWGKbGaaiykaiabgkHiTiaad6eadaWgaaWcbaGaamyq aaqabaGccaWGKbGaeyOeI0IaamivamaaBaaaleaacaWGcbaabeaaki aadIgacqGH9aqpcaaIWaaaaaa@53E7@

We have four unknowns MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  the reaction components N A , N B , T B MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtamaaBa aaleaacaWGbbaabeaakiaacYcacaWGobWaaSbaaSqaaiaadkeaaeqa aOGaaiilaiaadsfadaWgaaWcbaGaamOqaaqabaaaaa@3CC1@  and the acceleration a so we need another equation.   The missing equation is the friction law

T B =μ N B MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBa aaleaacaWGcbaabeaakiabg2da9iabeY7aTjaad6eadaWgaaWcbaGa amOqaaqabaaaaa@3C4E@

 

5.      Solution: (tedious algebra MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  you could avoid this by using Matlab)

 

The third equation and the friction law show that

N B (Ldμh) N A d=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtamaaBa aaleaacaWGcbaabeaakiaacIcacaWGmbGaeyOeI0IaamizaiabgkHi TiabeY7aTjaadIgacaGGPaGaeyOeI0IaamOtamaaBaaaleaacaWGbb aabeaakiaadsgacqGH9aqpcaaIWaaaaa@44BB@

Multiply ( N A + N B W)=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaad6 eadaWgaaWcbaGaamyqaaqabaGccqGHRaWkcaWGobWaaSbaaSqaaiaa dkeaaeqaaOGaeyOeI0Iaam4vaiaacMcacqGH9aqpcaaIWaaaaa@3F59@  by (Ldμh) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaacIcacaWGmb GaeyOeI0IaamizaiabgkHiTiabeY7aTjaadIgacaGGPaaaaa@3DCA@  and subtract it from this equation:

N A d(Ldμh)( N A W)=0 N A =W Ldμh Lμh MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsi slcaWGobWaaSbaaSqaaiaadgeaaeqaaOGaamizaiabgkHiTiaacIca caWGmbGaeyOeI0IaamizaiabgkHiTiabeY7aTjaadIgacaGGPaGaai ikaiaad6eadaWgaaWcbaGaamyqaaqabaGccqGHsislcaWGxbGaaiyk aiabg2da9iaaicdaaeaacqGHshI3caWGobWaaSbaaSqaaiaadgeaae qaaOGaeyypa0Jaam4vamaalaaabaGaamitaiabgkHiTiaadsgacqGH sislcqaH8oqBcaWGObaabaGaamitaiabgkHiTiabeY7aTjaadIgaaa aaaaa@5986@

 

We are interested in finding what makes the reaction force at A go to zero (that’s when the bike is about to tip).  So

N A =W Ldμh Lμh 0μ(Ld)/h MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtamaaBa aaleaacaWGbbaabeaakiabg2da9iaadEfadaWcaaqaaiaadYeacqGH sislcaWGKbGaeyOeI0IaeqiVd0MaamiAaaqaaiaadYeacqGHsislcq aH8oqBcaWGObaaaiabgsMiJkaaicdacqGHshI3cqaH8oqBcqGHKjYO caGGOaGaamitaiabgkHiTiaadsgacaGGPaGaai4laiaadIgaaaa@5226@

 

 

 

 

 

 

This tells us that the bike will tip if the friction coefficient exceeds a critical magnitude, which depends on the geometry of the bike.  The simplest way to design a tip-resistant bike is to make the height of the center of mass h small, and the distance (L-d) between the front wheel and the COM  as large as possible. 

 

A `recumbent’ bike is one way to achieve this MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  the figure (from http://en.wikipedia.org/wiki/Recumbent_bicycle) shows an example. The recumbent design offers many other significant advantages over the classic bicycle besides tipping resistance.

 

 

 

Example 4: A stupid problem that you might find in the FE professional engineering exam.  The purpose of this problem is to show what you need to do to solve problems involving more than one particle.

 

Two weights of mass m A MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGTbWaaSbaaSqaaiaadgeaaeqaaa aa@34A9@  and m B MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGTbWaaSbaaSqaaiaadkeaaeqaaa aa@34AA@  are connected by a cable passing over two freely rotating pulleys as shown.  They are released, and the system begins to move.  Find an expression for the tension in the cable connecting the two weights.

 

 

1.      Idealization MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  The masses will be idealized as particles; the cable is inextensible and the mass of the pulleys is neglected.  This means the internal forces in the cable, and the forces acting between cables/pulleys must satisfy  F=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaahcdaaaa@3884@  and M=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytaiabg2 da9iaahcdaaaa@388B@ , and we can treat them as though they were in static equilibrium. 

 

2.      FBD MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  we have to draw a separate FBD for each particle.  Since the pulleys and cable are massless, the tension T in the cable is constant.

 

3.      Kinematics  We know that both masses must move in the j direction.  We also know that the masses always move at the same speed but in opposite directions.  Therefore, their accelerations must be equal and opposite.  We can express this mathematically as

a A j= a B j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGHbWaaSbaaSqaaiaadgeaaeqaaO GaaCOAaiabg2da9iabgkHiTiaadggadaWgaaWcbaGaamOqaaqabaGc caWHQbGaaGPaVdaa@3BEE@

 

4.      EOM: We must write down two equations of motion, as there are two masses

(T m A g)j= m A a A j (T m B g)j= m B a B j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiaacIcacaWGubGaeyOeI0Iaam yBamaaBaaaleaacaWGbbaabeaakiaadEgacaGGPaGaaCOAaiabg2da 9iaad2gadaWgaaWcbaGaamyqaaqabaGccaWGHbWaaSbaaSqaaiaadg eaaeqaaOGaaCOAaiaaykW7aeaacaGGOaGaamivaiabgkHiTiaad2ga daWgaaWcbaGaamOqaaqabaGccaWGNbGaaiykaiaahQgacqGH9aqpca WGTbWaaSbaaSqaaiaadkeaaeqaaOGaamyyamaaBaaaleaacaWGcbaa beaakiaahQgaaaaa@4DC4@

We now have three equations for three unknowns (the unknowns are a A , a B MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGHbWaaSbaaSqaaiaadgeaaeqaaO GaaiilaiaaykW7caaMc8UaaGPaVlaadggadaWgaaWcbaGaamOqaaqa baaaaa@3BD1@  and T).

 

5.      Solution:  More algebra.   We can eliminate a B MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadggadaWgaa WcbaGaamOqaaqabaaaaa@3813@  so that the last two equations are:

(T m A g)= m A a A (T m B g)= m B a A MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiaacIcacaWGubGaeyOeI0Iaam yBamaaBaaaleaacaWGbbaabeaakiaadEgacaGGPaGaeyypa0JaamyB amaaBaaaleaacaWGbbaabeaakiaadggadaWgaaWcbaGaamyqaaqaba aakeaacaGGOaGaamivaiabgkHiTiaad2gadaWgaaWcbaGaamOqaaqa baGccaWGNbGaaiykaiabg2da9iabgkHiTiaad2gadaWgaaWcbaGaam OqaaqabaGccaWGHbWaaSbaaSqaaiaadgeaaeqaaaaaaa@494F@

Now we can multiply the first equation by m B MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2gadaWgaa WcbaGaamOqaaqabaaaaa@381F@  and the second by m A MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad2gadaWgaa WcbaGaamyqaaqabaaaaa@381E@  and add them

(T m A +T m B 2 m A m B g)=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaGGOaGaamivaiaad2gadaWgaaWcba GaamyqaaqabaGccqGHRaWkcaWGubGaamyBamaaBaaaleaacaWGcbaa beaakiabgkHiTiaaikdacaWGTbWaaSbaaSqaaiaadgeaaeqaaOGaam yBamaaBaaaleaacaWGcbaabeaakiaadEgacaGGPaGaeyypa0JaaGim aaaa@42C1@

So

T= 2 m A m B m A + m B g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGubGaeyypa0ZaaSaaaeaacaaIYa GaamyBamaaBaaaleaacaWGbbaabeaakiaad2gadaWgaaWcbaGaamOq aaqabaaakeaacaWGTbWaaSbaaSqaaiaadgeaaeqaaOGaey4kaSIaam yBamaaBaaaleaacaWGcbaabeaaaaGccaWGNbaaaa@3EF8@

We pass!

 

 

Example 5: Another stupid FE exam problem: The figure shows a small block on a rotating bar.  The contact between the block and the bar has friction coefficient μ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH8oqBaaa@347B@ .  The bar rotates at constant angular speed ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHjpWDaaa@3492@ .  Find the critical angular velocity that will just make the block start to slip when θ=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH4oqCcqGH9aqpcaaIWaaaaa@363B@ .  Which way does the block slide?

 

The general approach to this problem is the same as for the Magic trick example MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  we will calculate the reaction force exerted by the bar on the block, and see when the forces are large enough to cause slip at the contact.  We analyze the motion assuming the slip does not occur, and then find out the conditions where this can no longer be the case.

 

1.      Idealization MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  We will idealize the block as a particle.  This is dangerous, because the block is clearly rotating.  We hope that because it rotates at constant rate, the rotation will not have a significant effect MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  but we can only check this once we know how to deal with rotational motion.

2.      FBD: The figure shows a free body diagram for the block.  The block is subjected to a vertical gravitational force, and reaction forces at the contact with the bar.  Since we have assumed that the contact is not slipping, we can choose the direction of the tangential component of the reaction force arbitrarily.  The resultant force on the block is

F=( TcosθNsinθ )i+(Ncosθ+Tsinθmg)j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9maabmaabaGaamivaiGacogacaGGVbGaai4CaiabeI7aXjabgkHi Tiaad6eaciGGZbGaaiyAaiaac6gacqaH4oqCaiaawIcacaGLPaaaca WHPbGaey4kaSIaaiikaiaad6eaciGGJbGaai4BaiaacohacqaH4oqC cqGHRaWkcaWGubGaci4CaiaacMgacaGGUbGaeqiUdeNaeyOeI0Iaam yBaiaadEgacaGGPaGaaCOAaaaa@5794@

3.      Kinematics  We can use the circular motion formula to write down the acceleration of tbe block (see section 3.1.3)

a=r ω 2 (cosθi+sinθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabg2 da9iabgkHiTiaadkhacqaHjpWDdaahaaWcbeqaaiaaikdaaaGccaGG OaGaci4yaiaac+gacaGGZbGaeqiUdeNaaCyAaiabgUcaRiGacohaca GGPbGaaiOBaiabeI7aXjaahQgacaGGPaaaaa@4AC8@

4.      EOM: The equation of motion is

( TcosθNsinθ )i+(Ncosθmg)j=mr ω 2 (cosθi+sinθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca WGubGaci4yaiaac+gacaGGZbGaeqiUdeNaeyOeI0IaamOtaiGacoha caGGPbGaaiOBaiabeI7aXbGaayjkaiaawMcaaiaahMgacqGHRaWkca GGOaGaamOtaiGacogacaGGVbGaai4CaiabeI7aXjabgkHiTiaad2ga caWGNbGaaiykaiaahQgacqGH9aqpcqGHsislcaWGTbGaamOCaiabeM 8a3naaCaaaleqabaGaaGOmaaaakiaacIcaciGGJbGaai4Baiaacoha cqaH4oqCcaWHPbGaey4kaSIaci4CaiaacMgacaGGUbGaeqiUdeNaaC OAaiaacMcaaaa@6349@

5.      Solution: The i and j components of the equation of motion can be solved for N and T .  Doing this by hand is a pain, but Matlab makes it painless

syms T N theta m r omega g real
eq1 = T*cos(theta)-N*sin(theta) == -m*r*omega^2*cos(theta);
eq2 = N*cos(theta)+T*sin(theta)-m*g == -m*r*omega^2*sin(theta);
[N,T] = (solve([eq1,eq2],[N,T]));
N = simplify(N)
T = simplify(T)

To find the point where the block just starts to slip, we use the friction law.  Recall that, at the point of slip

| T |=μN MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca WGubaacaGLhWUaayjcSdGaeyypa0JaeqiVd0MaamOtaaaa@3E87@

For the block to slip with θ=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUdeNaey ypa0JaaGimaaaa@3A73@

| r ω 2 |=μg MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaacq GHsislcaWGYbGaeqyYdC3aaWbaaSqabeaacaaIYaaaaaGccaGLhWUa ayjcSdGaeyypa0JaeqiVd0Maam4zaaaa@426B@

so the critical angular velocity is ω= μg/r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyYdCNaey ypa0ZaaOaaaeaacqaH8oqBcaWGNbGaai4laiaadkhaaSqabaaaaa@3E37@ .  Since the tangential traction T is negative, and the friction force must oppose sliding, the block must slide outwards, i.e. r is increasing during slip.

 

 

Alternative method of solution using normal-tangential coordinates

 

We will solve this problem again, but this time we’ll use the short-cuts described in Section 3.1.4 to write down the acceleration vector, and we’ll write down the vectors in Newton’s laws of motion in terms of the unit vectors n and t normal and tangent to the object’s path.

(i) Acceleration vector  If the block does not slip, it moves with speed V=ωr MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfacqGH9a qpcqaHjpWDcaWGYbaaaa@3ADF@  around a circular arc with radius r.   Its acceleration vector has magnitude V 2 /r MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfadaahaa WcbeqaaiaaikdaaaGccaGGVaGaamOCaaaa@39B2@  and direction parallel to the unit vector n.  

(ii) The force vector can be resolved into components parallel to n and t.  Simple trig on the free body diagram shows that

F=( Nmgcosθ )t+( mgsinθT )n MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpdaqadaqaaiaad6eacqGHsislcaWGTbGaam4zaiGacogacaGGVbGa ai4CaiabeI7aXbGaayjkaiaawMcaaiaahshacqGHRaWkdaqadaqaai aad2gacaWGNbGaci4CaiaacMgacaGGUbGaeqiUdeNaeyOeI0Iaamiv aaGaayjkaiaawMcaaiaah6gaaaa@4E50@

(iii) Newton’s laws then give

F=ma=( Nmgcosθ )t+( mgsinθT )n=m ω 2 rn MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcaWGTbGaaCyyaiabg2da9maabmaabaGaamOtaiabgkHiTiaad2ga caWGNbGaci4yaiaac+gacaGGZbGaeqiUdehacaGLOaGaayzkaaGaaC iDaiabgUcaRmaabmaabaGaamyBaiaadEgaciGGZbGaaiyAaiaac6ga cqaH4oqCcqGHsislcaWGubaacaGLOaGaayzkaaGaaCOBaiabg2da9i aad2gacqaHjpWDdaahaaWcbeqaaiaaikdaaaGccaWGYbGaaCOBaaaa @57D8@

The components of this vector equation parallel to t and n yield two equations, with solution

N=mgcosθT=mgsinθm ω 2 r MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad6eacqGH9a qpcaWGTbGaam4zaiGacogacaGGVbGaai4CaiabeI7aXjaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaadsfacqGH9aqpcaWGTbGaam4zaiGacohacaGG PbGaaiOBaiabeI7aXjabgkHiTiaad2gacqaHjpWDdaahaaWcbeqaai aaikdaaaGccaWGYbaaaa@5EDF@

This is the same solution as before.   Normal-tangential coordinates makes the equations and algebra much simpler, however.  

 

 

 

 

Example 6: Window blinds.  Have you ever wondered how window shades work?  You give the shade a little downward jerk, let it go, and it winds itself up.  If you pull the shade down slowly, it stays down.

 

The figure shows the mechanism (which probably only costs a few cents to manufacture) that achieves this remarkable feat of engineering.  It’s called an `inertial latch’ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  the same principle is used in the inertia reels on the seatbelts in your car.

 

The picture shows an enlarged end view of the window shade.  The hub, shown in brown, is fixed to the bracket supporting the shade and cannot rotate.  The drum, shown in peach, rotates as the shade is pulled up or down.  The drum is attached to a torsional spring, which tends to cause the drum to rotate counterclockwise, so winding up the shade.  The rotation is prevented by the small dogs, shown in red, which engage with the teeth on the hub.  You can pull the shade downwards freely, since the dogs allow the drum to rotate counterclockwise.

 

To raise the shade, you need to give the end of the shade a jerk downwards, and then release it.  When the drum rotates sufficiently quickly (we will calculate how quickly shortly) the dogs open up, as shown on the right.  They remain open until the drum slows down, at which point the topmost dog drops and engages with the teeth on the hub, thereby locking up the shade once more.

 

We will estimate the critical rotation rate required to free the rotating drum.

 

1.      Idealization MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  We will idealize the topmost dog as a particle on the end of a massless, inextensible rod, as shown in the figure.

a.         We will assume that the drum rotates at constant angular rate ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHjpWDaaa@3492@ .  Our goal is to calculate the critical speed where the dog is just on the point of dropping down to engage with the hub.

b.      When the drum spins fast, the particle is contacts the outer rim of the drum MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  a normal force acts at the contact.  When the dog is on the point of dropping this contact force goes to zero.  So our goal is to calculate the contact force, and then to find the critical rotation rate where the force will drop to zero.

c.       We neglect friction.

 

2.      FBD. The figure shows a free body diagram for the particle. The particle is subjected to: (i) a reaction force N where it contacts the rim; (ii) a tension T in the link, and (iii) gravity.  The resultant force is

F=( Tcos(ϕθ)Ncosθ )i+(Nsinθ+Tsin(ϕθ)mg)j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9maabmaabaGaeyOeI0IaamivaiGacogacaGGVbGaai4CaiaacIca cqaHvpGzcqGHsislcqaH4oqCcaGGPaGaeyOeI0IaamOtaiGacogaca GGVbGaai4CaiabeI7aXbGaayjkaiaawMcaaiaahMgacqGHRaWkcaGG OaGaeyOeI0IaamOtaiGacohacaGGPbGaaiOBaiabeI7aXjabgUcaRi aadsfaciGGZbGaaiyAaiaac6gacaGGOaGaeqy1dyMaeyOeI0IaeqiU deNaaiykaiabgkHiTiaad2gacaWGNbGaaiykaiaahQgaaaa@618A@

 

3.      Kinematics  We can use the circular motion formula to write down the acceleration of the particle(see section 3.1.3)

a=R ω 2 (cosθi+sinθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabg2 da9iabgkHiTiaadkfacqaHjpWDdaahaaWcbeqaaiaaikdaaaGccaGG OaGaci4yaiaac+gacaGGZbGaeqiUdeNaaCyAaiabgUcaRiGacohaca GGPbGaaiOBaiabeI7aXjaahQgacaGGPaaaaa@4AA8@

4.      EOM: The equation of motion is

( Tcos(ϕθ)Ncosθ )i+(Nsinθ+Tsin(ϕθ)mg)j=a=R ω 2 (cosθi+sinθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq GHsislcaWGubGaci4yaiaac+gacaGGZbGaaiikaiabew9aMjabgkHi TiabeI7aXjaacMcacqGHsislcaWGobGaci4yaiaac+gacaGGZbGaeq iUdehacaGLOaGaayzkaaGaaCyAaiabgUcaRiaacIcacqGHsislcaWG obGaci4CaiaacMgacaGGUbGaeqiUdeNaey4kaSIaamivaiGacohaca GGPbGaaiOBaiaacIcacqaHvpGzcqGHsislcqaH4oqCcaGGPaGaeyOe I0IaamyBaiaadEgacaGGPaGaaCOAaiabg2da9iaahggacqGH9aqpcq GHsislcaWGsbGaeqyYdC3aaWbaaSqabeaacaaIYaaaaOGaaiikaiGa cogacaGGVbGaai4CaiabeI7aXjaahMgacqGHRaWkciGGZbGaaiyAai aac6gacqaH4oqCcaWHQbGaaiykaaaa@7466@

5.      Solution: The i and j components of the equation of motion can be solved for N and T MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  MAPLE makes this painless

syms T N theta m R omega g phi real
eq1 = -T*cos(phi-theta) - N*cos(theta) == -R*omega^2*cos(theta);
eq2 = -N*sin(theta)+T*sin(phi-theta)-m*g == -R*omega^2*sin(theta);
[T,N] = solve([eq1,eq2],[T,N]);
T = simplify(T)
N = simplify(N)

 

The normal reaction force is therefore

N=mgcos(θϕ)/sinϕ+mR ω 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtaiabg2 da9iabgkHiTiaad2gacaWGNbGaci4yaiaac+gacaGGZbGaaiikaiab eI7aXjabgkHiTiabew9aMjaacMcacaGGVaGaci4CaiaacMgacaGGUb Gaeqy1dyMaey4kaSIaamyBaiaadkfacqaHjpWDdaahaaWcbeqaaiaa ikdaaaaaaa@4EEC@

We are looking for the point where this can first become zero or negative.  Note that  max{cos(θϕ)}=1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciyBaiaacg gacaGG4bGaai4EaiGacogacaGGVbGaai4CaiaacIcacqaH4oqCcqGH sislcqaHvpGzcaGGPaGaaiyFaiabg2da9iaaigdaaaa@4629@  at the point where θϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUdeNaey OeI0Iaeqy1dygaaa@3B68@  =0.  The smallest value of N therefore occurs at this point, and has magnitude

N min =mg/sinϕ+mR ω 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtamaaBa aaleaaciGGTbGaaiyAaiaac6gaaeqaaOGaeyypa0JaeyOeI0IaamyB aiaadEgacaGGVaGaci4CaiaacMgacaGGUbGaeqy1dyMaey4kaSIaam yBaiaadkfacqaHjpWDdaahaaWcbeqaaiaaikdaaaaaaa@495D@

The critical speed where N=0 follows as

ω= g/(Rsinϕ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyYdCNaey ypa0ZaaOaaaeaacaWGNbGaai4laiaacIcacaWGsbGaci4CaiaacMga caGGUbGaeqy1dyMaaiykaaWcbeaaaaa@425A@

Changing the angle ϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqy1dygaaa@38C5@  and the radius R gives a convenient way to control the critical speed in designing an inertial latch.

 

 

Alternative solution using polar coordinates

 

We’ll work through the same problem again, but this time handle the vectors using polar coordinates.

 

1.      FBD. The figure shows a free body diagram for the particle. The particle is subjected to: (i) a reaction force N where it contacts the rim; (ii) a tension T in the link, and (iii) gravity.  The resultant force is

F=(N+Tcosϕ+mgsinθ) e r +(Tsinϕmgcosθ) e θ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iabgkHiTiaacIcacaWGobGaey4kaSIaamivaiGacogacaGGVbGa ai4Caiabew9aMjabgUcaRiaad2gacaWGNbGaci4CaiaacMgacaGGUb GaeqiUdeNaaiykaiaahwgadaWgaaWcbaGaamOCaaqabaGccqGHRaWk caGGOaGaamivaiGacohacaGGPbGaaiOBaiabew9aMjabgkHiTiaad2 gacaWGNbGaci4yaiaac+gacaGGZbGaeqiUdeNaaiykaiaahwgadaWg aaWcbaGaeqiUdehabeaaaaa@5C7B@

 

2.      Kinematics  The acceleration vector is now

a=R ω 2 e r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabg2 da9iabgkHiTiaadkfacqaHjpWDdaahaaWcbeqaaiaaikdaaaGccaWH LbWaaSbaaSqaaiaadkhaaeqaaaaa@3F82@

3.      EOM: The equation of motion is

(N+Tcosϕ+mgsinθ) e r +(Tsinϕmgcosθ) e θ =R ω 2 e r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Iaai ikaiaad6eacqGHRaWkcaWGubGaci4yaiaac+gacaGGZbGaeqy1dyMa ey4kaSIaamyBaiaadEgaciGGZbGaaiyAaiaac6gacqaH4oqCcaGGPa GaaCyzamaaBaaaleaacaWGYbaabeaakiabgUcaRiaacIcacaWGubGa ci4CaiaacMgacaGGUbGaeqy1dyMaeyOeI0IaamyBaiaadEgaciGGJb Gaai4BaiaacohacqaH4oqCcaGGPaGaaCyzamaaBaaaleaacqaH4oqC aeqaaOGaeyypa0JaeyOeI0IaamOuaiabeM8a3naaCaaaleqabaGaaG OmaaaakiaahwgadaWgaaWcbaGaamOCaaqabaaaaa@624B@

4.      Solution: The e r , e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyzamaaBa aaleaacaWGYbaabeaakiaacYcacaWHLbWaaSbaaSqaaiabeI7aXbqa baaaaa@3C99@  components of the equation of motion can be solved for N and T .  If we use polar coordinates we can do this by hand MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  the e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahwgadaWgaa WcbaGaeqiUdehabeaaaaa@390A@  component shows that

T=mgcosθ/sinθ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsfacqGH9a qpcaWGTbGaam4zaiGacogacaGGVbGaai4CaiabeI7aXjaac+caciGG ZbGaaiyAaiaac6gacqaH4oqCaaa@43C1@

We can substitute this back into the e r MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahwgadaWgaa WcbaGaamOCaaqabaaaaa@384B@  component to get

N=mgcos(θϕ)/sinϕ+mR ω 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtaiabg2 da9iabgkHiTiaad2gacaWGNbGaci4yaiaac+gacaGGZbGaaiikaiab eI7aXjabgkHiTiabew9aMjaacMcacaGGVaGaci4CaiaacMgacaGGUb Gaeqy1dyMaey4kaSIaamyBaiaadkfacqaHjpWDdaahaaWcbeqaaiaa ikdaaaaaaa@4EEC@

We are looking for the point where this can first become zero or negative.  Note that  max{cos(θϕ)}=1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciyBaiaacg gacaGG4bGaai4EaiGacogacaGGVbGaai4CaiaacIcacqaH4oqCcqGH sislcqaHvpGzcaGGPaGaaiyFaiabg2da9iaaigdaaaa@4629@  at the point where θϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUdeNaey OeI0Iaeqy1dygaaa@3B68@  =0.  The smallest value of N therefore occurs at this point, and has magnitude

N min =mg/sinϕ+mR ω 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtamaaBa aaleaaciGGTbGaaiyAaiaac6gaaeqaaOGaeyypa0JaeyOeI0IaamyB aiaadEgacaGGVaGaci4CaiaacMgacaGGUbGaeqy1dyMaey4kaSIaam yBaiaadkfacqaHjpWDdaahaaWcbeqaaiaaikdaaaaaaa@495D@

The critical speed where N=0 follows as

ω= g/(Rsinϕ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyYdCNaey ypa0ZaaOaaaeaacaWGNbGaai4laiaacIcacaWGsbGaci4CaiaacMga caGGUbGaeqy1dyMaaiykaaWcbeaaaaa@425A@

Changing the angle ϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqy1dygaaa@38C5@  and the radius R gives a convenient way to control the critical speed in designing an inertial latch.

 

 

Example 7: Aircraft Dynamics Aircraft performing certain instrument approach procedures (such as holding patterns or procedure turns) are required to make all turns at a standard rate, so that a complete 360 degree turn takes 2 minutes.  All turns must be made at constant altitude and constant speed, V

 

People who design instrument approach procedures need to know the radius of the resulting turn, to make sure the aircraft won’t hit anything.   Engineers designing the aircraft are interested in the forces needed to complete the turn MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  specifically, the load factor, which is the ratio of the lift force on the aircraft to its weight.

 

In this problem we will calculate the radius of the turn R and the bank angle required, as well as the load factor caused by the maneuver, as a function of the aircraft speed V.

 

Before starting the calculation, it is helpful to understand what makes an aircraft travel in a circular path. Recall that

1.      If an object travels at constant speed around a circle, its acceleration vector has constant magnitude, and has direction towards the center of the circle

2.      A force must act on the aircraft to produce this acceleration MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  i.e. the resultant force on the aircraft must act towards the center of the circle.  The necessary force comes from the horizontal component of the lift force MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  the pilot banks the wings, so that the lift acts at an angle to the vertical.

 

With this insight, we expect to be able to use the equations of motion to calculate the forces.

 

1.      Idealization MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  The aircraft is idealized as a particle MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  it’s not obvious that this is accurate, because the aircraft clearly rotates as it travels around the curve.  However, the forces we wish to calculate turn out to be fully determined by F=ma and are not influenced by the rotational motion.

 

2.      FBD. The figure shows a free body diagram for the aircraft.  It is subjected to (i) a gravitational force (mg); (ii) a thrust from the engines F T MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGubaabeaaaaa@38CD@ , (iii) a drag force F D MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGebaabeaaaaa@38BD@ , acting perpendicular to the direction of motion, and (iv) a lift force F L MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGmbaabeaaaaa@38C5@ , acting perpendicular to the plane of the wings.

 

The resultant force is

[ ( F T F D )cosθ F L sinαsinθ ]i+[ ( F D F T )sinθ F L sinαcosθ ]j+( F L cosαmg )k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaaca GGOaGaamOramaaBaaaleaacaWGubaabeaakiabgkHiTiaadAeadaWg aaWcbaGaamiraaqabaGccaGGPaGaci4yaiaac+gacaGGZbGaeqiUde NaeyOeI0IaamOramaaBaaaleaacaWGmbaabeaakiGacohacaGGPbGa aiOBaiabeg7aHjGacohacaGGPbGaaiOBaiabeI7aXbGaay5waiaaw2 faaiaahMgacqGHRaWkdaWadaqaaiaacIcacaWGgbWaaSbaaSqaaiaa dseaaeqaaOGaeyOeI0IaamOramaaBaaaleaacaWGubaabeaakiaacM caciGGZbGaaiyAaiaac6gacqaH4oqCcqGHsislcaWGgbWaaSbaaSqa aiaadYeaaeqaaOGaci4CaiaacMgacaGGUbGaeqySdeMaci4yaiaac+ gacaGGZbGaeqiUdehacaGLBbGaayzxaaGaaCOAaiabgUcaRmaabmaa baGaamOramaaBaaaleaacaWGmbaabeaakiGacogacaGGVbGaai4Cai abeg7aHjabgkHiTiaad2gacaWGNbaacaGLOaGaayzkaaGaaC4Aaaaa @7684@

(you may find the components of the lift force difficult to visualize MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  to see where these come from, note that the lift force can be projected onto components along OR and the k direction as F L = F L sinα RO + F L cosαk MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOramaaBa aaleaacaWGmbaabeaakiabg2da9iaadAeadaWgaaWcbaGaamitaaqa baGcciGGZbGaaiyAaiaac6gacqaHXoqydaWhcaqaaiaadkfacaWGpb aacaGLxdcacqGHRaWkcaWGgbWaaSbaaSqaaiaadYeaaeqaaOGaci4y aiaac+gacaGGZbGaeqySdeMaaC4Aaaaa@4B9B@ .  Then note that  RO =sinθicosθj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca WGsbGaam4taaGaay51GaGaeyypa0JaeyOeI0Iaci4CaiaacMgacaGG UbGaeqiUdeNaaCyAaiabgkHiTiGacogacaGGVbGaai4CaiabeI7aXj aahQgaaaa@4838@ .) 

 

3.      Kinematics  

a.       The aircraft moves at constant speed around a circle, so the angle θ=ωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUdeNaey ypa0JaeqyYdCNaamiDaaaa@3C7F@ , where ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyYdChaaa@38CA@  is the (constant) angular speed of the line OP.   Since the aircraft completes a turn in two minutes, we know that   ω=2π/(2×60)=π/60 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyYdCNaey ypa0JaaGOmaiabec8aWjaac+cacaGGOaGaaGOmaiabgEna0kaaiAda caaIWaGaaiykaiabg2da9iabec8aWjaac+cacaaI2aGaaGimaaaa@4792@  rad/sec

b.      The position vector of the plane is

r=Rsinωti+Rcosωtj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadkfaciGGZbGaaiyAaiaac6gacqaHjpWDcaWG0bGaaCyAaiab gUcaRiaadkfaciGGJbGaai4BaiaacohacqaHjpWDcaWG0bGaaCOAaa aa@48AA@

We can differentiate this expression with respect to time to find the velocity

v=Rω(cosωtisinωt)j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9iaadkfacqaHjpWDcaGGOaGaci4yaiaac+gacaGGZbGaeqyYdCNa amiDaiaahMgacqGHsislciGGZbGaaiyAaiaac6gacqaHjpWDcaWG0b GaaiykaiaahQgaaaa@4B08@

c.       The magnitude of the velocity is V=Rω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 da9iaadkfacqaHjpWDaaa@3B82@ , so if the aircraft flies at speed V, the radius of the turn must be R=V/ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 da9iaadAfacaGGVaGaeqyYdChaaa@3C35@

d.      Differentiating the velocity gives the acceleration

a=R ω 2 (sinωti+cosωt)j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabg2 da9iabgkHiTiaadkfacqaHjpWDdaahaaWcbeqaaiaaikdaaaGccaGG OaGaci4CaiaacMgacaGGUbGaeqyYdCNaamiDaiaahMgacqGHRaWkci GGJbGaai4BaiaacohacqaHjpWDcaWG0bGaaiykaiaahQgaaaa@4CC8@

 

4.      EOM: The equation of motion is

[ ( F T F D )cosθ F L sinαsinθ ]i+[ ( F D F T )sinθ F L sinαcosθ ]j+( F L cosαmg )k =mR ω 2 (sinθi+cosθj) =mVω(sinθi+cosθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWada qaaiaacIcacaWGgbWaaSbaaSqaaiaadsfaaeqaaOGaeyOeI0IaamOr amaaBaaaleaacaWGebaabeaakiaacMcaciGGJbGaai4Baiaacohacq aH4oqCcqGHsislcaWGgbWaaSbaaSqaaiaadYeaaeqaaOGaci4Caiaa cMgacaGGUbGaeqySdeMaci4CaiaacMgacaGGUbGaeqiUdehacaGLBb GaayzxaaGaaCyAaiabgUcaRmaadmaabaGaaiikaiaadAeadaWgaaWc baGaamiraaqabaGccqGHsislcaWGgbWaaSbaaSqaaiaadsfaaeqaaO GaaiykaiGacohacaGGPbGaaiOBaiabeI7aXjabgkHiTiaadAeadaWg aaWcbaGaamitaaqabaGcciGGZbGaaiyAaiaac6gacqaHXoqyciGGJb Gaai4BaiaacohacqaH4oqCaiaawUfacaGLDbaacaWHQbGaey4kaSYa aeWaaeaacaWGgbWaaSbaaSqaaiaadYeaaeqaaOGaci4yaiaac+gaca GGZbGaeqySdeMaeyOeI0IaamyBaiaadEgaaiaawIcacaGLPaaacaWH RbaabaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaey ypa0JaeyOeI0IaamyBaiaadkfacqaHjpWDdaahaaWcbeqaaiaaikda aaGccaGGOaGaci4CaiaacMgacaGGUbGaeqiUdeNaaCyAaiabgUcaRi GacogacaGGVbGaai4CaiabeI7aXjaahQgacaGGPaaabaGaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7cqGH9aqpcqGHsislcaWGTbGaam OvaiabeM8a3jaacIcaciGGZbGaaiyAaiaac6gacqaH4oqCcaWHPbGa ey4kaSIaci4yaiaac+gacaGGZbGaeqiUdeNaaCOAaiaacMcaaaaa@0DC8@

 

5.      Solution: The i j and k components of the equation of motion give three equations that can be solved for F T MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGubaabeaaaaa@38CD@ , F L MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGmbaabeaaaaa@38C5@  and α MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySdegaaa@389C@ .  We assume that the drag force is known, since this is a function of the aircraft’s speed.

syms FT FD FL alpha theta R m omega g V real
eq1 = (FT-FD)*cos(theta)-FL*sin(alpha)*sin(theta)==-m*V*omega*sin(theta);
eq2 = (FT-FD)*cos(theta)-FL*sin(alpha)*cos(theta)==-m*V*omega*cos(theta);
eq3 = FL*cos(alpha)-m*g ==0;
[FL,alpha,FT] = solve([eq1,eq2,eq3],[FL,alpha,FT]);
FL = simplify(FL)
alpha = simplify(alpha)
FT = simplify(FT)

 

(The two solutions are a bit weird, but to a mathematician having the airplane fly upside down and generate negative lift is a perfectly acceptable solution)

α=2 tan 1 ((g V 2 ω 2 + g 2 )/Vω) F L =mg 1+ V 2 ω 2 / g 2 F T = F D MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySdeMaey ypa0JaeyOeI0IaaGOmaiGacshacaGGHbGaaiOBamaaCaaaleqabaGa eyOeI0IaaGymaaaakiaacIcacaGGOaGaam4zaiabgkHiTmaakaaaba GaamOvamaaCaaaleqabaGaaGOmaaaakiabeM8a3naaCaaaleqabaGa aGOmaaaakiabgUcaRiaadEgadaahaaWcbeqaaiaaikdaaaaabeaaki aacMcacaGGVaGaamOvaiabeM8a3jaacMcacaaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamOram aaBaaaleaacaWGmbaabeaakiabg2da9iaad2gacaWGNbWaaOaaaeaa caaIXaGaey4kaSIaamOvamaaCaaaleqabaGaaGOmaaaakiabeM8a3n aaCaaaleqabaGaaGOmaaaakiaac+cacaWGNbWaaWbaaSqabeaacaaI YaaaaaqabaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamOramaa BaaaleaacaWGubaabeaakiabg2da9iaadAeadaWgaaWcbaGaamiraa qabaaaaa@84E0@

 

We can calculate values of α MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySdegaaa@389C@ , R=V/ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 da9iaadAfacaGGVaGaeqyYdChaaa@3C35@  and the load factor F L /mg MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGmbaabeaakiaac+cacaWGTbGaam4zaaaa@3B60@  for a few aircraft

a.       Cessna 150 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  V=70knots (36 m/s) : α= 11 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySdeMaey ypa0JaaGymaiaaigdadaahaaWcbeqaaiaaicdaaaaaaa@3BFF@  R=690m, F L /mg=1.02 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGmbaabeaakiaac+cacaWGTbGaam4zaiabg2da9iaaigda caGGUaGaaGimaiaaikdaaaa@3F49@

b.      Boeing 747: V=200 knots (102 m/s) α= 28 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySdeMaey ypa0JaaGOmaiaaiIdadaahaaWcbeqaaiaaicdaaaaaaa@3C07@  R=1950m, F L /mg=1.14 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGmbaabeaakiaac+cacaWGTbGaam4zaiabg2da9iaaigda caGGUaGaaGymaiaaisdaaaa@3F4C@

c.       F111  V=300 knots (154 m/s) α= 39 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySdeMaey ypa0JaaG4maiaaiMdadaahaaWcbeqaaiaaicdaaaaaaa@3C09@  R=2950m, F L /mg=1.3 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGmbaabeaakiaac+cacaWGTbGaam4zaiabg2da9iaaigda caGGUaGaaG4maaaa@3E90@

 

 

Alternative solution using normal-tangential coordinates

 

This problem can also be solved rather more quickly using normal and tangential basis vectors.

 

(i) Acceleration vector.  The aircraft travels around a circular path at constant speed, so its acceleration is

a= V 2 R n=Vωn MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahggacqGH9a qpdaWcaaqaaiaadAfadaahaaWcbeqaaiaaikdaaaaakeaacaWGsbaa aiaah6gacqGH9aqpcaWGwbGaeqyYdCNaaCOBaaaa@407B@

where n is a unit vector pointing towards the center of the circle.

 

(ii) Force vector. The force vector can be written in terms of the unit vectors n,t,k as

F=( F T F D )t+ F L sinαn+( F L cosαmg)k MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcaGGOaGaamOramaaBaaaleaacaWGubaabeaakiabgkHiTiaadAea daWgaaWcbaGaamiraaqabaGccaGGPaGaaCiDaiabgUcaRiaadAeada WgaaWcbaGaamitaaqabaGcciGGZbGaaiyAaiaac6gacqaHXoqycaWH UbGaey4kaSIaaiikaiaadAeadaWgaaWcbaGaamitaaqabaGcciGGJb Gaai4BaiaacohacqaHXoqycqGHsislcaWGTbGaam4zaiaacMcacaWH Rbaaaa@5356@

 

(iii) Newton’s law F=( F T F D )t+ F L sinαn+( F L cosαmg)k=mVωn MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcaGGOaGaamOramaaBaaaleaacaWGubaabeaakiabgkHiTiaadAea daWgaaWcbaGaamiraaqabaGccaGGPaGaaCiDaiabgUcaRiaadAeada WgaaWcbaGaamitaaqabaGcciGGZbGaaiyAaiaac6gacqaHXoqycaWH UbGaey4kaSIaaiikaiaadAeadaWgaaWcbaGaamitaaqabaGcciGGJb Gaai4BaiaacohacqaHXoqycqGHsislcaWGTbGaam4zaiaacMcacaWH RbGaeyypa0JaamyBaiaadAfacqaHjpWDcaWHUbaaaa@58ED@

 

The n, t and k components of this equation give three equations that can be solved for F T MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGubaabeaaaaa@38CD@ , F L MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGmbaabeaaaaa@38C5@  and α MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySdegaaa@389C@ .  We can easily do this by hand

α= tan 1 (Vω/g) F L =mg 1+ V 2 ω 2 / g 2 F T = F D MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySdeMaey ypa0JaciiDaiaacggacaGGUbWaaWbaaSqabeaacqGHsislcaaIXaaa aOGaaiikaiaadAfacqaHjpWDcaGGVaGaam4zaiaacMcacaaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaamOramaaBaaaleaacaWGmbaabeaakiabg2da9iaad2gacaWGNb WaaOaaaeaacaaIXaGaey4kaSIaamOvamaaCaaaleqabaGaaGOmaaaa kiabeM8a3naaCaaaleqabaGaaGOmaaaakiaac+cacaWGNbWaaWbaaS qabeaacaaIYaaaaaqabaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaamOramaaBaaaleaacaWGubaabeaakiabg2da9iaadAeadaWgaaWc baGaamiraaqabaaaaa@7992@

 

3.3 Deriving and solving equations of motion for systems of particles

 

We next turn to the more difficult problem of predicting the motion of a system that is subjected to a set of forces. 

 

 

3.3.1 General procedure for deriving and solving equations of motion for systems of particles

 

It is very straightforward to analyze the motion of systems of particles.   You should always use the following procedure

1.      Introduce a set of variables that can describe the motion of the system.  Don’t worry if this sounds vague MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  it will be clear what this means when we solve specific examples.

2.      Write down the position vector of each particle in the system in terms of these variables

3.      Differentiate the position vector(s), to calculate the velocity and acceleration of each particle in terms of your variables;

4.      Draw a free body diagram showing the forces acting on each particle.  You may need to introduce variables to describe reaction forces.  Write down the resultant force vector.

5.      Write down Newton’s law F=ma MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbaaaa@39A7@  for each particle.  This will generate up to 3 equations of motion (one for each vector component) for each particle.

6.      If you wish, you can eliminate any unknown reaction forces from Newton’s laws. If you are trying to solve the equations by hand, you should always do this; of you are using MATLAB, it’s not usually necessary MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  you can have MATLAB calculate the reactions for you. The result will be a set of differential equations for the variables defined in step (1)

7.      If you find you have fewer equations than unknown variables, you should look for any constraints that restrict the motion of the particles.  The constraints must be expressed in terms of the unknown accelerations.

8.      Identify the initial conditions for the variables defined in (1).  These are usually the values of the unknown variables, their time derivatives, at time t=0. If you happen to know the values of the variables at some other instant in time, you can use that too.   If you don’t know their values at all, you should just introduce new (unknown) variables to denote the initial conditions.

9.      Solve the differential equations, subject to the initial conditions.

 

Steps (3) (6) and (8) can usually be done on the computer, so you don’t actually have to do much calculus or math. 

 

Sometimes, you can avoid solving the equations of motion completely, by using conservation laws MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  conservation of energy, or conservation of momentum MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  to calculate quantities of interest.  These short-cuts will be discussed in the next chapter.

 

3.3.2 Simple examples of equations of motion and their solutions

 

The general process described in the preceding section can be illustrated using simple examples.  In this section, we derive equations of motion for a number of simple systems, and find their solutions.

 

The purpose of these examples is to illustrate the straightforward, step-by-step procedure for analyzing motion in a system.   Although we solve several problems of practical interest, we will simply set up and solve the equations of motion with some arbitrary values for system parameter, and won’t attempt to explore their behavior in detail.   More detailed discussions of the behavior of dynamical systems will follow in later chapters.

 

Example 1: Trajectory of a particle near the earth’s surface (no air resistance)

 

At time t=0, a projectile with mass m is launched from a position X 0 =Xi+Yj+Zk MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiwamaaBa aaleaacaaIWaaabeaakiabg2da9iaadIfacaWHPbGaey4kaSIaamyw aiaahQgacqGHRaWkcaWGAbGaaC4Aaaaa@410B@  with initial velocity vector   V 0 = V x i+ V y j+ V z k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOvamaaBa aaleaacaaIWaaabeaakiabg2da9iaadAfadaWgaaWcbaGaamiEaaqa baGccaWHPbGaey4kaSIaamOvamaaBaaaleaacaWG5baabeaakiaahQ gacqGHRaWkcaWGwbWaaSbaaSqaaiaadQhaaeqaaOGaaC4Aaaaa@449C@ .  Calculate its trajectory as a function of time.

 

 

1. Introduce variables to describe the motion: We can simply use the Cartesian coordinates of the particle  (x(t),y(t),z(t)) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadI hacaGGOaGaamiDaiaacMcacaGGSaGaamyEaiaacIcacaWG0bGaaiyk aiaacYcacaWG6bGaaiikaiaadshacaGGPaGaaiykaaaa@429F@

 

2. Write down the position vector in terms of these variables: r=x(t)i+y(t)j+z(t)k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadIhacaGGOaGaamiDaiaacMcacaWHPbGaey4kaSIaamyEaiaa cIcacaWG0bGaaiykaiaahQgacqGHRaWkcaWG6bGaaiikaiaadshaca GGPaGaaC4Aaaaa@4684@

 

3. Differentiate the position vector with respect to time to find the acceleration. For this example, this is trivial

v= dx dt i+ dy dt j+ dz dt ka= d 2 x d t 2 i+ d 2 y d t 2 j+ d 2 z d t 2 k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9maalaaabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacaWHPbGa ey4kaSYaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG0baaaiaahQ gacqGHRaWkdaWcaaqaaiaadsgacaWG6baabaGaamizaiaadshaaaGa aC4AaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaahggacqGH9aqpdaWcaaqaaiaadsgadaahaaWcbeqa aiaaikdaaaGccaWG4baabaGaamizaiaadshadaahaaWcbeqaaiaaik daaaaaaOGaaCyAaiabgUcaRmaalaaabaGaamizamaaCaaaleqabaGa aGOmaaaakiaadMhaaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaa aaaaGccaWHQbGaey4kaSYaaSaaaeaacaWGKbWaaWbaaSqabeaacaaI YaaaaOGaamOEaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaa aakiaahUgaaaa@6DD0@

4. Draw a free body diagram.  The only force acting on the particle is gravity MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  the free body diagram is shown in the figure.  The force vector follows as F=mgk MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iabgkHiTiaad2gacaWGNbGaaC4Aaaaa@3B8A@ .

 

 

5. Write down Newton’s laws of motion. This is easy

F=mamgk=m( d 2 x d t 2 i+ d 2 y d t 2 j+ d 2 z d t 2 k ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbGaeyO0H4TaeyOeI0IaamyBaiaadEgacaWHRbGa eyypa0JaamyBamaabmaabaWaaSaaaeaacaWGKbWaaWbaaSqabeaaca aIYaaaaOGaamiEaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaa aaaakiaahMgacqGHRaWkdaWcaaqaaiaadsgadaahaaWcbeqaaiaaik daaaGccaWG5baabaGaamizaiaadshadaahaaWcbeqaaiaaikdaaaaa aOGaaCOAaiabgUcaRmaalaaabaGaamizamaaCaaaleqabaGaaGOmaa aakiaadQhaaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGc caWHRbaacaGLOaGaayzkaaaaaa@591E@

The vector equation actually represents three separate differential equations of motion

d 2 x d t 2 =0 d 2 y d t 2 =0 d 2 z d t 2 =g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaqaaiaadsgacaWG0bWa aWbaaSqabeaacaaIYaaaaaaakiabg2da9iaaicdacaaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7daWcaaqa aiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG5baabaGaamizaiaads hadaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0JaaGimaiaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7daWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG6baabaGa amizaiaadshadaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0JaeyOeI0 Iaam4zaaaa@6AE3@

 

6. Eliminate reactions MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  this is not needed in this example.

 

7. Identify initial conditionsThe initial conditions were given in this problem MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  we have that

{ x= X 0 dx dt = V x }{ y= Y 0 dy dt = V y }{ z= Z 0 dz dt = V z } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaaca WG4bGaeyypa0JaamiwamaaBaaaleaacaaIWaaabeaakiaaykW7caaM c8UaaGPaVlaaykW7daWcaaqaaiaadsgacaWG4baabaGaamizaiaads haaaGaeyypa0JaamOvamaaBaaaleaacaWG4baabeaakiaaykW7aiaa wUhacaGL9baacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVpaacmaabaGaamyEaiabg2da9iaadMfadaWgaaWcbaGa aGimaaqabaGccaaMc8UaaGPaVlaaykW7caaMc8+aaSaaaeaacaWGKb GaamyEaaqaaiaadsgacaWG0baaaiabg2da9iaadAfadaWgaaWcbaGa amyEaaqabaGccaaMc8oacaGL7bGaayzFaaGaaGPaVlaaykW7caaMc8 UaaGPaVpaacmaabaGaamOEaiabg2da9iaadQfadaWgaaWcbaGaaGim aaqabaGccaaMc8UaaGPaVlaaykW7caaMc8+aaSaaaeaacaWGKbGaam OEaaqaaiaadsgacaWG0baaaiabg2da9iaadAfadaWgaaWcbaGaamOE aaqabaGccaaMc8oacaGL7bGaayzFaaaaaa@8672@

 

8. Solve the equations of motion.  In general we will use MAPLE or matlab to do the rather tedious algebra necessary to solve the equations of motion.  Here, however, we will integrate the equations by hand, just to show that there is no magic in MAPLE.

 

The equations of motion are

d 2 x d t 2 =0 d 2 y d t 2 =0 d 2 z d t 2 =g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaqaaiaadsgacaWG0bWa aWbaaSqabeaacaaIYaaaaaaakiabg2da9iaaicdacaaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7daWcaaqa aiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG5baabaGaamizaiaads hadaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0JaaGimaiaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7daWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG6baabaGa amizaiaadshadaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0JaeyOeI0 Iaam4zaaaa@6AE3@

It is a bit easier to see how to solve these if we define

dx dt = v x dy dt = v y dz dt = v z MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbGaamiEaaqaaiaadsgacaWG0baaaiabg2da9iaadAhadaWgaaWc baGaamiEaaqabaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7daWcaaqaaiaadsgacaWG5baabaGaamiz aiaadshaaaGaeyypa0JaamODamaaBaaaleaacaWG5baabeaakiaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7daWcaaqaaiaadsgacaWG6baabaGaamizaiaadshaaaGaey ypa0JaamODamaaBaaaleaacaWG6baabeaaaaa@6867@

The equation of motion can be re-written in terms of ( v x , v y , v z ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadA hadaWgaaWcbaGaamiEaaqabaGccaGGSaGaamODamaaBaaaleaacaWG 5baabeaakiaaykW7caGGSaGaamODamaaBaaaleaacaWG6baabeaaki aacMcaaaa@40C7@  as

d v x dt =0 d v y dt =0 d v z dt =g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbGaamODamaaBaaaleaacaWG4baabeaaaOqaaiaadsgacaWG0baa aiabg2da9iaaicdacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7daWcaaqaaiaadsgacaWG2bWaaSbaaSqa aiaadMhaaeqaaaGcbaGaamizaiaadshaaaGaeyypa0JaaGimaiaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7daWcaaqaaiaadsgacaWG2bWaaSbaaSqaaiaadQhaaeqaaa GcbaGaamizaiaadshaaaGaeyypa0JaeyOeI0Iaam4zaaaa@68C4@

We can separate variables and integrate, using the initial conditions as limits of integration

V x v x d v x = 0 t 0dt V y v y d v x = 0 t 0dt V z v z d v z = 0 t gdt v x = V x v y = V y v z = V z gt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWdXb qaaiaadsgacaWG2bWaaSbaaSqaaiaadIhaaeqaaaqaaiaadAfadaWg aaadbaGaamiEaaqabaaaleaacaWG2bWaaSbaaWqaaiaadIhaaeqaaa qdcqGHRiI8aOGaeyypa0Zaa8qCaeaacaaIWaGaamizaiaadshaaSqa aiaaicdaaeaacaWG0baaniabgUIiYdGccaaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVpaa pehabaGaamizaiaadAhadaWgaaWcbaGaamiEaaqabaaabaGaamOvam aaBaaameaacaWG5baabeaaaSqaaiaadAhadaWgaaadbaGaamyEaaqa baaaniabgUIiYdGccqGH9aqpdaWdXbqaaiaaicdacaWGKbGaamiDaa WcbaGaaGimaaqaaiaadshaa0Gaey4kIipakiaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 +aa8qCaeaacaWGKbGaamODamaaBaaaleaacaWG6baabeaaaeaacaWG wbWaaSbaaWqaaiaadQhaaeqaaaWcbaGaamODamaaBaaameaacaWG6b aabeaaa0Gaey4kIipakiabg2da9maapehabaGaeyOeI0Iaam4zaiaa dsgacaWG0baaleaacaaIWaaabaGaamiDaaqdcqGHRiI8aaGcbaGaam ODamaaBaaaleaacaWG4baabeaakiabg2da9iaadAfadaWgaaWcbaGa amiEaaqabaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWG2bWaaS baaSqaaiaadMhaaeqaaOGaeyypa0JaamOvamaaBaaaleaacaWG5baa beaakiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaadAhadaWgaaWcbaGaamOEaaqabaGccqGH9aqpcaWGwbWaaSba aSqaaiaadQhaaeqaaOGaeyOeI0Iaam4zaiaadshaaaaa@E579@

Now we can re-write the velocity components in terms of (x,y,z) as

dx dt = V x dy dy = V y dz dt = V z gt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbGaamiEaaqaaiaadsgacaWG0baaaiabg2da9iaadAfadaWgaaWc baGaamiEaaqabaGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7daWcaa qaaiaadsgacaWG5baabaGaamizaiaadMhaaaGaeyypa0JaamOvamaa BaaaleaacaWG5baabeaakiaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVpaalaaabaGaamizaiaadQhaaeaacaWGKbGa amiDaaaacqGH9aqpcaWGwbWaaSbaaSqaaiaadQhaaeqaaOGaeyOeI0 Iaam4zaiaadshaaaa@9306@

Again, we can separate variables and integrate

X 0 x dx = 0 t V x dt Y 0 y dy = 0 t V y dt Z 0 z dz = 0 t ( V z gt )dt x= X 0 + V x ty= Y 0 + V y tz= Z 0 + V z t 1 2 g t 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWdXb qaaiaadsgacaWG4baaleaacaWGybWaaSbaaWqaaiaaicdaaeqaaaWc baGaamiEaaqdcqGHRiI8aOGaeyypa0Zaa8qCaeaacaWGwbWaaSbaaS qaaiaadIhaaeqaaOGaamizaiaadshaaSqaaiaaicdaaeaacaWG0baa niabgUIiYdGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVpaapehabaGaamizaiaadMha aSqaaiaadMfadaWgaaadbaGaaGimaaqabaaaleaacaWG5baaniabgU IiYdGccqGH9aqpdaWdXbqaaiaadAfadaWgaaWcbaGaamyEaaqabaGc caWGKbGaamiDaaWcbaGaaGimaaqaaiaadshaa0Gaey4kIipakiaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8+aa8qCaeaacaWGKbGaamOEaaWcbaGaamOwamaaBa aameaacaaIWaaabeaaaSqaaiaadQhaa0Gaey4kIipakiabg2da9maa pehabaWaaeWaaeaacaWGwbWaaSbaaSqaaiaadQhaaeqaaOGaeyOeI0 Iaam4zaiaadshaaiaawIcacaGLPaaacaWGKbGaamiDaaWcbaGaaGim aaqaaiaadshaa0Gaey4kIipaaOqaaiaadIhacqGH9aqpcaWGybWaaS baaSqaaiaaicdaaeqaaOGaey4kaSIaamOvamaaBaaaleaacaWG4baa beaakiaadshacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWG5bGaey ypa0JaamywamaaBaaaleaacaaIWaaabeaakiabgUcaRiaadAfadaWg aaWcbaGaamyEaaqabaGccaWG0bGaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaamOEaiabg2da9iaadQfadaWgaaWc baGaaGimaaqabaGccqGHRaWkcaWGwbWaaSbaaSqaaiaadQhaaeqaaO GaamiDaiabgkHiTmaalaaabaGaaGymaaqaaiaaikdaaaGaam4zaiaa dshadaahaaWcbeqaaiaaikdaaaaaaaa@EEFE@

so the position and velocity vectors are

r=( X+ V x t )i+( Y+ V y t )j+( Z+ V z t 1 2 g t 2 )k v= V x i+ V y j+( V z gt )k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWHYb Gaeyypa0ZaaeWaaeaacaWGybGaey4kaSIaamOvamaaBaaaleaacaWG 4baabeaakiaadshaaiaawIcacaGLPaaacaWHPbGaey4kaSYaaeWaae aacaWGzbGaey4kaSIaamOvamaaBaaaleaacaWG5baabeaakiaadsha aiaawIcacaGLPaaacaWHQbGaey4kaSYaaeWaaeaacaWGAbGaey4kaS IaamOvamaaBaaaleaacaWG6baabeaakiaadshacqGHsisldaWcaaqa aiaaigdaaeaacaaIYaaaaiaadEgacaWG0bWaaWbaaSqabeaacaaIYa aaaaGccaGLOaGaayzkaaGaaC4AaaqaaiaahAhacqGH9aqpcaWGwbWa aSbaaSqaaiaadIhaaeqaaOGaaCyAaiabgUcaRiaadAfadaWgaaWcba GaamyEaaqabaGccaWHQbGaey4kaSYaaeWaaeaacaWGwbWaaSbaaSqa aiaadQhaaeqaaOGaeyOeI0Iaam4zaiaadshaaiaawIcacaGLPaaaca WHRbaaaaa@670B@

 

 

Applications of trajectory problems: It is traditional in elementary physics and dynamics courses to solve vast numbers of problems involving particle trajectories.   These invariably involve being given some information about the trajectory, which you must then use to work out something else.  These problems are all somewhat tedious, but we will show a couple of examples to uphold the fine traditions of a 19th century education.

 

 Estimate how far you could throw a stone from the top of the Kremlin palace. 

 

Note that

1.      The horizontal and vertical components of velocity at time t=0 follow as

V x = v 0 cosθ V y =0 V z = v 0 sinθ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa aaleaacaWG4baabeaakiabg2da9iaadAhadaWgaaWcbaGaaGimaaqa baGcciGGJbGaai4BaiaacohacqaH4oqCcaaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWGwbWaaSbaaSqa aiaadMhaaeqaaOGaeyypa0JaaGimaiaaykW7caaMc8UaaGPaVlaayk W7caWGwbWaaSbaaSqaaiaadQhaaeqaaOGaeyypa0JaamODamaaBaaa leaacaaIWaaabeaakiGacohacaGGPbGaaiOBaiabeI7aXbaa@61F2@

2.      The components of the position of the particle at time t=0 are   X=0,Y=0,Z=H MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiwaiabg2 da9iaaicdacaGGSaGaamywaiabg2da9iaaicdacaGGSaGaamOwaiab g2da9iaadIeaaaa@404A@

3.      The trajectory of the particle follows as

r=( v 0 cosθt )i+( H+ v 0 sinθt 1 2 g t 2 )k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9maabmaabaGaamODamaaBaaaleaacaaIWaaabeaakiGacogacaGG VbGaai4CaiabeI7aXjaadshaaiaawIcacaGLPaaacaWHPbGaey4kaS YaaeWaaeaacaWGibGaey4kaSIaamODamaaBaaaleaacaaIWaaabeaa kiGacohacaGGPbGaaiOBaiabeI7aXjaadshacqGHsisldaWcaaqaai aaigdaaeaacaaIYaaaaiaadEgacaWG0bWaaWbaaSqabeaacaaIYaaa aaGccaGLOaGaayzkaaGaaC4Aaaaa@54B2@

4.      When the particle hits the ground, its position vector is r=Di MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadseacaWHPbaaaa@3AB9@ .  This must be on the trajectory, so

( v 0 cosθ t I )i+( H+ v 0 sinθ t I 1 2 g t I 2 )k=Di MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca WG2bWaaSbaaSqaaiaaicdaaeqaaOGaci4yaiaac+gacaGGZbGaeqiU deNaamiDamaaBaaaleaacaWGjbaabeaaaOGaayjkaiaawMcaaiaahM gacqGHRaWkdaqadaqaaiaadIeacqGHRaWkcaWG2bWaaSbaaSqaaiaa icdaaeqaaOGaci4CaiaacMgacaGGUbGaeqiUdeNaamiDamaaBaaale aacaWGjbaabeaakiabgkHiTmaalaaabaGaaGymaaqaaiaaikdaaaGa am4zaiaadshadaqhaaWcbaGaamysaaqaaiaaikdaaaaakiaawIcaca GLPaaacaWHRbGaeyypa0JaamiraiaahMgaaaa@5848@

where t I MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG0bWaaSbaaSqaaiaadMeaaeqaaa aa@34B9@  is the time of impact.  

5.      The two components of this vector equation gives us two equations for the two unknowns { t I ,D} MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaGG7bGaamiDamaaBaaaleaacaWGjb aabeaakiaacYcacaWGebGaaiyFaaaa@383C@ , which can be solved

clear all
syms v0 theta tI g D H real
eq1 = v0*cos(theta)*tI==D;
eq2 = H+v0*sin(theta)*tI - g*tI^2/2;
[D,tI] = solve([eq1,eq2],[D,tI]);
D = simplify(D)

For a rough estimate of the distance we can use the following numbers

1.      Height of Kremlin palace MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  71m

2.      Throwing velocity MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  maybe 25mph?  (pretty pathetic, I know - you can probably do better, especially if you are on the baseball/softball teams). 

3.      Throwing angle MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  45 degrees.

Substituting numbers gives 36m (118ft).

If you want to go wild, you can maximize D with respect to θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH4oqCaaa@347C@ , but this won’t improve your estimate much.

 

Silicon nanoparticles with radius 20nm are in thermal motion near a flat surface.  At the surface, they have an average velocity 2kT/m MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaGcaaqaaiaaikdacaWGRbGaamivai aac+cacaWGTbaaleqaaaaa@370B@ , where m is their mass, T is the temperature and k=1.3806503 × 10-23  is the Boltzmann constant. Estimate the maximum height above the surface that a typical particle can reach during its thermal motion, assuming that the only force acting on the particles is gravity

 

1.      The particle will reach its maximum height if it happens to be traveling vertically, and does not collide with any other particles.

2.      At time t=0 such a particle has position X=0,Y=0,Z=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiwaiabg2 da9iaaicdacaGGSaGaamywaiabg2da9iaaicdacaGGSaGaamOwaiab g2da9iaaicdaaaa@4037@  and velocity V x =0 V y =0 V z = 2kT/m MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa aaleaacaWG4baabeaakiabg2da9iaaicdacaaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWGwbWaaSbaaS qaaiaadMhaaeqaaOGaeyypa0JaaGimaiaaykW7caaMc8UaaGPaVlaa ykW7caWGwbWaaSbaaSqaaiaadQhaaeqaaOGaeyypa0ZaaOaaaeaaca aIYaGaam4AaiaadsfacaGGVaGaamyBaaWcbeaaaaa@5A04@

3.      For time t>0 the position vector of the particle follows as

r=( ( 2kT/m )t 1 2 g t 2 )k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9maabmaabaWaaeWaaeaadaGcaaqaaiaaikdacaWGRbGaamivaiaa c+cacaWGTbaaleqaaaGccaGLOaGaayzkaaGaamiDaiabgkHiTmaala aabaGaaGymaaqaaiaaikdaaaGaam4zaiaadshadaahaaWcbeqaaiaa ikdaaaaakiaawIcacaGLPaaacaWHRbaaaa@4798@

Its velocity is

v=( 2kT/m gt )k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9maabmaabaWaaOaaaeaacaaIYaGaam4AaiaadsfacaGGVaGaamyB aaWcbeaakiabgkHiTiaadEgacaWG0baacaGLOaGaayzkaaGaaC4Aaa aa@42A0@

4.      When the particle reaches its maximum height, its velocity must be equal to zero (if you don’t see this by visualizing the motion of the particle, you can use the mathematical statement that if   r=yk MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWHYbGaeyypa0JaamyEaiaahUgaaa a@36B9@  is a maximum, then dr/dt=v=(dy/dt)k=0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGKbGaaCOCaiaac+cacaWGKbGaam iDaiabg2da9iaahAhacqGH9aqpcaGGOaGaamizaiaadMhacaGGVaGa amizaiaadshacaGGPaGaaC4Aaiabg2da9iaaicdaaaa@42D3@  ).  Therefore, at the instant of maximum height v=( 2kT/m g t max )k=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9maabmaabaWaaOaaaeaacaaIYaGaam4AaiaadsfacaGGVaGaamyB aaWcbeaakiabgkHiTiaadEgacaWG0bWaaSbaaSqaaiGac2gacaGGHb GaaiiEaaqabaaakiaawIcacaGLPaaacaWHRbGaeyypa0JaaCimaaaa @4769@

5.      This shows that the instant of max height occurs at time t max =( 2kT/m )/g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaaciGGTbGaaiyyaiaacIhaaeqaaOGaeyypa0ZaaeWaaeaadaGc aaqaaiaaikdacaWGRbGaamivaiaac+cacaWGTbaaleqaaaGccaGLOa GaayzkaaGaai4laiaadEgaaaa@437D@

6.      Substituting this time back into the position vector shows that the position vector at max height is

r=( 2kT mg 1 2 2kT mg )k= kT mg k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9maabmaabaWaaSaaaeaacaaIYaGaam4AaiaadsfaaeaacaWGTbGa am4zaaaacqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaamaalaaaba GaaGOmaiaadUgacaWGubaabaGaamyBaiaadEgaaaaacaGLOaGaayzk aaGaaC4Aaiabg2da9maalaaabaGaam4AaiaadsfaaeaacaWGTbGaam 4zaaaacaWHRbaaaa@4C86@

7.      Si has a density of about 2330 kg/m^3. At room temperature (293K) we find that the distance is surprisingly large: 10mm or so.   Gravity is a very weak force at the nano-scale MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  surface forces acting between the particles, and the particles and the surface, are much larger.  

 

Example 2: Free vibration of a suspension system.

A vehicle suspension can be idealized as a mass m supported by a spring.  The spring has stiffness k and un-stretched length L 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGmbWaaSbaaSqaaiaaicdaaeqaaa aa@347D@ .  To test the suspension, the vehicle is constrained to move vertically, as shown in the figure. It is set in motion by stretching the spring to a length L MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGmbaaaa@3397@  and then releasing it (from rest).  Find an expression for the motion of the vehicle after it is released.

 

As an aside, it is worth noting that a particle idealization is usually too crude to model a vehicle MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  a rigid body approximation is much better.  In this case, however, we assume that the vehicle does not rotate. Under these conditions the equations of motion for a rigid body reduce to F=ma MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbaaaa@39A7@  and M=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytaiabg2 da9iaahcdaaaa@388B@ , and we shall find that we can analyze the system as if it were a particle.

 

 

1. Introduce variables to describe the motion: The length of the spring x(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiaacI cacaWG0bGaaiykaaaa@3945@  is a convenient way to describe motion. 

 

2. Write down the position vector in terms of these variables:  We can take the origin at O as shown in the figure.  The position vector of the center of mass of the block is then

r=[ x(t)+ b 2 ]j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9maadmaabaGaamiEaiaacIcacaWG0bGaaiykaiabgUcaRmaalaaa baGaamOyaaqaaiaaikdaaaaacaGLBbGaayzxaaGaaCOAaaaa@40C0@

 

3. Differentiate the position vector with respect to time to find the acceleration. For this example, this is trivial

v= dx dt ja= d 2 x d t 2 j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9maalaaabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacaWHQbGa aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaCyyaiabg2da9maalaaabaGaamiz amaaCaaaleqabaGaaGOmaaaakiaadIhaaeaacaWGKbGaamiDamaaCa aaleqabaGaaGOmaaaaaaGccaWHQbaaaa@57EB@

 

4. Draw a free body diagram.  The free body diagram is shown in the figure: the mass is subjected to the following forces

·         Gravity, acting at the center of mass of the vehicle

·         The force due to the spring

·         Reaction forces at each of the rollers that force the vehicle to move vertically.

Recall the spring force law, which says that the forces exerted by a spring act parallel to its length, tend to shorten the spring, and are proportional to the difference between the length of the spring and its un-stretched length. 

 

 

5. Write down Newton’s laws of motion. This is easy

F=ma( R Ax + R Bx )i( mg+k(x L 0 ) )j=m d 2 x d t 2 j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbGaeyO0H4TaaiikaiaadkfadaWgaaWcbaGaamyq aiaadIhaaeqaaOGaey4kaSIaamOuamaaBaaaleaacaWGcbGaamiEaa qabaGccaGGPaGaaCyAaiabgkHiTmaabmaabaGaamyBaiaadEgacqGH RaWkcaWGRbGaaiikaiaadIhacqGHsislcaWGmbWaaSbaaSqaaiaaic daaeqaaOGaaiykaaGaayjkaiaawMcaaiaahQgacqGH9aqpcaWGTbWa aSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaqaaiaads gacaWG0bWaaWbaaSqabeaacaaIYaaaaaaakiaahQgaaaa@5A9F@

The i and j components give two scalar equations of motion

( R Ax + R Bx )=0 d 2 x d t 2 =( g+ k m (x L 0 ) ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaGGOa GaamOuamaaBaaaleaacaWGbbGaamiEaaqabaGccqGHRaWkcaWGsbWa aSbaaSqaaiaadkeacaWG4baabeaakiaacMcacqGH9aqpcaaIWaaaba WaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaqaaiaa dsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaakiabg2da9iabgkHiTm aabmaabaGaam4zaiabgUcaRmaalaaabaGaam4Aaaqaaiaad2gaaaGa aiikaiaadIhacqGHsislcaWGmbWaaSbaaSqaaiaaicdaaeqaaOGaai ykaaGaayjkaiaawMcaaaaaaa@529E@

6. Eliminate reactions MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  this is not needed in this example.

 

7. Identify initial conditionsThe initial conditions were given in this problem MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  at time t=0, we know that x=L MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 da9iaadYeaaaa@38CA@  and dx/dt=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadI hacaGGVaGaamizaiaadshacqGH9aqpcaaIWaaaaa@3C31@

 

8. Solve the equations of motion. Again, we will first integrate the equations of motion by hand, and then repeat the calculation with Matlab.  The equation of motion is

d 2 x d t 2 =( g+ k m (x L 0 ) ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaqaaiaadsgacaWG0bWa aWbaaSqabeaacaaIYaaaaaaakiabg2da9iabgkHiTmaabmaabaGaam 4zaiabgUcaRmaalaaabaGaam4Aaaqaaiaad2gaaaGaaiikaiaadIha cqGHsislcaWGmbWaaSbaaSqaaiaaicdaaeqaaOGaaiykaaGaayjkai aawMcaaaaa@48FB@

We can re-write this in terms of

dx dt = v x MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbGaamiEaaqaaiaadsgacaWG0baaaiabg2da9iaadAhadaWgaaWc baGaamiEaaqabaaaaa@3CF8@

This gives

d v x dt = d v x dx dx dt = v x d v x dx =( g+ k m (x L 0 ) ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbGaamODamaaBaaaleaacaWG4baabeaaaOqaaiaadsgacaWG0baa aiabg2da9maalaaabaGaamizaiaadAhadaWgaaWcbaGaamiEaaqaba aakeaacaWGKbGaamiEaaaadaWcaaqaaiaadsgacaWG4baabaGaamiz aiaadshaaaGaeyypa0JaamODamaaBaaaleaacaWG4baabeaakmaala aabaGaamizaiaadAhadaWgaaWcbaGaamiEaaqabaaakeaacaWGKbGa amiEaaaacqGH9aqpcqGHsisldaqadaqaaiaadEgacqGHRaWkdaWcaa qaaiaadUgaaeaacaWGTbaaaiaacIcacaWG4bGaeyOeI0Iaamitamaa BaaaleaacaaIWaaabeaakiaacMcaaiaawIcacaGLPaaaaaa@5A72@

We can separate variables and integrate

0 v x v x d v x = L x ( g+ k m (x L 0 ) ) dx 1 2 v x 2 =g(xL) k 2m ( x 2 L 2 )+ k m L 0 (xL) v x = k m [ ( L L 0 + mg k ) 2 ( x L 0 + mg k ) 2 ] MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWdXb qaaiaadAhadaWgaaWcbaGaamiEaaqabaaabaGaaGimaaqaaiaadAha daWgaaadbaGaamiEaaqabaaaniabgUIiYdGccaWGKbGaamODamaaBa aaleaacaWG4baabeaakiabg2da9maapehabaGaeyOeI0YaaeWaaeaa caWGNbGaey4kaSYaaSaaaeaacaWGRbaabaGaamyBaaaacaGGOaGaam iEaiabgkHiTiaadYeadaWgaaWcbaGaaGimaaqabaGccaGGPaaacaGL OaGaayzkaaaaleaacaWGmbaabaGaamiEaaqdcqGHRiI8aOGaamizai aadIhaaeaacqGHshI3daWcaaqaaiaaigdaaeaacaaIYaaaaiaadAha daqhaaWcbaGaamiEaaqaaiaaikdaaaGccqGH9aqpcqGHsislcaWGNb GaaiikaiaadIhacqGHsislcaWGmbGaaiykaiabgkHiTmaalaaabaGa am4AaaqaaiaaikdacaWGTbaaamaabmaabaGaamiEamaaCaaaleqaba GaaGOmaaaakiabgkHiTiaadYeadaahaaWcbeqaaiaaikdaaaaakiaa wIcacaGLPaaacqGHRaWkdaWcaaqaaiaadUgaaeaacaWGTbaaaiaadY eadaWgaaWcbaGaaGimaaqabaGccaGGOaGaamiEaiabgkHiTiaadYea caGGPaaabaGaeyO0H4TaamODamaaBaaaleaacaWG4baabeaakiabg2 da9maakaaabaWaaSaaaeaacaWGRbaabaGaamyBaaaadaWadaqaamaa bmaabaGaamitaiabgkHiTiaadYeadaWgaaWcbaGaaGimaaqabaGccq GHRaWkdaWcaaqaaiaad2gacaWGNbaabaGaam4AaaaaaiaawIcacaGL PaaadaahaaWcbeqaaiaaikdaaaGccqGHsisldaqadaqaaiaadIhacq GHsislcaWGmbWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSYaaSaaaeaa caWGTbGaam4zaaqaaiaadUgaaaaacaGLOaGaayzkaaWaaWbaaSqabe aacaaIYaaaaaGccaGLBbGaayzxaaaaleqaaaaaaa@923D@

Don’t worry if the last line looks mysterious MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  writing the solution in this form just makes the algebra a bit simpler.  We can now integrate the velocity to find x

v x = dx dt = k m [ ( L L 0 + mg k ) 2 ( x L 0 + mg k ) 2 ] L x dx k m [ ( L L 0 + mg k ) 2 ( x L 0 + mg k ) 2 ] = 0 t dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG2b WaaSbaaSqaaiaadIhaaeqaaOGaeyypa0ZaaSaaaeaacaWGKbGaamiE aaqaaiaadsgacaWG0baaaiabg2da9maakaaabaWaaSaaaeaacaWGRb aabaGaamyBaaaadaWadaqaamaabmaabaGaamitaiabgkHiTiaadYea daWgaaWcbaGaaGimaaqabaGccqGHRaWkdaWcaaqaaiaad2gacaWGNb aabaGaam4AaaaaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGc cqGHsisldaqadaqaaiaadIhacqGHsislcaWGmbWaaSbaaSqaaiaaic daaeqaaOGaey4kaSYaaSaaaeaacaWGTbGaam4zaaqaaiaadUgaaaaa caGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGccaGLBbGaayzxaa aaleqaaaGcbaGaeyO0H49aa8qCaeaadaWcaaqaaiaadsgacaWG4baa baWaaOaaaeaadaWcaaqaaiaadUgaaeaacaWGTbaaamaadmaabaWaae WaaeaacaWGmbGaeyOeI0IaamitamaaBaaaleaacaaIWaaabeaakiab gUcaRmaalaaabaGaamyBaiaadEgaaeaacaWGRbaaaaGaayjkaiaawM caamaaCaaaleqabaGaaGOmaaaakiabgkHiTmaabmaabaGaamiEaiab gkHiTiaadYeadaWgaaWcbaGaaGimaaqabaGccqGHRaWkdaWcaaqaai aad2gacaWGNbaabaGaam4AaaaaaiaawIcacaGLPaaadaahaaWcbeqa aiaaikdaaaaakiaawUfacaGLDbaaaSqabaaaaaqaaiaadYeaaeaaca WG4baaniabgUIiYdGccqGH9aqpdaWdXbqaaiaadsgacaWG0baaleaa caaIWaaabaGaamiDaaqdcqGHRiI8aaaaaa@7FC9@

The integral on the left can be evaluated using the substitution

( x L 0 +mg/k ) ( L L 0 +mg/k ) =cosθ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaada qadaqaaiaadIhacqGHsislcaWGmbWaaSbaaSqaaiaaicdaaeqaaOGa ey4kaSIaamyBaiaadEgacaGGVaGaam4AaaGaayjkaiaawMcaaaqaam aabmaabaGaamitaiabgkHiTiaadYeadaWgaaWcbaGaaGimaaqabaGc cqGHRaWkcaWGTbGaam4zaiaac+cacaWGRbaacaGLOaGaayzkaaaaai abg2da9iGacogacaGGVbGaai4CaiabeI7aXbaa@4F9E@

so that

0 θ 0 sinθdθ k m [ 1 cos 2 θ ] = 0 t dt θ 0 = cos 1 ( x L 0 +mg/k ) ( L L 0 +mg/k ) θ 0 =t k m ( x L 0 +mg/k ) ( L L 0 +mg/k ) =cos( t k m ) x=( L L 0 +mg/k )cos( t k m )+ L 0 mg/k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWdXb qaamaalaaabaGaeyOeI0Iaci4CaiaacMgacaGGUbGaeqiUdeNaamiz aiabeI7aXbqaamaakaaabaWaaSaaaeaacaWGRbaabaGaamyBaaaada WadaqaaiaaigdacqGHsislciGGJbGaai4BaiaacohadaahaaWcbeqa aiaaikdaaaGccqaH4oqCaiaawUfacaGLDbaaaSqabaaaaaqaaiaaic daaeaacqaH4oqCdaWgaaadbaGaaGimaaqabaaaniabgUIiYdGccqGH 9aqpdaWdXbqaaiaadsgacaWG0baaleaacaaIWaaabaGaamiDaaqdcq GHRiI8aOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlabeI7aXnaaBaaaleaacaaIWaaabeaaki abg2da9iGacogacaGGVbGaai4CamaaCaaaleqabaGaeyOeI0IaaGym aaaakmaalaaabaWaaeWaaeaacaWG4bGaeyOeI0IaamitamaaBaaale aacaaIWaaabeaakiabgUcaRiaad2gacaWGNbGaai4laiaadUgaaiaa wIcacaGLPaaaaeaadaqadaqaaiaadYeacqGHsislcaWGmbWaaSbaaS qaaiaaicdaaeqaaOGaey4kaSIaamyBaiaadEgacaGGVaGaam4AaaGa ayjkaiaawMcaaaaaaeaacqGHshI3cqaH4oqCdaWgaaWcbaGaaGimaa qabaGccqGH9aqpcqGHsislcaWG0bWaaOaaaeaadaWcaaqaaiaadUga aeaacaWGTbaaaaWcbeaaaOqaaiabgkDiEpaalaaabaWaaeWaaeaaca WG4bGaeyOeI0IaamitamaaBaaaleaacaaIWaaabeaakiabgUcaRiaa d2gacaWGNbGaai4laiaadUgaaiaawIcacaGLPaaaaeaadaqadaqaai aadYeacqGHsislcaWGmbWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSIa amyBaiaadEgacaGGVaGaam4AaaGaayjkaiaawMcaaaaacqGH9aqpci GGJbGaai4BaiaacohadaqadaqaaiabgkHiTiaadshadaGcaaqaamaa laaabaGaam4Aaaqaaiaad2gaaaaaleqaaaGccaGLOaGaayzkaaaaba GaeyO0H4TaamiEaiabg2da9maabmaabaGaamitaiabgkHiTiaadYea daWgaaWcbaGaaGimaaqabaGccaaMc8Uaey4kaSIaamyBaiaadEgaca GGVaGaam4AaaGaayjkaiaawMcaaiGacogacaGGVbGaai4Camaabmaa baGaeyOeI0IaamiDamaakaaabaWaaSaaaeaacaWGRbaabaGaamyBaa aaaSqabaaakiaawIcacaGLPaaacaaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8Uaey4kaSIaamitamaaBaaaleaacaaIWaaabe aakiabgkHiTiaad2gacaWGNbGaai4laiaadUgaaaaa@D45F@

Here’s the Matlab solution

clear all
syms g k m L L0 real
assume(k>0); assume(m>0);
syms x(t) v(t)
eq = diff(x(t),t,2) == -(g+k*(x(t)-L0)/m);
v(t) = diff(x(t),t);
IC = [x(0)==L, v(0)==0];
x(t) = simplify(dsolve([eq,IC]))

It doesn’t quite look the same as the hand calculation MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  but of course cosθ=cos(θ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4yaiaac+ gacaGGZbGaeqiUdeNaeyypa0Jaci4yaiaac+gacaGGZbGaaiikaiab gkHiTiabeI7aXjaacMcaaaa@435B@  so they really are the same.

 

The solution is plotted in the figure.  The behavior of vibrating systems will be discussed in more detail later in this course, but it is worth noting some features of the solution:

1.      The average position of the mass is x ¯ = L 0 mg/k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmiEayaara Gaeyypa0JaamitamaaBaaaleaacaaIWaaabeaakiabgkHiTiaad2ga caWGNbGaai4laiaadUgaaaa@3F47@ .  Here, mg/k is the static deflection of the spring i.e. the deflection of the spring due to the weight of the vehicle (without motion). This means that the car vibrates symmetrically about its static deflection.

2.      The amplitude of vibration is L L 0 +mg/k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamitaiabgk HiTiaadYeadaWgaaWcbaGaaGimaaqabaGccaaMc8Uaey4kaSIaamyB aiaadEgacaGGVaGaam4Aaaaa@406A@ .  This corresponds to the distance of the mass above its average position at time t=0.

3.      The period of oscillation (the time taken for one complete cycle of vibration) is T=2π m/k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 da9iaaikdacqaHapaCdaGcaaqaaiaad2gacaGGVaGaam4AaaWcbeaa aaa@3E05@

4.      The frequency of oscillation (the number of cycles per second) is f= 1 2π k m MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiabg2 da9maalaaabaGaaGymaaqaaiaaikdacqaHapaCaaWaaOaaaeaadaWc aaqaaiaadUgaaeaacaWGTbaaaaWcbeaaaaa@3E3F@  (note f=1/T).  Frequency is also sometimes quoted as angular frequency, which is related to f  by ω=2πf= k/m MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyYdCNaey ypa0JaaGOmaiabec8aWjaadAgacqGH9aqpdaGcaaqaaiaadUgacaGG VaGaamyBaaWcbeaaaaa@40EA@ .  Angular frequency is in radians per second.

 

An interesting feature of these results is that the static deflection is related to the frequency of oscillation - so if you measure the static deflection δ=mg/k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMaey ypa0JaamyBaiaadEgacaGGVaGaam4Aaaaa@3D29@ , you can calculate the (angular) vibration frequency as ω= g/δ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyYdCNaey ypa0ZaaOaaaeaacaWGNbGaai4laiabes7aKbWcbeaaaaa@3D2F@

 

 

 

Example 3: Silly FE exam problem

 

This example shows how polar coordinates can be used to analyze motion. 

 

The rod shown in the picture rotates at constant angular speed in the horizontal plane.  The interface between block and rod has friction coefficient μ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiVd0gaaa@38B4@ .  The rod pushes a block of mass m, which starts at r=0 with radial speed V.   Find an expression for r(t).

 

 

1.      Introduce variables to describe the motion MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  the polar coordinates r,θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOCaiaacY cacqaH4oqCaaa@3A5B@  work for this problem

 

2.      Write down the position vector and differentiate to find acceleration MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  we don’t need to do this MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  we can just write down the standard result for polar coordinates

a=( d 2 r d t 2 r ω 2 ) e r +2 dr dt ω e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabg2 da9maabmaabaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGa amOCaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaakiabgk HiTiaadkhacqaHjpWDdaahaaWcbeqaaiaaikdaaaaakiaawIcacaGL PaaacaWHLbWaaSbaaSqaaiaadkhaaeqaaOGaey4kaSIaaGOmamaala aabaGaamizaiaadkhaaeaacaWGKbGaamiDaaaacqaHjpWDcaWHLbWa aSbaaSqaaiabeI7aXbqabaaaaa@50FB@

 

3.      Draw a free body diagram MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqabKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E7@  shown in the figure MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  note that it is important to draw the friction force in the correct direction.  The block will slide radially outwards, and friction opposes the slip.

 

4.      Write down Newton’s law

 

T e r +N e θ =m{ ( d 2 r d t 2 r ω 2 ) e r +2 dr dt ω e θ } MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Iaam ivaiaahwgadaWgaaWcbaGaamOCaaqabaGccqGHRaWkcaWGobGaaCyz amaaBaaaleaacqaH4oqCaeqaaOGaeyypa0JaamyBamaacmaabaWaae WaaeaadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWGYbaa baGaamizaiaadshadaahaaWcbeqaaiaaikdaaaaaaOGaeyOeI0Iaam OCaiabeM8a3naaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaiaa hwgadaWgaaWcbaGaamOCaaqabaGccqGHRaWkcaaIYaWaaSaaaeaaca WGKbGaamOCaaqaaiaadsgacaWG0baaaiabeM8a3jaahwgadaWgaaWc baGaeqiUdehabeaaaOGaay5Eaiaaw2haaaaa@5BAE@

 

5.      Eliminate reactions

 

The e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahwgadaWgaa WcbaGaeqiUdehabeaaaaa@390A@  component of F=ma shows that

N=2 dr dt ω MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad6eacqGH9a qpcaaIYaWaaSaaaeaacaWGKbGaamOCaaqaaiaadsgacaWG0baaaiab eM8a3baa@3E6E@

The friction law gives T=μN MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 da9iabeY7aTjaad6eaaaa@3B66@ .  Substituting this into the e r MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahwgadaWgaa WcbaGaamOCaaqabaaaaa@384B@  component of F=ma and simplifying shows that

 

d 2 r d t 2 +2μω dr dt r ω 2 =0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbWaaWbaaSqabeaacaaIYaaaaOGaamOCaaqaaiaadsgacaWG0bWa aWbaaSqabeaacaaIYaaaaaaakiabgUcaRiaaikdacqaH8oqBcqaHjp WDdaWcaaqaaiaadsgacaWGYbaabaGaamizaiaadshaaaGaeyOeI0Ia amOCaiabeM8a3naaCaaaleqabaGaaGOmaaaakiabg2da9iaaicdaaa a@4C0D@

 

6.      Identify initial conditions Here, r=0  dr/dt=V at time t=0.

 

7.      Solve the equation: If you’ve taken AM33 you will know how to solve this equation…   If not you can use Matlab:

 

clear all
syms mu omega V0 real
syms r(t) v(t)
diffeq = diff(r(t),t,2) + 2*mu*omega*diff(r(t),t) - r(t)*omega^2 ==0;
v(t) = diff(r(t),t);
IC = [r(0)==0,v(0)==V0];
r(t) = simplify(dsolve(diffeq,IC))

This can be simplified slightly by hand:

r(t)= V ω 1+ μ 2 e μωt sinh( 1+ μ 2 ωt) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOCaiaacI cacaWG0bGaaiykaiabg2da9maalaaabaGaamOvaaqaaiabeM8a3naa kaaabaGaaGymaiabgUcaRiabeY7aTnaaCaaaleqabaGaaGOmaaaaae qaaaaakiaadwgadaahaaWcbeqaaiabgkHiTiabeY7aTjabeM8a3jaa dshaaaGcciGGZbGaaiyAaiaac6gacaGGObGaaiikamaakaaabaGaaG ymaiabgUcaRiabeY7aTnaaCaaaleqabaGaaGOmaaaaaeqaaOGaeqyY dCNaamiDaiaacMcaaaa@551E@

 

 

 

3.3.3 Numerical solutions to equations of motion using MATLAB

 

In the preceding section, we were able to solve all our equations of motion exactly, and hence to find formulas that describe the motion of the system.  This should give you a warm and fuzzy feeling MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  it appears that with very little work, you can predict everything about the motion of the system.  You may even have visions of running a consulting business from your yacht in the Caribbean, with nothing more than your chef, your masseur (or masseuse) and a laptop with a copy of MAPLE.

 

Unfortunately real life is not so simple.   Equations of motion for most engineering systems cannot be solved exactly.  Even very simple problems, such as calculating the effects of air resistance on the trajectory of a particle, cannot be solved exactly.

 

For nearly all practical problems, the equations of motion need to be solved numerically, by using a computer to calculate values for the position, velocity and acceleration of the system as functions of time. Vast numbers of computer programs have been written for this purpose MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  some focus on very specialized applications, such as calculating orbits for spacecraft (STK); calculating motion of atoms in a material (LAMMPS); solving fluid flow problems (e.g. fluent, CFDRC); or analyzing deformation in solids (e.g. ABAQUS, ANSYS, NASTRAN, DYNA); others are more general purpose equation solving programs.

 

In this course we will use a general purpose program called MATLAB, which is widely used in all engineering applications.  You should complete the MATLAB tutorial before proceeding any further.

 

In the remainder of this section, we provide a number of examples that illustrate how MATLAB can be used to solve dynamics problems.   Each example illustrates one or more important technique for setting up or solving equations of motion.

 

 

 

Example 1: Trajectory of a particle near the earth’s surface (with air resistance)

 

As a simple example we set up MATLAB to solve the particle trajectory problem discussed in the preceding section.  We will extend the calculation to account for the effects of air resistance, however.   We will assume that our projectile is spherical, with diameter D, and we will assume that there is no wind.  You may find it helpful to review the discussion of aerodynamic drag forces in Section 2.1.7 before proceeding with this example.

 

1. Introduce variables to describe the motion: We can simply use the Cartesian coordinates of the particle  (x(t),y(t),z(t)) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadI hacaGGOaGaamiDaiaacMcacaGGSaGaamyEaiaacIcacaWG0bGaaiyk aiaacYcacaWG6bGaaiikaiaadshacaGGPaGaaiykaaaa@429F@

 

2. Write down the position vector in terms of these variables: r=x(t)i+y(t)j+z(t)k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadIhacaGGOaGaamiDaiaacMcacaWHPbGaey4kaSIaamyEaiaa cIcacaWG0bGaaiykaiaahQgacqGHRaWkcaWG6bGaaiikaiaadshaca GGPaGaaC4Aaaaa@4684@

 

3. Differentiate the position vector with respect to time to find the acceleration. Simple calculus gives

 

v= dx dt i+ dy dt j+ dz dt ka= d 2 x d t 2 i+ d 2 y d t 2 j+ d 2 z d t 2 k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9maalaaabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacaWHPbGa ey4kaSYaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG0baaaiaahQ gacqGHRaWkdaWcaaqaaiaadsgacaWG6baabaGaamizaiaadshaaaGa aC4AaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa hggacqGH9aqpdaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGcca WG4baabaGaamizaiaadshadaahaaWcbeqaaiaaikdaaaaaaOGaaCyA aiabgUcaRmaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadM haaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGccaWHQbGa ey4kaSYaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaamOEaa qaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaakiaahUgaaaa@7712@

 

4. Draw a free body diagram.  The particle is now subjected to two forces, as shown in the picture.

 

Gravity MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  as always we have F g =mgk MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOramaaBa aaleaacaWGNbaabeaakiabg2da9iabgkHiTiaad2gacaWGNbGaaC4A aaaa@3DB3@ .

 

 

Air resistance. 

 

The magnitude of the air drag force is given by F D = 1 2 ρ C D D V 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGebaabeaakiabg2da9maalaaabaGaaGymaaqaaiaaikda aaGaeqyWdiNaam4qamaaBaaaleaacaWGebaabeaakiaadseacaWGwb WaaWbaaSqabeaacaaIYaaaaaaa@4168@ , where

·           ρ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdihaaa@38BD@  is the air density,

·         C D MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaBa aaleaacaWGebaabeaaaaa@38BA@  is the drag coefficient,

·         D is the projectile’s diameter, and

·         V MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaaaa@37D8@  is the magnitude of the projectile’s velocity relative to the air. Since we assumed the air is stationary, V is simply the magnitude of the particle’s velocity, i.e.

V= ( dx dt ) 2 + ( dy dt ) 2 + ( dz dt ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 da9maakaaabaWaaeWaaeaadaWcaaqaaiaadsgacaWG4baabaGaamiz aiaadshaaaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaey 4kaSYaaeWaaeaadaWcaaqaaiaadsgacaWG5baabaGaamizaiaadsha aaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaey4kaSYaae WaaeaadaWcaaqaaiaadsgacaWG6baabaGaamizaiaadshaaaaacaGL OaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaqabaGccaaMc8oaaa@4F3C@

 

The Direction of the air drag force is always opposite to the direction of motion of the projectile through the air.  In this case the air is stationary, so the drag force is simply opposite to the direction of the particle’s velocity.   Note that v/V MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiaac+ cacaWGwbaaaa@398A@  is a unit vector parallel to the particle’s velocity.  The drag force vector is therefore

F D = 1 2 ρ C D π D 2 4 V 2 v V = 1 2 ρ C D π D 2 4 V( dx dt i+ dy dt j+ dz dt k ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOramaaBa aaleaacaWGebaabeaakiabg2da9iabgkHiTmaalaaabaGaaGymaaqa aiaaikdaaaGaeqyWdiNaam4qamaaBaaaleaacaWGebaabeaakmaala aabaGaeqiWdaNaamiramaaCaaaleqabaGaaGOmaaaaaOqaaiaaisda aaGaamOvamaaCaaaleqabaGaaGOmaaaakmaalaaabaGaaCODaaqaai aadAfaaaGaeyypa0JaeyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaaaa cqaHbpGCcaWGdbWaaSbaaSqaaiaadseaaeqaaOWaaSaaaeaacqaHap aCcaWGebWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGinaaaacaWGwbWa aeWaaeaadaWcaaqaaiaadsgacaWG4baabaGaamizaiaadshaaaGaaC yAaiabgUcaRmaalaaabaGaamizaiaadMhaaeaacaWGKbGaamiDaaaa caWHQbGaey4kaSYaaSaaaeaacaWGKbGaamOEaaqaaiaadsgacaWG0b aaaiaahUgaaiaawIcacaGLPaaaaaa@659F@

 

The total force vector is therefore

F=mgk 1 2 ρ C D π D 2 4 V( dx dt i+ dy dt j+ dz dt k ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iabgkHiTiaad2gacaWGNbGaaC4AaiabgkHiTmaalaaabaGaaGym aaqaaiaaikdaaaGaeqyWdiNaam4qamaaBaaaleaacaWGebaabeaakm aalaaabaGaeqiWdaNaamiramaaCaaaleqabaGaaGOmaaaaaOqaaiaa isdaaaGaamOvamaabmaabaWaaSaaaeaacaWGKbGaamiEaaqaaiaads gacaWG0baaaiaahMgacqGHRaWkdaWcaaqaaiaadsgacaWG5baabaGa amizaiaadshaaaGaaCOAaiabgUcaRmaalaaabaGaamizaiaadQhaae aacaWGKbGaamiDaaaacaWHRbaacaGLOaGaayzkaaaaaa@595F@

 

 

5. Write down Newton’s laws of motion.

F=mamgk 1 2 ρ C D π D 2 4 V( dx dt i+ dy dt j+ dz dt k )=m( d 2 x d t 2 i+ d 2 y d t 2 j+ d 2 z d t 2 k ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbGaeyO0H4TaeyOeI0IaamyBaiaadEgacaWHRbGa eyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaaaacqaHbpGCcaWGdbWaaS baaSqaaiaadseaaeqaaOWaaSaaaeaacqaHapaCcaWGebWaaWbaaSqa beaacaaIYaaaaaGcbaGaaGinaaaacaWGwbWaaeWaaeaadaWcaaqaai aadsgacaWG4baabaGaamizaiaadshaaaGaaCyAaiabgUcaRmaalaaa baGaamizaiaadMhaaeaacaWGKbGaamiDaaaacaWHQbGaey4kaSYaaS aaaeaacaWGKbGaamOEaaqaaiaadsgacaWG0baaaiaahUgaaiaawIca caGLPaaacqGH9aqpcaWGTbWaaeWaaeaadaWcaaqaaiaadsgadaahaa WcbeqaaiaaikdaaaGccaWG4baabaGaamizaiaadshadaahaaWcbeqa aiaaikdaaaaaaOGaaCyAaiabgUcaRmaalaaabaGaamizamaaCaaale qabaGaaGOmaaaakiaadMhaaeaacaWGKbGaamiDamaaCaaaleqabaGa aGOmaaaaaaGccaWHQbGaey4kaSYaaSaaaeaacaWGKbWaaWbaaSqabe aacaaIYaaaaOGaamOEaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaI YaaaaaaakiaahUgaaiaawIcacaGLPaaaaaa@75EC@

It is helpful to simplify the equation by defining a specific drag coefficient c= π 8m ρ D 2 C D MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yaiabg2 da9maalaaabaGaeqiWdahabaGaaGioaiaad2gaaaGaeqyWdiNaamir amaaCaaaleqabaGaaGOmaaaakiaadoeadaWgaaWcbaGaamiraaqaba aaaa@409E@ , so that

gkcV( dx dt i+ dy dt j+ dz dt k )=( d 2 x d t 2 i+ d 2 y d t 2 j+ d 2 z d t 2 k ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Iaam 4zaiaahUgacqGHsislcaWGJbGaamOvamaabmaabaWaaSaaaeaacaWG KbGaamiEaaqaaiaadsgacaWG0baaaiaahMgacqGHRaWkdaWcaaqaai aadsgacaWG5baabaGaamizaiaadshaaaGaaCOAaiabgUcaRmaalaaa baGaamizaiaadQhaaeaacaWGKbGaamiDaaaacaWHRbaacaGLOaGaay zkaaGaeyypa0ZaaeWaaeaadaWcaaqaaiaadsgadaahaaWcbeqaaiaa ikdaaaGccaWG4baabaGaamizaiaadshadaahaaWcbeqaaiaaikdaaa aaaOGaaCyAaiabgUcaRmaalaaabaGaamizamaaCaaaleqabaGaaGOm aaaakiaadMhaaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaa GccaWHQbGaey4kaSYaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaa aOGaamOEaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaaki aahUgaaiaawIcacaGLPaaaaaa@658D@

The vector equation actually represents three separate differential equations of motion

d 2 x d t 2 =cV dx dt d 2 y d t 2 =cV dy dt d 2 z d t 2 =gcV dz dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaqaaiaadsgacaWG0bWa aWbaaSqabeaacaaIYaaaaaaakiabg2da9iabgkHiTiaadogacaWGwb WaaSaaaeaacaWGKbGaamiEaaqaaiaadsgacaWG0baaaiaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVpaala aabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadMhaaeaacaWGKbGa amiDamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpcqGHsislcaWGJb GaamOvamaalaaabaGaamizaiaadMhaaeaacaWGKbGaamiDaaaacaaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVpaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaa kiaadQhaaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGccq GH9aqpcqGHsislcaWGNbGaeyOeI0Iaam4yaiaadAfadaWcaaqaaiaa dsgacaWG6baabaGaamizaiaadshaaaGaaGPaVdaa@8020@

 

6. Eliminate reactions MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  this is not needed in this example.

 

7. Identify initial conditionsThe initial conditions were given in this problem - we have that

{ x= X 0 dx dt = V x }{ y= Y 0 dy dt = V y }{ z= Z 0 dz dt = V z } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaaca WG4bGaeyypa0JaamiwamaaBaaaleaacaaIWaaabeaakiaaykW7caaM c8UaaGPaVlaaykW7daWcaaqaaiaadsgacaWG4baabaGaamizaiaads haaaGaeyypa0JaamOvamaaBaaaleaacaWG4baabeaakiaaykW7aiaa wUhacaGL9baacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVpaacmaabaGaamyEaiabg2da9iaadMfadaWgaaWcbaGa aGimaaqabaGccaaMc8UaaGPaVlaaykW7caaMc8+aaSaaaeaacaWGKb GaamyEaaqaaiaadsgacaWG0baaaiabg2da9iaadAfadaWgaaWcbaGa amyEaaqabaGccaaMc8oacaGL7bGaayzFaaGaaGPaVlaaykW7caaMc8 UaaGPaVpaacmaabaGaamOEaiabg2da9iaadQfadaWgaaWcbaGaaGim aaqabaGccaaMc8UaaGPaVlaaykW7caaMc8+aaSaaaeaacaWGKbGaam OEaaqaaiaadsgacaWG0baaaiabg2da9iaadAfadaWgaaWcbaGaamOE aaqabaGccaaMc8oacaGL7bGaayzFaaaaaa@8672@

 

8. Solve the equations of motion.  We can’t use the magic ‘dsolve’ command in MAPLE to solve this equation MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  it has no known exact solution.  So instead, we use MATLAB to generate a numerical solution.

 

This takes two steps.  First, we must turn the equations of motion into a form that MATLAB can use.  This means we must convert the equations into first-order vector valued differential equation of the general form dy dt =f(t,y) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbGaaCyEaaqaaiaadsgacaWG0baaaiabg2da9iaadAgacaGGOaGa amiDaiaacYcacaWH5bGaaiykaaaa@3FC8@ .  Then, we must write a MATLAB script to integrate the equations of motion.

 

Converting the equations of motion:  We can’t solve directly for (x,y,z), because these variables get differentiated more than once with respect to time.   To fix this, we introduce the time derivatives of (x,y,z) as new unknown variables.   In other words, we will solve for (x,y,z, v x , v y , v z ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadI hacaGGSaGaamyEaiaacYcacaWG6bGaaiilaiaadAhadaWgaaWcbaGa amiEaaqabaGccaGGSaGaamODamaaBaaaleaacaWG5baabeaakiaacY cacaWG2bWaaSbaaSqaaiaadQhaaeqaaOGaaiykaaaa@454D@ , where

v x = dx dt v y = dx dt v z = dx dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaWG4baabeaakiabg2da9maalaaabaGaamizaiaadIhaaeaa caWGKbGaamiDaaaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaamODamaaBaaaleaacaWG5baabeaakiabg2da9maalaaa baGaamizaiaadIhaaeaacaWGKbGaamiDaaaacaaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaamODamaaBaaaleaacaWG6baabeaakiabg2da9m aalaaabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaaaaa@6B00@

These definitions are three new equations of motion relating our unknown variables.   In addition, we can re-write our original equations of motion as

d v x dt =cV v x d v y dt =cV v y d v z dt =gcV v z MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbGaamODamaaBaaaleaacaWG4baabeaaaOqaaiaadsgacaWG0baa aiabg2da9iabgkHiTiaadogacaWGwbGaamODamaaBaaaleaacaWG4b aabeaakiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVpaalaaabaGaamizaiaadAhadaWgaaWcbaGaamyEaa qabaaakeaacaWGKbGaamiDaaaacqGH9aqpcqGHsislcaWGJbGaamOv aiaadAhadaWgaaWcbaGaamyEaaqabaGccaaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVpaa laaabaGaamizaiaadAhadaWgaaWcbaGaamOEaaqabaaakeaacaWGKb GaamiDaaaacqGH9aqpcqGHsislcaWGNbGaeyOeI0Iaam4yaiaadAfa caWG2bWaaSbaaSqaaiaadQhaaeqaaOGaaGPaVdaa@7A0A@

So, expressed as a vector valued differential equation, our equations of motion are

d dt [ x y z v x v y v z ]=[ v x v y v z cV v x cV v y gcV v z ] MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbaabaGaamizaiaadshaaaWaamWaaeaafaqabeGbbaaaaeaacaWG 4baabaGaamyEaaqaaiaadQhaaeaacaWG2bWaaSbaaSqaaiaadIhaae qaaaGcbaGaamODamaaBaaaleaacaWG5baabeaaaOqaaiaadAhadaWg aaWcbaGaamOEaaqabaaaaaGccaGLBbGaayzxaaGaeyypa0ZaamWaae aafaqabeGbbaaaaeaacaWG2bWaaSbaaSqaaiaadIhaaeqaaaGcbaGa amODamaaBaaaleaacaWG5baabeaaaOqaaiaadAhadaWgaaWcbaGaam OEaaqabaaakeaacqGHsislcaWGJbGaamOvaiaadAhadaWgaaWcbaGa amiEaaqabaaakeaacqGHsislcaWGJbGaamOvaiaadAhadaWgaaWcba GaamyEaaqabaaakeaacqGHsislcaWGNbGaeyOeI0Iaam4yaiaadAfa caWG2bWaaSbaaSqaaiaadQhaaeqaaaaaaOGaay5waiaaw2faaiaayk W7aaa@6101@

 

MATLAB script.  The procedure for solving these equations is discussed in the MATLAB tutorial.  A basic MATLAB script is listed below.

 

function trajectory_3d

% Function to plot trajectory of a projectile

% launched from position X0 with velocity V0

% with specific air drag coefficient c

% stop_time specifies the end of the calculation

 

g = 9.81; % gravitational accel

c=0.5; % The constant c

X0=0; Y0=0; Z0=0; % The initial position

VX0=10; VY0=10; VZ0=20;

stop_time = 5;

 

initial_w = [X0,Y0,Z0,VX0,VY0,VZ0]; % The solution at t=0

 

[times,sols] = ode45(@(t,w) eom(t,w,c,g),[0,stop_time],initial_w);

 

plot3(sols(:,1),sols(:,2),sols(:,3)) % Plot the trajectory

 

end

function dwdt = eom(t,w,c,g)

% Variables stored as follows w = [x,y,z,vx,vy,vz]

% i.e. x = w(1), y=w(2), z=w(3), etc

   x = w(1); y=w(2); z=w(3);

   vx = w(4); vy = w(5); vz = w(6);

   vmag = sqrt(vx^2+vy^2+vz^2);

   dxdt = vx; dydt = vy; dzdt = vz;

   dvxdt = -c*vmag*vx;

   dvydt = -c*vmag*vy;

   dvzdt = -c*vmag*vz-g;

   dwdt = [dxdt;dydt;dzdt;dvxdt;dvydt;dvzdt];

end

 

 

This produces a plot that looks like this (the plot’s been edited to add the grid,etc)

 

 

Example 2: Simple satellite orbit calculation

 

The figure shows satellite with mass m orbiting a planet with mass M.  At time t=0 the satellite has position vector r=Ri MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadkfacaWHPbaaaa@39C0@  and velocity vector v=Vj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9iaadAfacaWHQbaaaa@39C9@ .  The planet’s motion may be neglected (this is accurate as long as M>>m). Calculate and plot the orbit of the satellite.

 

1. Introduce variables to describe the motion: We will use the (x,y) coordinates of the satellite.

 

2. Write down the position vector in terms of these variables: r=xi+yj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadIhacaWHPbGaey4kaSIaamyEaiaahQgaaaa@3CB9@

 

3. Differentiate the position vector with respect to time to find the acceleration.

v= dx dt i+ dy dt ja= d 2 x d t 2 i+ d 2 y d t 2 j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9maalaaabaGaamizaiaadIhaaeaacaWGKbGaamiDaaaacaWHPbGa ey4kaSYaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG0baaaiaahQ gacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaahggacqGH9a qpdaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG4baabaGa amizaiaadshadaahaaWcbeqaaiaaikdaaaaaaOGaaCyAaiabgUcaRm aalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadMhaaeaacaWG KbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGccaWHQbaaaa@6841@

 

4. Draw a free body diagram.  The satellite is subjected to a gravitational force.

The magnitude of the force is F g = GMm r 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGNbaabeaakiabg2da9maalaaabaGaam4raiaad2eacaWG TbaabaGaamOCamaaCaaaleqabaGaaGOmaaaaaaaaaa@3E70@ , where

·         G MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4raaaa@36C2@  is the gravitational constant, and

·         r= x 2 + y 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOCaiabg2 da9maakaaabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaa dMhadaahaaWcbeqaaiaaikdaaaaabeaaaaa@3DC3@  is the distance between the planet and the satellite

 

The direction of the force is always towards the origin:  r/r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaC OCaiaac+cacaWGYbaaaa@3A8F@  is therefore a unit vector parallel to the direction of the force.   The total force acting on the satellite is therefore

F= GMm r 2 r r = GMm r 3 ( xi+yj ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iabgkHiTmaalaaabaGaam4raiaad2eacaWGTbaabaGaamOCamaa CaaaleqabaGaaGOmaaaaaaGcdaWcaaqaaiaahkhaaeaacaWGYbaaai abg2da9iabgkHiTmaalaaabaGaam4raiaad2eacaWGTbaabaGaamOC amaaCaaaleqabaGaaG4maaaaaaGcdaqadaqaaiaadIhacaWHPbGaey 4kaSIaamyEaiaahQgaaiaawIcacaGLPaaaaaa@4D14@

 

5. Write down Newton’s laws of motion.

F=ma GMm r 3 ( xi+yj )=m( d 2 x d t 2 i+ d 2 y d t 2 j ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbGaeyO0H4TaeyOeI0YaaSaaaeaacaWGhbGaamyt aiaad2gaaeaacaWGYbWaaWbaaSqabeaacaaIZaaaaaaakmaabmaaba GaamiEaiaahMgacqGHRaWkcaWG5bGaaCOAaaGaayjkaiaawMcaaiab g2da9iaad2gadaqadaqaamaalaaabaGaamizamaaCaaaleqabaGaaG OmaaaakiaadIhaaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaa aaGccaWHPbGaey4kaSYaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYa aaaOGaamyEaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaa kiaahQgaaiaawIcacaGLPaaaaaa@598C@

The vector equation represents two separate differential equations of motion

d 2 x d t 2 = GM r 3 x d 2 y d t 2 = GM r 3 y MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaqaaiaadsgacaWG0bWa aWbaaSqabeaacaaIYaaaaaaakiabg2da9iabgkHiTmaalaaabaGaam 4raiaad2eaaeaacaWGYbWaaWbaaSqabeaacaaIZaaaaaaakiaadIha caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7daWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG5baa baGaamizaiaadshadaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0Jaey OeI0YaaSaaaeaacaWGhbGaamytaaqaaiaadkhadaahaaWcbeqaaiaa iodaaaaaaOGaamyEaaaa@5C69@

 

6. Eliminate reactions MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  this is not needed in this example.

 

7. Identify initial conditionsThe initial conditions were given in this problem - we have that

{ x=R dx dt =0 }{ y=0 dy dt =V } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaaca WG4bGaeyypa0JaamOuaiaaykW7caaMc8UaaGPaVlaaykW7daWcaaqa aiaadsgacaWG4baabaGaamizaiaadshaaaGaeyypa0JaaGimaiaayk W7aiaawUhacaGL9baacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVpaacmaabaGaamyEaiabg2da9iaaicdacaaMc8 UaaGPaVlaaykW7caaMc8+aaSaaaeaacaWGKbGaamyEaaqaaiaadsga caWG0baaaiabg2da9iaadAfaaiaawUhacaGL9baaaaa@657D@

 

8. Solve the equations of motion.  We follow the usual procedure: (i) convert the equations into MATLAB form; and (ii) code a MATLAB script to solve them.

 

Converting the equations of motion:  We introduce the time derivatives of (x,y) as new unknown variables.   In other words, we will solve for (x,y, v x , v y ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadI hacaGGSaGaamyEaiaacYcacaWG2bWaaSbaaSqaaiaadIhaaeqaaOGa aiilaiaadAhadaWgaaWcbaGaamyEaaqabaGccaGGPaaaaa@40BE@ , where

v x = dx dt v y = dx dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaWG4baabeaakiabg2da9maalaaabaGaamizaiaadIhaaeaa caWGKbGaamiDaaaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaamODamaaBaaaleaacaWG5baabeaakiabg2da9maalaaa baGaamizaiaadIhaaeaacaWGKbGaamiDaaaacaaMc8oaaa@516E@

These definitions are new equations of motion relating our unknown variables.   In addition, we can re-write our original equations of motion as

d v x dt = GM r 3 x d v y dt = GM r 3 y MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbGaamODamaaBaaaleaacaWG4baabeaaaOqaaiaadsgacaWG0baa aiabg2da9iabgkHiTmaalaaabaGaam4raiaad2eaaeaacaWGYbWaaW baaSqabeaacaaIZaaaaaaakiaadIhacaaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8+aaSaaaeaaca WGKbGaamODamaaBaaaleaacaWG5baabeaaaOqaaiaadsgacaWG0baa aiabg2da9iabgkHiTmaalaaabaGaam4raiaad2eaaeaacaWGYbWaaW baaSqabeaacaaIZaaaaaaakiaadMhacaaMc8oaaa@5F1C@

So, expressed as a vector valued differential equation, our equations of motion are

d dt [ x y v x v y ]=[ v x v y GMx/ r 3 GMy/ r 3 ] MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbaabaGaamizaiaadshaaaWaamWaaeaafaqabeabbaaaaeaacaWG 4baabaGaamyEaaqaaiaadAhadaWgaaWcbaGaamiEaaqabaaakeaaca WG2bWaaSbaaSqaaiaadMhaaeqaaaaaaOGaay5waiaaw2faaiabg2da 9maadmaabaqbaeqabqqaaaaabaGaamODamaaBaaaleaacaWG4baabe aaaOqaaiaadAhadaWgaaWcbaGaamyEaaqabaaakeaacqGHsislcaWG hbGaamytaiaadIhacaGGVaGaamOCamaaCaaaleqabaGaaG4maaaaaO qaaiabgkHiTiaadEeacaWGnbGaamyEaiaac+cacaWGYbWaaWbaaSqa beaacaaIZaaaaaaaaOGaay5waiaaw2faaiaaykW7caaMc8oaaa@58FC@

 

Matlab script: Here’s a simple script to solve these equations.

 

function satellite_orbit

% Function to plot orbit of a satellite

% launched from position (R,0) with velocity (0,V)

 

GM=1;

R=1;

V=1;

Time=100;

w0 = [R,0,0,V]; % Initial conditions

 

[t_values,w_values] = ode45(@(t,w) odefunc(t,w,GM),[0,time],w0);

 

plot(w_values(:,1),w_values(:,2))

 

 

end

function dwdt = odefunc(t,w,GM)

   x=w(1); y=w(2); vx=w(3); vy=w(4);
   r = sqrt(x^2+y^2);

   dwdt = [vx;vy;-GM*x/r^3;-GM*y/r^3];

end

 

 

Running the script produces the result shown (the plot was annotated by hand)

 

Do we believe this result?   It is a bit surprising MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  the satellite seems to be spiraling in towards the planet.   Most satellites don’t do this MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  so the result is a bit suspicious.   The First Law of Scientific Computing states that ` if a computer simulation predicts a result that surprises you, it is probably wrong.’

So how can we test our computation?   There are two good tests:

1.      Look for any features in the simulation that you can predict without computation, and compare your predictions with those of the computer.

2.      Try to find a special choice of system parameters for which you can derive an exact solution to your problem, and compare your result with the computer

We can use both these checks here.

 

1. Conserved quantities  For this particular problem, we know that (i) the total energy of the system should be constant; and (ii) the angular momentum of the system about the planet should be constant (these conservation laws will be discussed in the next chapter MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  for now, just take this as given).  The total energy of the system consists of the potential energy and kinetic energy of the satellite, and can be calculated from the formula

E= GMm r + 1 2 m v 2 = GMm r + 1 2 m( v x 2 + v y 2 ) E m = GM r + 1 2 ( v x 2 + v y 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGfb Gaeyypa0JaeyOeI0YaaSaaaeaacaWGhbGaamytaiaad2gaaeaacaWG YbaaaiabgUcaRiaaykW7daWcaaqaaiaaigdaaeaacaaIYaaaaiaad2 gacaWG2bWaaWbaaSqabeaacaaIYaaaaOGaeyypa0JaeyOeI0YaaSaa aeaacaWGhbGaamytaiaad2gaaeaacaWGYbaaaiabgUcaRiaaykW7da WcaaqaaiaaigdaaeaacaaIYaaaaiaad2gadaqadaqaaiaadAhadaqh aaWcbaGaamiEaaqaaiaaikdaaaGccqGHRaWkcaWG2bWaa0baaSqaai aadMhaaeaacaaIYaaaaaGccaGLOaGaayzkaaaabaGaeyO0H49aaSaa aeaacaWGfbaabaGaamyBaaaacqGH9aqpcqGHsisldaWcaaqaaiaadE eacaWGnbaabaGaamOCaaaacqGHRaWkcaaMc8+aaSaaaeaacaaIXaaa baGaaGOmaaaadaqadaqaaiaadAhadaqhaaWcbaGaamiEaaqaaiaaik daaaGccqGHRaWkcaWG2bWaa0baaSqaaiaadMhaaeaacaaIYaaaaaGc caGLOaGaayzkaaaaaaa@6BD5@

The total angular momentum of the satellite (about the origin) can be calculated from the formula

H=r×mv=( xi+yj )×m( v x i+ v y j)=m( x v y y v x )k | H | m =( x v y y v x ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWHib Gaeyypa0JaaCOCaiabgEna0kaad2gacaWH2bGaeyypa0ZaaeWaaeaa caWG4bGaaCyAaiabgUcaRiaadMhacaWHQbaacaGLOaGaayzkaaGaey 41aqRaamyBaiaacIcacaWG2bWaaSbaaSqaaiaadIhaaeqaaOGaaCyA aiabgUcaRiaadAhadaWgaaWcbaGaamyEaaqabaGccaWHQbGaaiykai abg2da9iaad2gadaqadaqaaiaadIhacaWG2bWaaSbaaSqaaiaadMha aeqaaOGaeyOeI0IaamyEaiaadAhadaWgaaWcbaGaamiEaaqabaaaki aawIcacaGLPaaacaWHRbaabaGaeyO0H49aaSaaaeaadaabdaqaaiaa hIeaaiaawEa7caGLiWoaaeaacaWGTbaaaiabg2da9maabmaabaGaam iEaiaadAhadaWgaaWcbaGaamyEaaqabaGccqGHsislcaWG5bGaamOD amaaBaaaleaacaWG4baabeaaaOGaayjkaiaawMcaaaaaaa@6D95@

(If you don’t know these formulas, don’t panic MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqacKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E8@  we will discuss energy and angular momentum in the next part of the course)

 

We can have MATLAB plot E/m and | H |/m MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca WHibaacaGLhWUaayjcSdGaai4laiaad2gaaaa@3C95@ , and see if these are really conserved.  The energy and momentum can be calculated by adding these lines to the MATLAB script

 

 

for i =1:length(t)

     r = sqrt(w_values(i,1)^2 + w_values(i,2)^2)

     vmag = sqrt(w_values(i,1)^2 + w_values(i,2)^2)

energy(i) = -GM/r + vmag^2/2;

angularm(i) = w_values(i,1)*w_values(i,4)-w_values(i,2)*w_values(i,3);

end

 

You can then plot the results (e.g. plot(t_values,energy)).  The results are shown below.

These results look really bad MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqacKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E8@  neither energy, nor angular momentum, are conserved in the simulation.  Something is clearly very badly wrong.

 

Comparison to exact solution: It is not always possible to find a convenient exact solution, but in this case, we might guess that some special initial conditions could set the satellite moving on a circular path.  A circular path might be simple enough to analyze by hand.  So let’s assume that the path is circular, and try to find the necessary initial conditions. If you still remember the circular motion formulas, you could use them to do this.  But only morons use formulas MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  here we will derive the solution from scratch. Note that, for a circular path

(a) the particle’s radius r=constant.  In fact, we know r=R, from the position at time t=0.

(b) The satellite must move at constant speed, and the angle θ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUdehaaa@37AC@  must increase linearly with time, i.e. θ=ωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUdeNaey ypa0JaeqyYdCNaamiDaaaa@3B78@  where θ=ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUdeNaey ypa0JaeqyYdChaaa@3A7F@  is a constant (see section 3.1.3 to review motion at constant speed around a circle).

With this information we can solve the equations of motion.   Recall that the position, velocity and acceleration vectors for a particle traveling at constant speed around a circle are

r=Rcosθ(t)i+Rsinθ(t)j v=Rω(sinθ(t)i+cosθ(t)j) a=R ω 2 ( cosθ(t)i+Rsinθ(t)j ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWHYb Gaeyypa0JaamOuaiGacogacaGGVbGaai4CaiabeI7aXjaacIcacaWG 0bGaaiykaiaahMgacqGHRaWkcaWGsbGaci4CaiaacMgacaGGUbGaeq iUdeNaaiikaiaadshacaGGPaGaaCOAaaqaaiaahAhacqGH9aqpcaWG sbGaeqyYdCNaaiikaiabgkHiTiGacohacaGGPbGaaiOBaiabeI7aXj aacIcacaWG0bGaaiykaiaahMgacqGHRaWkciGGJbGaai4Baiaacoha cqaH4oqCcaGGOaGaamiDaiaacMcacaWHQbGaaiykaaqaaiaahggacq GH9aqpcqGHsislcaWGsbGaeqyYdC3aaWbaaSqabeaacaaIYaaaaOWa aeWaaeaaciGGJbGaai4BaiaacohacqaH4oqCcaGGOaGaamiDaiaacM cacaWHPbGaey4kaSIaamOuaiGacohacaGGPbGaaiOBaiabeI7aXjaa cIcacaWG0bGaaiykaiaahQgaaiaawIcacaGLPaaaaaaa@7AF6@

We know that | v |=V MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca WH2baacaGLhWUaayjcSdGaeyypa0JaamOvaaaa@3BF8@  from the initial conditions, and | v | MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca WH2baacaGLhWUaayjcSdaaaa@3A17@  is constant. This tells us that

V=Rω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 da9iaadkfacqaHjpWDaaa@3A7B@

Finally, we can substitute this into Newton’s law

F=ma GMm R 2 r R =ma GMm R 2 (cosθi+sinθj)=m V 2 R (cosθi+sinθj) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbGaeyO0H4TaeyOeI0YaaSaaaeaacaWGhbGaamyt aiaad2gaaeaacaWGsbWaaWbaaSqabeaacaaIYaaaaaaakmaalaaaba GaaCOCaaqaaiaadkfaaaGaeyypa0JaamyBaiaahggacqGHshI3cqGH sisldaWcaaqaaiaadEeacaWGnbGaamyBaaqaaiaadkfadaahaaWcbe qaaiaaikdaaaaaaOGaaiikaiGacogacaGGVbGaai4CaiabeI7aXjaa hMgacqGHRaWkciGGZbGaaiyAaiaac6gacqaH4oqCcaWHQbGaaiykai abg2da9iabgkHiTiaad2gadaWcaaqaaiaadAfadaahaaWcbeqaaiaa ikdaaaaakeaacaWGsbaaaiaacIcaciGGJbGaai4BaiaacohacqaH4o qCcaWHPbGaey4kaSIaci4CaiaacMgacaGGUbGaeqiUdeNaaCOAaiaa cMcaaaa@6EE2@

Both components of the equation of motion are satisfied if we choose

GM R 2 = V 2 R MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGhbGaamytaaqaaiaadkfadaahaaWcbeqaaiaaikdaaaaaaOGaeyyp a0ZaaSaaaeaacaWGwbWaaWbaaSqabeaacaaIYaaaaaGcbaGaamOuaa aaaaa@3E30@

So, if we choose initial values of GM,V,R MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4raiaad2 eacaGGSaGaamOvaiaacYcacaWGsbaaaa@3BAD@  satisfying this equation, the orbit will be circular.  In fact, our original choice, GM=1,V=1,R=1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4raiaad2 eacqGH9aqpcaaIXaGaaiilaiaadAfacqGH9aqpcaaIXaGaaiilaiaa dkfacqGH9aqpcaaIXaaaaa@40F0@  should have given a circular orbit.  It did not.  Again, this means our computer generated solution is totally wrong.

 

 

 

Fixing the problem:  In general, when computer predictions are suspect, we need to check the following

1.      Is there an error in our MATLAB program?  This is nearly always the cause of the problem.  In this case, however, the program is correct (it’s too simple to get wrong, even for me).

2.      There may be something wrong with our equations of motion (because we made a mistake in the derivation).  This would not explain the discrepancy between the circular orbit we predict and the simulation, since we used the same equations in both cases.

3.      Is the MATLAB solution sufficiently accurate?  Remember that by default the ODE solver tries to give a solution that has 0.1% error.  This may not be good enough.  So we can try solving the problem again, but with better accuracy.  We can do this by modifying the MATLAB call to the equation solver as follows

      options = odeset('RelTol',0.00001);

      [t_values,w_vlues] = ode45(@(t,w) odefunc(t,w,GM),[0,time],w0,options);

  1. Is there some feature of the equation of motion that makes them especially difficult to solve? In this case we might have to try a different equation solver, or try a different way to set up the problem.

 

The figure on the right shows the orbit predicted with the better accuracy.  You can see there is no longer any problem MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  the orbit is perfectly circular.   The figures below plot the energy and angular momentum predicted by the computer.

 

There is a small change in energy and angular momentum but the rate of change has been reduced dramatically.   We can make the error smaller still by using improving the tolerances further, if this is needed.  But the changes in energy and angular momentum are only of order 0.01% over a large number of orbits: this would be sufficiently accurate for most practical applications.

 

Most ODE solvers are purposely designed to lose a small amount of energy as the simulation proceeds, because this helps to make them more stable.  But in some applications this is unacceptable MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  for example in a molecular dynamic simulation where we are trying to predict the entropic response of a polymer, or a free vibration problem where we need to run the simulation for an extended period of time.  There are special ODE solvers that will conserve energy exactly.

 

 

 

Example 3: Earthquake response of a 2-storey building

 

The figure shows a very simple idealization of a 2-storey building.   The roof and second floor are idealized as blocks with mass m.  They are supported by structural columns, which can be idealized as springs with stiffness k and unstretched length L.   At time t=0 the floors are at rest and the columns have lengths l 1 =Lmg/k l 2 =Lmg/2k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiBamaaBa aaleaacaaIXaaabeaakiabg2da9iaadYeacqGHsislcaWGTbGaam4z aiaac+cacaWGRbGaaGPaVlaaykW7caaMc8UaaGPaVlaadYgadaWgaa WcbaGaaGOmaaqabaGccqGH9aqpcaWGmbGaeyOeI0IaamyBaiaadEga caGGVaGaaGOmaiaadUgaaaa@4E34@  (can you show this?).  We will neglect the thickness of the floors themselves, to keep things simple.

 

For time t>0, an earthquake makes the ground vibrate vertically. The ground motion can be described using the equation d= d 0 sinωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiabg2 da9iaadsgadaWgaaWcbaGaaGimaaqabaGcciGGZbGaaiyAaiaac6ga cqaHjpWDcaWG0baaaa@4063@ .  Horizontal motion may be neglected. Our goal is to calculate the motion of the first and second floor of the building.

It is worth noting a few points about this problem:

1.      You may be skeptical that the floor of a building can be idealized as a particle (then again, maybe you couldn’t care less…).    If so, you are right MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  it certainly is not a `small’ object.   However, because the floors move vertically without rotation, the rigid body equations of motion simply reduce to   F=ma MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbaaaa@3AAE@  and M=0, where the moments are taken about the center of mass of the block.  The floors behave as though they are particles, even though they are very large.

2.      Real earthquakes involve predominantly horizontal, not vertical motion of the ground.  In addition, structural columns resist extensional loading much more strongly than transverse loading.  So we should really be analyzing horizontal motion of the building rather than vertical motion.  However, the free body diagrams for horizontal motion are messy (see if you can draw them) and the equations of motion for vertical and horizontal motion turn out to be the same, so we consider vertical motion to keep things simple.

3.      This problem could be solved analytically (e.g. using the `dsolve’ feature of MAPLE) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  a numerical solution is not necessary.  Try this for yourself.

 

1. Introduce variables to describe the motion: We will use the height of each floor ( y 1 , y 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadM hadaWgaaWcbaGaaGymaaqabaGccaGGSaGaamyEamaaBaaaleaacaaI YaaabeaakiaacMcaaaa@3CE5@  as the variables.

 

2. Write down the position vector in terms of these variables: We now have to worry about two masses, and must write down the position vector of both

r 1 = y 1 j r 2 = y 2 j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCamaaBa aaleaacaaIXaaabeaakiabg2da9iaadMhadaWgaaWcbaGaaGymaaqa baGccaWHQbGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaCOCamaaBaaaleaa caaIYaaabeaakiabg2da9iaadMhadaWgaaWcbaGaaGOmaaqabaGcca WHQbGaaGPaVdaa@56B6@

Note that we must measure the position of each mass from a fixed point.

 

 

 

3. Differentiate the position vector with respect to time to find the acceleration.

v 1 =( d y 1 dt )j a 1 =( d y 1 2 d t 2 )j v 2 =( d y 2 dt )j a 2 =( d y 2 2 d t 2 )j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWH2b WaaSbaaSqaaiaaigdaaeqaaOGaeyypa0ZaaeWaaeaadaWcaaqaaiaa dsgacaWG5bWaaSbaaSqaaiaaigdaaeqaaaGcbaGaamizaiaadshaaa aacaGLOaGaayzkaaGaaCOAaiaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaahggadaWgaaWcbaGaaGymaa qabaGccqGH9aqpdaqadaqaamaalaaabaGaamizaiaadMhadaqhaaWc baGaaGymaaqaaiaaikdaaaaakeaacaWGKbGaamiDamaaCaaaleqaba GaaGOmaaaaaaaakiaawIcacaGLPaaacaWHQbGaaGPaVdqaaiaahAha daWgaaWcbaGaaGOmaaqabaGccqGH9aqpdaqadaqaamaalaaabaGaam izaiaadMhadaWgaaWcbaGaaGOmaaqabaaakeaacaWGKbGaamiDaaaa aiaawIcacaGLPaaacaWHQbGaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaCyyamaaBaaaleaacaaIYaaabeaakiabg2da9maabmaa baWaaSaaaeaacaWGKbGaamyEamaaDaaaleaacaaIYaaabaGaaGOmaa aaaOqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaaaOGaayjk aiaawMcaaiaahQgacaaMc8oaaaa@7D69@

 

4. Draw a free body diagram. We must draw a free body diagram for each mass. The resultant force acting on the bottom and top masses, respectively, are

   F 1 ={ mg2k( l 1 L)+2k( l 2 L) }j F 2 =mg2k( l 2 L)j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOramaaBa aaleaacaaIXaaabeaakiabg2da9maacmaabaGaeyOeI0IaamyBaiaa dEgacqGHsislcaaIYaGaam4AaiaacIcacaWGSbWaaSbaaSqaaiaaig daaeqaaOGaeyOeI0IaamitaiaacMcacqGHRaWkcaaIYaGaam4Aaiaa cIcacaWGSbWaaSbaaSqaaiaaikdaaeqaaOGaeyOeI0IaamitaiaacM caaiaawUhacaGL9baacaWHQbGaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaC OramaaBaaaleaacaaIYaaabeaakiabg2da9iabgkHiTiaad2gacaWG NbGaeyOeI0IaaGOmaiaadUgacaGGOaGaamiBamaaBaaaleaacaaIYa aabeaakiabgkHiTiaadYeacaGGPaGaaCOAaiaaykW7aaa@70F3@

 

We will have to find the spring lengths l 1 , l 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiBamaaBa aaleaacaaIXaaabeaakiaacYcacaWGSbWaaSbaaSqaaiaaikdaaeqa aaaa@3B69@  in terms of our coordinates y 1 , y 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEamaaBa aaleaacaaIXaaabeaakiaacYcacaWG5bWaaSbaaSqaaiaaikdaaeqa aaaa@3B83@  to solve the problem.   Geometry shows that

l 2 = y 2 y 1 l 1 = y 1 d 0 sinωt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiBamaaBa aaleaacaaIYaaabeaakiabg2da9iaadMhadaWgaaWcbaGaaGOmaaqa baGccqGHsislcaWG5bWaaSbaaSqaaiaaigdaaeqaaOGaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaadYgadaWgaaWcbaGaaGymaaqabaGccq GH9aqpcaWG5bWaaSbaaSqaaiaaigdaaeqaaOGaeyOeI0Iaamizamaa BaaaleaacaaIWaaabeaakiGacohacaGGPbGaaiOBaiabeM8a3jaads haaaa@5FFD@

 

5. Write down Newton’s laws of motion.  F=ma for each mass gives

{ mg2k( l 1 L)+2k( l 2 L) }j=m d 2 y 1 d t 2 j { mg2k( l 2 L) }j=m d 2 y 2 d t 2 j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaGada qaaiabgkHiTiaad2gacaWGNbGaeyOeI0IaaGOmaiaadUgacaGGOaGa amiBamaaBaaaleaacaaIXaaabeaakiabgkHiTiaadYeacaGGPaGaey 4kaSIaaGOmaiaadUgacaGGOaGaamiBamaaBaaaleaacaaIYaaabeaa kiabgkHiTiaadYeacaGGPaaacaGL7bGaayzFaaGaaCOAaiabg2da9i aad2gadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG5bWa aSbaaSqaaiaaigdaaeqaaaGcbaGaamizaiaadshadaahaaWcbeqaai aaikdaaaaaaOGaaCOAaaqaamaacmaabaGaeyOeI0IaamyBaiaadEga cqGHsislcaaIYaGaam4AaiaacIcacaWGSbWaaSbaaSqaaiaaikdaae qaaOGaeyOeI0IaamitaiaacMcaaiaawUhacaGL9baacaWHQbGaeyyp a0JaamyBamaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadM hadaWgaaWcbaGaaGOmaaqabaaakeaacaWGKbGaamiDamaaCaaaleqa baGaaGOmaaaaaaGccaWHQbaaaaa@6CC6@

This is two equations of motion MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  we can substitute for l 1 , l 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiBamaaBa aaleaacaaIXaaabeaakiaacYcacaWGSbWaaSbaaSqaaiaaikdaaeqa aaaa@3B69@  and rearrange them as

d y 1 2 d t 2 =( g2 k m ( y 1 d 0 sinωtL)+2 k m ( y 2 y 1 L) ) d y 2 2 d t 2 =( g2 k m ( y 2 y 1 L) ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWcaa qaaiaadsgacaWG5bWaa0baaSqaaiaaigdaaeaacaaIYaaaaaGcbaGa amizaiaadshadaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0ZaaeWaae aacqGHsislcaWGNbGaeyOeI0IaaGOmamaalaaabaGaam4Aaaqaaiaa d2gaaaGaaiikaiaadMhadaWgaaWcbaGaaGymaaqabaGccqGHsislca WGKbWaaSbaaSqaaiaaicdaaeqaaOGaci4CaiaacMgacaGGUbGaeqyY dCNaamiDaiabgkHiTiaadYeacaGGPaGaey4kaSIaaGOmamaalaaaba Gaam4Aaaqaaiaad2gaaaGaaiikaiaadMhadaWgaaWcbaGaaGOmaaqa baGccqGHsislcaWG5bWaaSbaaSqaaiaaigdaaeqaaOGaeyOeI0Iaam itaiaacMcaaiaawIcacaGLPaaaaeaadaWcaaqaaiaadsgacaWG5bWa a0baaSqaaiaaikdaaeaacaaIYaaaaaGcbaGaamizaiaadshadaahaa WcbeqaaiaaikdaaaaaaOGaeyypa0ZaaeWaaeaacqGHsislcaWGNbGa eyOeI0IaaGOmamaalaaabaGaam4Aaaqaaiaad2gaaaGaaiikaiaadM hadaWgaaWcbaGaaGOmaaqabaGccqGHsislcaWG5bWaaSbaaSqaaiaa igdaaeqaaOGaeyOeI0IaamitaiaacMcaaiaawIcacaGLPaaaaaaa@74BF@

 

6. Eliminate reactions MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  this is not needed in this example.

 

7. Identify initial conditions. We know that, at time t=0

y 1 =Lmg/k d y 1 dt =0 y 2 =2L3mg/2k d y 2 dt =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEamaaBa aaleaacaaIXaaabeaakiabg2da9iaadYeacqGHsislcaWGTbGaam4z aiaac+cacaWGRbGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVpaalaaabaGaamizaiaadMhadaWgaaWcbaGaaGymaaqabaaa keaacaWGKbGaamiDaaaacqGH9aqpcaaIWaGaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaamyEamaaBaaaleaacaaIYaaabeaakiabg2da9iaaikdaca WGmbGaeyOeI0IaaG4maiaad2gacaWGNbGaai4laiaaikdacaWGRbGa aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVpaalaaaba GaamizaiaadMhadaWgaaWcbaGaaGOmaaqabaaakeaacaWGKbGaamiD aaaacqGH9aqpcaaIWaaaaa@7ECE@

 

8. Solve the equations of motion. We need to (i) reduce the equations to the standard MATLAB form and (ii) write a MATLAB script to solve them.

 

Converting the equations.  We now need to do two things: (a) remove the second derivatives with respect to time, by introducing new variables; and (b) rearrange the equations into the form dy/dt=f(t,y) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaahM hacaGGVaGaamizaiaadshacqGH9aqpcaWHMbGaaiikaiaadshacaGG SaGaaCyEaiaacMcaaaa@4176@ .  We remove the derivatives by introducing v 1 = d l 1 dt v 2 = d l 2 dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaaIXaaabeaakiabg2da9iaaykW7daWcaaqaaiaadsgacaWG SbWaaSbaaSqaaiaaigdaaeqaaaGcbaGaamizaiaadshaaaGaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaadAhadaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcaaMc8UaaG PaVpaalaaabaGaamizaiaadYgadaWgaaWcbaGaaGOmaaqabaaakeaa caWGKbGaamiDaaaaaaa@5A6C@  as additional unknown variables, in the usual way.   Our equations of motion can then be expressed as

d y 1 dt = v 1 d y 2 dt = v 2 d v 1 d t 2 =( g2 k m ( y 1 d 0 sinωtL)+2 k m ( y 2 y 1 L) ) d v 2 d t 2 =( g2 k m ( y 2 y 1 L) ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWcaa qaaiaadsgacaWG5bWaaSbaaSqaaiaaigdaaeqaaaGcbaGaamizaiaa dshaaaGaeyypa0JaamODamaaBaaaleaacaaIXaaabeaaaOqaamaala aabaGaamizaiaadMhadaWgaaWcbaGaaGOmaaqabaaakeaacaWGKbGa amiDaaaacqGH9aqpcaWG2bWaaSbaaSqaaiaaikdaaeqaaaGcbaWaaS aaaeaacaWGKbGaamODamaaBaaaleaacaaIXaaabeaaaOqaaiaadsga caWG0bWaaWbaaSqabeaacaaIYaaaaaaakiabg2da9maabmaabaGaey OeI0Iaam4zaiabgkHiTiaaikdadaWcaaqaaiaadUgaaeaacaWGTbaa aiaacIcacaWG5bWaaSbaaSqaaiaaigdaaeqaaOGaeyOeI0Iaamizam aaBaaaleaacaaIWaaabeaakiGacohacaGGPbGaaiOBaiabeM8a3jaa dshacqGHsislcaWGmbGaaiykaiabgUcaRiaaikdadaWcaaqaaiaadU gaaeaacaWGTbaaaiaacIcacaWG5bWaaSbaaSqaaiaaikdaaeqaaOGa eyOeI0IaamyEamaaBaaaleaacaaIXaaabeaakiabgkHiTiaadYeaca GGPaaacaGLOaGaayzkaaaabaWaaSaaaeaacaWGKbGaamODamaaBaaa leaacaaIYaaabeaaaOqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYa aaaaaakiabg2da9maabmaabaGaeyOeI0Iaam4zaiabgkHiTiaaikda daWcaaqaaiaadUgaaeaacaWGTbaaaiaacIcacaWG5bWaaSbaaSqaai aaikdaaeqaaOGaeyOeI0IaamyEamaaBaaaleaacaaIXaaabeaakiab gkHiTiaadYeacaGGPaaacaGLOaGaayzkaaaaaaa@82BB@

 

We can now code MATLAB to solve these equations directly for dy/dt.  A script (which plots the position of each floor as a function of time) is shown below.

 

 

function building

%

k=100;

m=1;

omega=9;

d=0.1;

L=10;

time=20;

g = 9.81;

w0 = [L-m*g/k,2*L-3*m*g/(2*k),0,0];

[t_values,w_values] = ode45(@(t,w) eom(k,m,L,d,omega,g),[0,time],w0);

plot(t_values,w_values(:,1:2));

end

function dwdt = eom(t,w,k,m,L,d,omega,g)

      y1=w(1);

      y2=w(2);

     v1=w(3);

     v2=w(4);

 

      dwdt = [v1;v2;...
     -2*k*(y1-d*sin(omega*t)-L)/m+2*k*(y2-y1-L)/m;...

    -g-2*k*(y2-y1-L)/m];

end

 

 

The figures below plot the height of each floor as a function of time, for various earthquake frequencies.  For special earthquake frequencies (near the two resonant frequencies of the structure) the building vibrations are very severe.   As long as the structure is designed so that its resonant frequencies are well away from the frequency of a typical earthquake, it will be safe.

 

We will discuss vibrations in much more detail later in this course.

         

        

 

 

 

Final remark: we made this calculation a bit more complicated than necessary by solving for the heights y 1 , y 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadMhadaWgaa WcbaGaaGymaaqabaGccaGGSaGaamyEamaaBaaaleaacaaIYaaabeaa aaa@3ABF@  .    It is better to solve for the deflections of the floors instead of their heights.   Define

z 1 = y 1 (Lmg/k) z 2 = y 2 (2L3mg/2k) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadQhadaWgaa WcbaGaaGymaaqabaGccqGH9aqpcaWG5bWaaSbaaSqaaiaaigdaaeqa aOGaeyOeI0IaaiikaiaadYeacqGHsislcaWGTbGaam4zaiaac+caca WGRbGaaiykaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaadQhadaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcaWG5bWaaSba aSqaaiaaikdaaeqaaOGaeyOeI0IaaiikaiaaikdacaWGmbGaeyOeI0 IaaG4maiaad2gacaWGNbGaai4laiaaikdacaWGRbGaaiykaaaa@686A@

If we substitute for y 1 , y 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadMhadaWgaa WcbaGaaGymaaqabaGccaGGSaGaamyEamaaBaaaleaacaaIYaaabeaa aaa@3ABF@  in our equations we find that

d z 1 dt = v 1 d z 2 dt = v 2 d v 1 d t 2 =( 2 k m ( z 1 d 0 sinωt)+2 k m ( z 2 z 1 ) ) d v 2 d t 2 =( 2 k m ( z 2 z 1 ) ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWcaa qaaiaadsgacaWG6bWaaSbaaSqaaiaaigdaaeqaaaGcbaGaamizaiaa dshaaaGaeyypa0JaamODamaaBaaaleaacaaIXaaabeaaaOqaamaala aabaGaamizaiaadQhadaWgaaWcbaGaaGOmaaqabaaakeaacaWGKbGa amiDaaaacqGH9aqpcaWG2bWaaSbaaSqaaiaaikdaaeqaaaGcbaWaaS aaaeaacaWGKbGaamODamaaBaaaleaacaaIXaaabeaaaOqaaiaadsga caWG0bWaaWbaaSqabeaacaaIYaaaaaaakiabg2da9maabmaabaGaey OeI0IaaGOmamaalaaabaGaam4Aaaqaaiaad2gaaaGaaiikaiaadQha daWgaaWcbaGaaGymaaqabaGccqGHsislcaWGKbWaaSbaaSqaaiaaic daaeqaaOGaci4CaiaacMgacaGGUbGaeqyYdCNaamiDaiaacMcacqGH RaWkcaaIYaWaaSaaaeaacaWGRbaabaGaamyBaaaacaGGOaGaamOEam aaBaaaleaacaaIYaaabeaakiabgkHiTiaadQhadaWgaaWcbaGaaGym aaqabaGccaGGPaaacaGLOaGaayzkaaaabaWaaSaaaeaacaWGKbGaam ODamaaBaaaleaacaaIYaaabeaaaOqaaiaadsgacaWG0bWaaWbaaSqa beaacaaIYaaaaaaakiabg2da9maabmaabaGaeyOeI0IaaGOmamaala aabaGaam4Aaaqaaiaad2gaaaGaaiikaiaadQhadaWgaaWcbaGaaGOm aaqabaGccqGHsislcaWG6bWaaSbaaSqaaiaaigdaaeqaaOGaaiykaa GaayjkaiaawMcaaaaaaa@79D6@

These are a lot simpler, and more importantly, tell us that the motion of the system does not depend on the spring length L or gravity.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3.4 Summary of main equations and definitions

 

Position-velocity-acceleration relations in a Cartesian Frame

 

r(t)=x(t)i+y(t)j+z(t)k v(t)= v x (t)i+ v y (t)j+ v z (t)k = dx dt i+ dy dt j+ dz dt k a(t)= a x (t)i+ a y (t)j+ a z (t)k = d v x dt i+ d v y dt j+ d v z dt k= d 2 x d t 2 i+ d 2 y d t 2 j+ d 2 z d t 2 k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaaCOCai aacIcacaWG0bGaaiykaiabg2da9iaadIhacaGGOaGaamiDaiaacMca caWHPbGaey4kaSIaamyEaiaacIcacaWG0bGaaiykaiaahQgacqGHRa WkcaWG6bGaaiikaiaadshacaGGPaGaaC4AaaqaaiaahAhacaGGOaGa amiDaiaacMcacqGH9aqpcaWG2bWaaSbaaSqaaiaadIhaaeqaaOGaai ikaiaadshacaGGPaGaaCyAaiabgUcaRiaadAhadaWgaaWcbaGaamyE aaqabaGccaGGOaGaamiDaiaacMcacaWHQbGaey4kaSIaamODamaaBa aaleaacaWG6baabeaakiaacIcacaWG0bGaaiykaiaahUgaaeaacaaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlabg2da9maalaaabaGaamizaiaadIhaaeaacaWG KbGaamiDaaaacaWHPbGaey4kaSYaaSaaaeaacaWGKbGaamyEaaqaai aadsgacaWG0baaaiaahQgacqGHRaWkdaWcaaqaaiaadsgacaWG6baa baGaamizaiaadshaaaGaaC4AaaqaaiaahggacaGGOaGaamiDaiaacM cacqGH9aqpcaWGHbWaaSbaaSqaaiaadIhaaeqaaOGaaiikaiaadsha caGGPaGaaCyAaiabgUcaRiaadggadaWgaaWcbaGaamyEaaqabaGcca GGOaGaamiDaiaacMcacaWHQbGaey4kaSIaamyyamaaBaaaleaacaWG 6baabeaakiaacIcacaWG0bGaaiykaiaahUgaaeaacaaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua eyypa0ZaaSaaaeaacaWGKbGaamODamaaBaaaleaacaWG4baabeaaaO qaaiaadsgacaWG0baaaiaahMgacqGHRaWkdaWcaaqaaiaadsgacaWG 2bWaaSbaaSqaaiaadMhaaeqaaaGcbaGaamizaiaadshaaaGaaCOAai abgUcaRmaalaaabaGaamizaiaadAhadaWgaaWcbaGaamOEaaqabaaa keaacaWGKbGaamiDaaaacaWHRbGaeyypa0ZaaSaaaeaacaWGKbWaaW baaSqabeaacaaIYaaaaOGaamiEaaqaaiaadsgacaWG0bWaaWbaaSqa beaacaaIYaaaaaaakiaahMgacqGHRaWkdaWcaaqaaiaadsgadaahaa WcbeqaaiaaikdaaaGccaWG5baabaGaamizaiaadshadaahaaWcbeqa aiaaikdaaaaaaOGaaCOAaiabgUcaRmaalaaabaGaamizamaaCaaale qabaGaaGOmaaaakiaadQhaaeaacaWGKbGaamiDamaaCaaaleqabaGa aGOmaaaaaaGccaWHRbaaaaa@D2F3@

 

The direction of the velocity vector is tangent to its path.

The magnitude of the velocity vector v x 2 + v y 2 + v z 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaakaaabaGaam ODamaaDaaaleaacaWG4baabaGaaGOmaaaakiabgUcaRiaadAhadaqh aaWcbaGaamyEaaqaaiaaikdaaaGccqGHRaWkcaWG2bWaa0baaSqaai aadQhaaeaacaaIYaaaaaqabaaaaa@40C8@  is the distance traveled along the path per  unit time (speed).

 

A unit vector tangent to the path can be found as

t= v x (t)i+ v y (t)j+ v z (t)k v x 2 + v y 2 + v z 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahshacqGH9a qpdaWcaaqaaiaadAhadaWgaaWcbaGaamiEaaqabaGccaGGOaGaamiD aiaacMcacaWHPbGaey4kaSIaamODamaaBaaaleaacaWG5baabeaaki aacIcacaWG0bGaaiykaiaahQgacqGHRaWkcaWG2bWaaSbaaSqaaiaa dQhaaeqaaOGaaiikaiaadshacaGGPaGaaC4AaaqaamaakaaabaGaam ODamaaDaaaleaacaWG4baabaGaaGOmaaaakiabgUcaRiaadAhadaqh aaWcbaGaamyEaaqaaiaaikdaaaGccqGHRaWkcaWG2bWaa0baaSqaai aadQhaaeaacaaIYaaaaaqabaaaaaaa@54FA@

 

Straight line motion with constant acceleration

 

r=[ X 0 + V 0 t+ 1 2 a t 2 ]iv=( V 0 +at )ia=ai MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWHYbGaeyypa0ZaamWaaeaacaWGyb WaaSbaaSqaaiaaicdaaeqaaOGaey4kaSIaamOvamaaBaaaleaacaaI WaaabeaakiaadshacqGHRaWkdaWcaaqaaiaaigdaaeaacaaIYaaaai aadggacaWG0bWaaWbaaSqabeaacaaIYaaaaOGaaGPaVdGaay5waiaa w2faaiaahMgacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaahAhacqGH9aqpdaqadaqaaiaadAfadaWgaaWcbaGa aGimaaqabaGccqGHRaWkcaWGHbGaamiDaaGaayjkaiaawMcaaiaahM gacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaCyyaiabg2da9iaadggacaWHPbaaaa@6C7D@

Here, a is the (constant) acceleration; X 0 , V 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfadaWgaa WcbaGaaGimaaqabaGccaGGSaGaamOvamaaBaaaleaacaaIWaaabeaa aaa@3A78@  are the position and speed at time t=0.

 

 

Straight line motion with time/position dependent acceleration

 

Acceleration given as a function of time: r=( X 0 + 0 t v(t)dt )iv=( V 0 + 0 t a(t)dt )i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWHYbGaeyypa0ZaaeWaaeaacaWGyb WaaSbaaSqaaiaaicdaaeqaaOGaey4kaSYaa8qCaeaacaWG2bGaaiik aiaadshacaGGPaGaamizaiaadshaaSqaaiaaicdaaeaacaWG0baani abgUIiYdaakiaawIcacaGLPaaacaWHPbGaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWH2bGaeyypa0ZaaeWaae aacaWGwbWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSYaa8qCaeaacaWG HbGaaiikaiaadshacaGGPaGaamizaiaadshaaSqaaiaaicdaaeaaca WG0baaniabgUIiYdaakiaawIcacaGLPaaacaWHPbGaaGPaVlaaykW7 caaMc8oaaa@64AB@

Acceleration given as a function of position V 0 v(t) vdv = 0 x(t) a(x)dx MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7daWd XbqaaiaadAhacaWGKbGaamODaaWcbaGaamOvamaaBaaameaacaaIWa aabeaaaSqaaiaadAhacaGGOaGaamiDaiaacMcaa0Gaey4kIipakiab g2da9maapehabaGaamyyaiaacIcacaWG4bGaaiykaiaadsgacaWG4b aaleaacaaIWaaabaGaamiEaiaacIcacaWG0bGaaiykaaqdcqGHRiI8 aaaa@5C4F@

 

Separation of variables for one-dimensional motion

a= dv dt = g(t) f(v) V 0 v f(v) dv= 0 t g(t)dt v= dx dt = g(t) f(v) X 0 x(t) f(x) dv= 0 t v(t)dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamyyai abg2da9maalaaabaGaamizaiaadAhaaeaacaWGKbGaamiDaaaacqGH 9aqpdaWcaaqaaiaadEgacaGGOaGaamiDaiaacMcaaeaacaWGMbGaai ikaiaadAhacaGGPaaaaiabgkDiEpaapehabaGaamOzaiaacIcacaWG 2bGaaiykaaWcbaGaamOvamaaBaaameaacaaIWaaabeaaaSqaaiaadA haa0Gaey4kIipakiaadsgacaWG2bGaeyypa0Zaa8qCaeaacaWGNbGa aiikaiaadshacaGGPaGaamizaiaadshaaSqaaiaaicdaaeaacaWG0b aaniabgUIiYdaakeaacaWG2bGaeyypa0ZaaSaaaeaacaWGKbGaamiE aaqaaiaadsgacaWG0baaaiabg2da9maalaaabaGaam4zaiaacIcaca WG0bGaaiykaaqaaiaadAgacaGGOaGaamODaiaacMcaaaGaeyO0H49a a8qCaeaacaWGMbGaaiikaiaadIhacaGGPaaaleaacaWGybWaaSbaaW qaaiaaicdaaeqaaaWcbaGaamiEaiaacIcacaWG0bGaaiykaaqdcqGH RiI8aOGaamizaiaadAhacqGH9aqpdaWdXbqaaiaadAhacaGGOaGaam iDaiaacMcacaWGKbGaamiDaaWcbaGaaGimaaqaaiaadshaa0Gaey4k Iipaaaaa@816D@

 

Simple Harmonic Motion shm              

 

r=[ X 0 +ΔXsin(2πt/T) ]i v=Vcos(2πt/T)i a=Asin(2πt/T)i V= 2πΔX T A= 2πV T = 4 π 2 ΔX T 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiaahkhacqGH9aqpdaWadaqaai aadIfadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcqqHuoarcaWGybGa ci4CaiaacMgacaGGUbGaaiikaiaaikdacqaHapaCcaWG0bGaai4lai aadsfacaGGPaGaaGPaVdGaay5waiaaw2faaiaahMgacaaMc8oabaGa aCODaiabg2da9iaadAfaciGGJbGaai4BaiaacohacaGGOaGaaGOmai abec8aWjaadshacaGGVaGaamivaiaacMcacaWHPbaabaGaaCyyaiab g2da9iabgkHiTiaadgeaciGGZbGaaiyAaiaac6gacaGGOaGaaGOmai abec8aWjaadshacaGGVaGaamivaiaacMcacaWHPbaabaGaamOvaiab g2da9maalaaabaGaaGOmaiabec8aWjabfs5aejaadIfaaeaacaWGub aaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadgeacqGH9aqpdaWc aaqaaiaaikdacqaHapaCcaWGwbaabaGaamivaaaacqGH9aqpdaWcaa qaaiaaisdacqaHapaCdaahaaWcbeqaaiaaikdaaaGccqqHuoarcaWG ybaabaGaamivamaaCaaaleqabaGaaGOmaaaaaaaaaaa@9A29@

 

 

Circular Motion at Constant Speed

r=R( cosθi+sinθj ) v=ωR( sinθi+cosθj )=Vt a= ω 2 R(cosθi+sinθj)= ω 2 Rn= V 2 R n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiaahkhacqGH9aqpcaWGsbWaae WaaeaaciGGJbGaai4BaiaacohacqaH4oqCcaWHPbGaey4kaSIaci4C aiaacMgacaGGUbGaeqiUdeNaaCOAaaGaayjkaiaawMcaaaqaaiaahA hacqGH9aqpcqaHjpWDcaWGsbWaaeWaaeaacqGHsislciGGZbGaaiyA aiaac6gacqaH4oqCcaWHPbGaey4kaSIaci4yaiaac+gacaGGZbGaeq iUdeNaaCOAaaGaayjkaiaawMcaaiaaykW7cqGH9aqpcaWGwbGaaCiD aaqaaiaahggacqGH9aqpcqGHsislcqaHjpWDdaahaaWcbeqaaiaaik daaaGccaWGsbGaaiikaiGacogacaGGVbGaai4CaiabeI7aXjaahMga cqGHRaWkciGGZbGaaiyAaiaac6gacqaH4oqCcaWHQbGaaiykaiaayk W7cqGH9aqpcaaMc8UaaGPaVlabeM8a3naaCaaaleqabaGaaGOmaaaa kiaadkfacaWHUbGaeyypa0ZaaSaaaeaacaWGwbWaaWbaaSqabeaaca aIYaaaaaGcbaGaamOuaaaacaWHUbaaaaa@7D04@

 

 

General Circular Motion

 

ω=dθ/dtα=dω/dt= d 2 θ/d t 2 s=RθV=ds/dt=Rω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiabeM8a3jabg2da9iaadsgacq aH4oqCcaGGVaGaamizaiaadshacaaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlabeg7aHjabg2da9iaadsgacqaHjpWDcaGGVaGaamizaiaads hacqGH9aqpcaWGKbWaaWbaaSqabeaacaaIYaaaaOGaeqiUdeNaai4l aiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaGcbaGaam4Caiabg2 da9iaadkfacqaH4oqCcaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caWGwbGaeyypa0JaamizaiaadohacaGGVaGaamizaiaadshacq GH9aqpcaWGsbGaeqyYdChaaaa@6A68@

 

r=R( cosθi+sinθj ) v=ωR( sinθi+cosθj )=Vt a=Rα(sinθi+cosθj)R ω 2 (cosθi+sinθj) =αRt+ ω 2 Rn= dV dt t+ V 2 R n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiaahkhacqGH9aqpcaWGsbWaae WaaeaaciGGJbGaai4BaiaacohacqaH4oqCcaWHPbGaey4kaSIaci4C aiaacMgacaGGUbGaeqiUdeNaaCOAaaGaayjkaiaawMcaaaqaaiaahA hacqGH9aqpcqaHjpWDcaWGsbWaaeWaaeaacqGHsislciGGZbGaaiyA aiaac6gacqaH4oqCcaWHPbGaey4kaSIaci4yaiaac+gacaGGZbGaeq iUdeNaaCOAaaGaayjkaiaawMcaaiaaykW7cqGH9aqpcaWGwbGaaCiD aaqaaiaahggacqGH9aqpcaWGsbGaeqySdeMaaiikaiabgkHiTiGaco hacaGGPbGaaiOBaiabeI7aXjaahMgacqGHRaWkciGGJbGaai4Baiaa cohacqaH4oqCcaWHQbGaaiykaiabgkHiTiaadkfacqaHjpWDdaahaa WcbeqaaiaaikdaaaGccaGGOaGaci4yaiaac+gacaGGZbGaeqiUdeNa aCyAaiabgUcaRiGacohacaGGPbGaaiOBaiabeI7aXjaahQgacaGGPa aabaGaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9iaaykW7caaMc8Ua eqySdeMaamOuaiaahshacqGHRaWkcqaHjpWDdaahaaWcbeqaaiaaik daaaGccaWGsbGaaCOBaiabg2da9maalaaabaGaamizaiaadAfaaeaa caWGKbGaamiDaaaacaWH0bGaey4kaSYaaSaaaeaacaWGwbWaaWbaaS qabeaacaaIYaaaaaGcbaGaamOuaaaacaWHUbaaaaa@9C2A@

 

Note that the straight-line motion relations can be used to relate θ,ω,α MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeI7aXjaacY cacqaHjpWDcaGGSaGaeqySdegaaa@3CBC@ , by exchanging xθ,vω,aα MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacqGHsg IRcqaH4oqCcaGGSaGaamODaiabgkziUkabeM8a3jaacYcacaWGHbGa eyOKH4QaeqySdegaaa@4561@

 

 

 

 

 

 

 

 

 

 

 

 

 

Motion along an arbitrary path in normal-tangential coordinates

 

r=x(s)i+y(s)j t= dr ds n=R dt ds R= 1 ( d 2 x d s 2 ) 2 + ( d 2 y d s 2 ) 2 v= ds dt t=Vt a= d 2 s d t 2 t+ 1 R ( ds dt ) 2 n= dV dt t+ V 2 R n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiaahkhacqGH9aqpcaWG4bGaai ikaiaadohacaGGPaGaaCyAaiabgUcaRiaadMhacaGGOaGaam4Caiaa cMcacaWHQbaabaGaaCiDaiabg2da9maalaaabaGaamizaiaahkhaae aacaWGKbGaam4CaaaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaah6gacqGH9aqpca WGsbWaaSaaaeaacaWGKbGaaCiDaaqaaiaadsgacaWGZbaaaiaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadkfacqGH9aqp daWcaaqaaiaaigdaaeaadaGcaaqaamaabmaabaWaaSaaaeaacaWGKb WaaWbaaSqabeaacaaIYaaaaOGaamiEaaqaaiaadsgacaWGZbWaaWba aSqabeaacaaIYaaaaaaaaOGaayjkaiaawMcaamaaCaaaleqabaGaaG OmaaaakiabgUcaRmaabmaabaWaaSaaaeaacaWGKbWaaWbaaSqabeaa caaIYaaaaOGaamyEaaqaaiaadsgacaWGZbWaaWbaaSqabeaacaaIYa aaaaaaaOGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaeqaaaaa aOqaaiaahAhacqGH9aqpdaWcaaqaaiaadsgacaWGZbaabaGaamizai aadshaaaGaaCiDaiabg2da9iaadAfacaWH0baabaGaaCyyaiabg2da 9maalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadohaaeaaca WGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaaGccaWH0bGaey4kaSYa aSaaaeaacaaIXaaabaGaamOuaaaadaqadaqaamaalaaabaGaamizai aadohaaeaacaWGKbGaamiDaaaaaiaawIcacaGLPaaadaahaaWcbeqa aiaaikdaaaGccaWHUbGaeyypa0ZaaSaaaeaacaWGKbGaamOvaaqaai aadsgacaWG0baaaiaahshacqGHRaWkdaWcaaqaaiaadAfadaahaaWc beqaaiaaikdaaaaakeaacaWGsbaaaiaah6gaaaaa@AD4B@

 

 

 

 

Position-velocity-acceleration relations in polar-coordinates

 

e r =cosθi+sinθj e θ =sinθi+cosθj i=cosθ e r sinθ e θ j=sinθ e r +cosθ e θ x=rcosθy=rsinθ r= x 2 + y 2 θ= tan 1 y/x MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaaCyzam aaBaaaleaacaWGYbaabeaakiabg2da9iGacogacaGGVbGaai4Caiab eI7aXjaahMgacqGHRaWkciGGZbGaaiyAaiaac6gacqaH4oqCcaWHQb GaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaCyzamaaBaaaleaacqaH4oqCaeqaaOGaeyypa0JaeyOeI0 Iaci4CaiaacMgacaGGUbGaeqiUdeNaaCyAaiabgUcaRiGacogacaGG VbGaai4CaiabeI7aXjaahQgaaeaacaWHPbGaeyypa0Jaci4yaiaac+ gacaGGZbGaeqiUdeNaaCyzamaaBaaaleaacaWGYbaabeaakiabgkHi TiGacohacaGGPbGaaiOBaiabeI7aXjaahwgadaWgaaWcbaGaeqiUde habeaakiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caWHQbGaeyypa0Jaci4CaiaacMgacaGGUbGaeqiUdeNaaCyzamaaBa aaleaacaWGYbaabeaakiabgUcaRiGacogacaGGVbGaai4CaiabeI7a XjaahwgadaWgaaWcbaGaeqiUdehabeaaaOqaaiaadIhacqGH9aqpca WGYbGaci4yaiaac+gacaGGZbGaeqiUdeNaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadMhacq GH9aqpcaWGYbGaci4CaiaacMgacaGGUbGaeqiUdehabaGaamOCaiab g2da9maakaaabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRi aadMhadaahaaWcbeqaaiaaikdaaaaabeaakiaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaeqiUdeNaeyypa0JaciiDaiaacggacaGG UbWaaWbaaSqabeaacqGHsislcaaIXaaaaOGaamyEaiaac+cacaWG4b aaaaa@D573@

 

 

r=r e r v= dr dt e r +r dθ dt e θ a=( d 2 r d t 2 r ( dθ dt ) 2 ) e r +( r d 2 θ d t 2 +2 dr dt dθ dt ) e θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaaCOCai abg2da9iaadkhacaWHLbWaaSbaaSqaaiaadkhaaeqaaaGcbaGaaCOD aiabg2da9maalaaabaGaamizaiaadkhaaeaacaWGKbGaamiDaaaaca WHLbWaaSbaaSqaaiaadkhaaeqaaOGaey4kaSIaamOCamaalaaabaGa amizaiabeI7aXbqaaiaadsgacaWG0baaaiaahwgadaWgaaWcbaGaeq iUdehabeaaaOqaaiaahggacqGH9aqpdaqadaqaamaalaaabaGaamiz amaaCaaaleqabaGaaGOmaaaakiaadkhaaeaacaWGKbGaamiDamaaCa aaleqabaGaaGOmaaaaaaGccqGHsislcaWGYbWaaeWaaeaadaWcaaqa aiaadsgacqaH4oqCaeaacaWGKbGaamiDaaaaaiaawIcacaGLPaaada ahaaWcbeqaaiaaikdaaaaakiaawIcacaGLPaaacaWHLbWaaSbaaSqa aiaadkhaaeqaaOGaey4kaSYaaeWaaeaacaWGYbWaaSaaaeaacaWGKb WaaWbaaSqabeaacaaIYaaaaOGaeqiUdehabaGaamizaiaadshadaah aaWcbeqaaiaaikdaaaaaaOGaey4kaSIaaGOmamaalaaabaGaamizai aadkhaaeaacaWGKbGaamiDaaaadaWcaaqaaiaadsgacqaH4oqCaeaa caWGKbGaamiDaaaaaiaawIcacaGLPaaacaWHLbWaaSbaaSqaaiabeI 7aXbqabaaaaaa@7772@

 

 

 

Newton’s laws

For a particle F=ma MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcaWGTbGaaCyyaaaa@39EB@

 

For a rigid body moving without rotation or rotating at fixed angular rate about a fixed axis M C =0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaah2eadaWgaa WcbaGaam4qaaqabaGccqGH9aqpcaaIWaaaaa@39CE@  (you must take moments about the center of mass)

 

 

 

 

 

 

Drawing free body diagrams:

 

1.      Decide which part of a system you will idealize as a particle (you may need more than one particle)

2.      Draw the part of the system you have idealized as a particle by itself (very important!).  It is important to make sure that your particle is isolated MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  it can’t be touching something else. You may need more than one drawing if you have more than one particle in your system.

3.      Draw on any of the following external forces that apply. Make sure you draw them acting in the correct direction, acting on the correct part of the body:

a.       gravity (at the COM);

b.      air resistance or lift forces (and sometimes moments) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  various conventions are used to locate these forces but in this course we usually put them at the COM and neglect moments;

c.       Buoyancy forces (act at the COM of the displaced fluid)

d.      Electrostatic or electromagnetic forces

4.      Draw the forces exerted by springs attached to the particle.   It is best to assume that springs always pull on the point they are connected to, and that the magnitude of the force in the spring is F s =k(l l 0 ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAeadaWgaa WcbaGaam4CaaqabaGccqGH9aqpcaWGRbGaaiikaiaadYgacqGHsisl caWGSbWaaSbaaSqaaiaaicdaaeqaaOGaaiykaaaa@3F41@ , where l is the length of the spring, and l 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadYgadaWgaa WcbaGaaGimaaqabaaaaa@3811@  is its unstretched length.

5.      Draw the forces exerted by dashpots or dampers (like springs, assume they pull on the object they are connected to, and exert a force magnitude F d =λdl/dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAeadaWgaa WcbaGaamizaaqabaGccqGH9aqpcqaH7oaBcaWGKbGaamiBaiaac+ca caWGKbGaamiDaaaa@3F4D@  where l is the length of the dashpot.

6.      Draw forces exerted by cables.  Cables always pull, and exert a force parallel to the direction of the cable.  The magnitude of the force has to be left as an unknown.

7.      Draw any unknown reaction forces, with the following rules:

a.       Reaction forces must act at any point on any point of the body that is touching something outside the particle (i.e. a part of your system that you did not include in your drawing in step 2).    

b.      If the connection between the two touching objects prevents them from rotating with respect to one another (or, like a motor, makes them rotate with some controllable angular speed), you will need to draw both reaction forces and moments.  (Reaction moments do sometimes come up in dynamics problems, but they are not very common, so think carefully before including them).

c.        If friction acts at the contact point, and you don’t know whether the two objects slide at the contact (or you know they do not slide), draw both a normal and a tangential force with unknown magnitudes N,T (or some suitable variable).  The direction of the friction force is not important.  DO NOT assume T=μN MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsfacqGH9a qpcqaH8oqBcaWGobaaaa@3AA2@ .

d.      If friction acts at the contact point, and you know the contact slips,  draw both a tangential and a normal force.  You must draw the tangential force so that it opposes the direction of sliding (ask a faculty member or TA if you don’t understand this).  If slip occurs you can assume T=μN MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsfacqGH9a qpcqaH8oqBcaWGobaaaa@3AA2@ .

e.       If the contact point is frictionless, draw only a normal force.

f.       If your particle is being touched by a two-force member (no, this is not a gender and sexuality class… a two force member is a massless rod, connected through freely rotating hinges at both ends.  A massless freely rotating wheel can also be idealized as a two-force member) you can assume the reaction force acts parallel to the two-force member.

g.       If you have more than one particle in your system, make sure that any forces exerted by one particle on the other have equal and opposite reactions.

 

 

Calculating unknown forces or accelerations using Newton’s laws:

 

1.      Decide how to idealize the system (what are the particles?)

2.      Draw a free body diagram showing the forces acting on each particle

3.      Consider the kinematics of the problem. The goal is to calculate the acceleration of each particle in the system MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  you may be able to start by writing down the position vector and differentiating it, or you may be able to relate the accelerations of two particles (eg if two particles move together, their accelerations must be equal).

4.      Write down F=ma for each particle.

5.      If you are solving a problem involving a massless frames (see, e.g. Example 3, involving a bicycle with negligible mass) you also need to write down M C =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytamaaBa aaleaacaWGdbaabeaakiabg2da9iaahcdaaaa@3989@  about the particle.

6.      Solve the resulting equations for any unknown components of force or acceleration (this is just like a statics problem, except the right hand side is not zero).

 

Problems like this will usually ask you to make some design prediction at the end, which might involve calculating critical conditions for something to slip, tip, break, etc.

·         At the onset of slip at a contact | T |=μ| N | MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaemaabaGaam ivaaGaay5bSlaawIa7aiabg2da9iabeY7aTnaaemaabaGaamOtaaGa ay5bSlaawIa7aaaa@40E6@

·         At the critical point where an object tips over, a reaction force somewhere will go to zero.  You will have to identify where this point is, find the reaction force, and set it to zero.

 

 

Deriving equations of motion for a system of particles

 

1.      Introduce a set of variables that can describe the motion of the system.  Don’t worry if this sounds vague MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  it will be clear what this means when we solve specific examples.

2.      Write down the position vector of each particle in the system in terms of these variables

3.      Differentiate the position vector(s), to calculate the velocity and acceleration of each particle in terms of your variables;

4.      Draw a free body diagram showing the forces acting on each particle.  You may need to introduce variables to describe reaction forces.  Write down the resultant force vector.

5.      Write down Newton’s law F=ma MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbaaaa@39A7@  for each particle.  This will generate up to 3 equations of motion (one for each vector component) for each particle.

6.      If you wish, you can eliminate any unknown reaction forces from Newton’s laws. If you are trying to solve the equations by hand, you should always do this; of you are using MATLAB, it’s not usually necessary MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzGfaeaa aaaaaaa8qacaWFtacaaa@37E6@  you can have MATLAB calculate the reactions for you. The result will be a set of differential equations for the variables defined in step (1)

7.      If you find you have fewer equations than unknown variables, you should look for any constraints that restrict the motion of the particles.  The constraints must be expressed in terms of the unknown accelerations.

8.      Identify the initial conditions for the variables defined in (1).  These are usually the values of the unknown variables, their time derivatives, at time t=0. If you happen to know the values of the variables at some other instant in time, you can use that too.   If you don’t know their values at all, you should just introduce new (unknown) variables to denote the initial conditions.

9.      Solve the differential equations, subject to the initial conditions.

 

 

 

 

 

 

Trajectory equations for particle moving near earth’s surface with no air resistance

 

r= X 0 i+ Y 0 j+ Z 0 k dr dt = V x0 i+ V y0 j+ V z0 k }t=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaGacaabaeqabaGaaCOCaiabg2da9i aadIfadaWgaaWcbaGaaGimaaqabaGccaWHPbGaey4kaSIaamywamaa BaaaleaacaaIWaaabeaakiaahQgacqGHRaWkcaWGAbWaaSbaaSqaai aaicdaaeqaaOGaaC4AaaqaamaalaaabaGaamizaiaahkhaaeaacaWG KbGaamiDaaaacqGH9aqpcaWGwbWaaSbaaSqaaiaadIhacaaIWaaabe aakiaahMgacqGHRaWkcaWGwbWaaSbaaSqaaiaadMhacaaIWaaabeaa kiaahQgacqGHRaWkcaWGwbWaaSbaaSqaaiaadQhacaaIWaaabeaaki aahUgaaaGaayzFaaGaamiDaiabg2da9iaaicdaaaa@547E@

 

r=( X 0 + V x0 t )i+( Y 0 + V y0 t )j+( Z 0 + V z0 t 1 2 g t 2 )k v=( V x0 )i+( V y0 )j+( V z0 gt )k a=gk MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiaahkhacqGH9aqpdaqadaqaai aadIfadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcaWGwbWaaSbaaSqa aiaadIhacaaIWaaabeaakiaadshaaiaawIcacaGLPaaacaWHPbGaey 4kaSYaaeWaaeaacaWGzbWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSIa amOvamaaBaaaleaacaWG5bGaaGimaaqabaGccaWG0baacaGLOaGaay zkaaGaaCOAaiabgUcaRmaabmaabaGaamOwamaaBaaaleaacaaIWaaa beaakiabgUcaRiaadAfadaWgaaWcbaGaamOEaiaaicdaaeqaaOGaam iDaiabgkHiTmaalaaabaGaaGymaaqaaiaaikdaaaGaam4zaiaadsha daahaaWcbeqaaiaaikdaaaaakiaawIcacaGLPaaacaWHRbaabaGaaC ODaiabg2da9maabmaabaGaamOvamaaBaaaleaacaWG4bGaaGimaaqa baaakiaawIcacaGLPaaacaWHPbGaey4kaSYaaeWaaeaacaWGwbWaaS baaSqaaiaadMhacaaIWaaabeaaaOGaayjkaiaawMcaaiaahQgacqGH RaWkdaqadaqaaiaadAfadaWgaaWcbaGaamOEaiaaicdaaeqaaOGaey OeI0Iaam4zaiaadshaaiaawIcacaGLPaaacaWHRbaabaGaaCyyaiab g2da9iabgkHiTiaadEgacaWHRbaaaaa@71CF@

 

 

Solving differential equations with Matlab:

 

Example: to solve

d 2 y d t 2 +2ζ ω n dy dt + ω n 2 y=0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam izamaaCaaaleqabaGaaGOmaaaakiaadMhaaeaacaWGKbGaamiDamaa CaaaleqabaGaaGOmaaaaaaGccqGHRaWkcaaIYaGaeqOTdONaeqyYdC 3aaSbaaSqaaiaad6gaaeqaaOWaaSaaaeaacaWGKbGaamyEaaqaaiaa dsgacaWG0baaaiabgUcaRiabeM8a3naaDaaaleaacaWGUbaabaGaaG OmaaaakiaadMhacqGH9aqpcaaIWaaaaa@4D76@

with initial conditions y= y 0 dy dt = v 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadMhacqGH9a qpcaWG5bWaaSbaaSqaaiaaicdaaeqaaOGaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7daWcaaqaaiaadsgacaWG5baabaGaamizaiaadshaaa Gaeyypa0JaamODamaaBaaaleaacaaIWaaabeaaaaa@48A3@  at time t=0

clear all
syms x(t) t omega_n zeta x0 v0
diffeq = (1/omega_n^2)*diff(x(t),t,2) + 2*zeta/omega_n * diff(x(t),t) + x(t)==0
Dx = diff(x)
initial_condition = [x(0)==x0, Dx(0)==v0]
x(t) = simplify(dsolve(diffeq,initial_condition))

 

 

Re-writing a second-order differential equation as a pair of first-order equations for MATLAB

 

Example: to solve

d 2 y d t 2 +2ζ ω n dy dt + ω n 2 y=0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam izamaaCaaaleqabaGaaGOmaaaakiaadMhaaeaacaWGKbGaamiDamaa CaaaleqabaGaaGOmaaaaaaGccqGHRaWkcaaIYaGaeqOTdONaeqyYdC 3aaSbaaSqaaiaad6gaaeqaaOWaaSaaaeaacaWGKbGaamyEaaqaaiaa dsgacaWG0baaaiabgUcaRiabeM8a3naaDaaaleaacaWGUbaabaGaaG OmaaaakiaadMhacqGH9aqpcaaIWaaaaa@4D76@

we introduce v=dy/dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhacqGH9a qpcaWGKbGaamyEaiaac+cacaWGKbGaamiDaaaa@3CB7@  as an additional variable.   This new equation, together with the original ODE can then be written in the following form

d dt [ y v ]=[ v 2ζ ω n v ω n 2 y ] MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam izaaqaaiaadsgacaWG0baaamaadmaabaqbaeqabiqaaaqaaiaadMha aeaacaWG2baaaaGaay5waiaaw2faaiabg2da9maadmaabaqbaeqabi qaaaqaaiaadAhaaeaacqGHsislcaaIYaGaeqOTdONaeqyYdC3aaSba aSqaaiaad6gaaeqaaOGaamODaiabgkHiTiabeM8a3naaDaaaleaaca WGUbaabaGaaGOmaaaakiaadMhaaaaacaGLBbGaayzxaaaaaa@4E02@

This is now in the form

dw dt =f(t,w)w=[ y v ] MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam izaiaahEhaaeaacaWGKbGaamiDaaaacqGH9aqpcaWGMbGaaiikaiaa dshacaGGSaGaaC4DaiaacMcacaaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caWH3bGaeyypa0ZaamWaaeaafaqabeGabaaaba GaamyEaaqaaiaadAhaaaaacaGLBbGaayzxaaaaaa@5D2A@

as required.