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  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  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 most of the problems we solve in this course, we will specify the position of a particle using the Cartesian components of its position vector with respect to a convenient origin.  This means

1.      We choose three, mutually perpendicular, fixed directions in space.  The three directions are described by unit vectors  

2.      We choose a convenient point to use as origin.

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

 

 

In dynamics problems, all three components can be functions of time.

 

 

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

 

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

 

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  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  are constant (independent of time) and so

 

This is really three equations  one for each velocity component, i.e.

 

 

 

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.

 

The acceleration is a vector, with Cartesian representation .

 

Like velocity, acceleration has magnitude and direction. Sometimes it may be possible to visualize an acceleration vector  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

 

 

 

 

3.1.3 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  the particle has speed  

At time  the particle has position vector  

The position, velocity acceleration vectors are then

 

 

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  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  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

 

Here  is the average length of the spring,  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  

 

The motion is plotted in the figure on the right.

 

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

 

 

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 , ,  denote the amplitudes of the displacement, velocity and acceleration, we have that

 

 

 

Example 3: Motion at constant speed around a circular path  Circular motion is also very common  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  increases at constant rate.  We can write , where  is the (constant) angular speed of the shaft, in radians/seconds

·       The speed of the particle is related to  by .   To see this, notice that the circumferential distance traveled by the particle is .  Therefore, .

 

For this example the position vector is

 

The velocity can be calculated by differentiating the position vector. 

 

Here, we have used the chain rule of differentiation, and noted that .

 

The acceleration vector follows as

 

Note that

 (i) The magnitude of the velocity is , and its direction is (obviously!) tangent to the path (to see this, visualize (using trig) the direction of the unit vector  

(ii) The magnitude of the acceleration is  and its direction is towards the center of the circle. To see this, visualize (using trig) the direction of the unit vector  

 

We can write these mathematically as

 

 

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  is now a general function of time. 

 

We can write down some useful scalar relations:

·       Angular rate:  

·       Angular acceleration  

·       Speed  

·       Rate of change of speed  

 

We can now calculate vector velocities and accelerations

 

The velocity can be calculated by differentiating the position vector. 

 

The acceleration vector follows as

 

 

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 , .  Then

 

 

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

 

 

Using MAPLE to differentiate position-velocity-acceleration relations

 

If you find that your calculus is a bit rusty you can use MAPLE to do the tedious work for you.  You already know how to differentiate and integrate in MAPLE  the only thing you may not know is how to tell MAPLE that a variable is a function of time.   Here’s how this works.  To differentiate the vector

 

you would type

 

It is essential to type in the (t) after x,y,and z  if you don’t do this, Mupad 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  is some arbitrary function of time, but don’t know what the function is

 

As you’ve already seen in EN3, 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  so keep taking those math classes…

 

3.1.4 Velocity and acceleration in normal-tangential and cylindrical polar 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  go ahead and use it.  If not, just use Cartesian coordinates  this will always work, and with MAPLE is not very hard.  The only benefit of using the special coordinate systems is to save a couple of lines of rather tedious trigonometric algebra  which can be extremely helpful when solving an exam question, but is generally insignificant when solving a real problem.

 

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

                         

In some problems, you might know the particle speed, and the x,y coordinates of the path (a car traveling along a road is a good example).   In this case it is often easiest to use normal-tangential coordinates to describe forces and motion.  For this purpose we

·       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

·       Introduce the radius of curvature of the path R.

 

If you happen to know the parametric equation of the path (i.e. the x,y coordinates are known in terms of some variable  ), then

 

The sign of n should be selected so that

 

The radius of curvature can be computed from

 

The radius of curvature is always positive.

 

The direction of the velocity vector of a particle is tangent to its path.  The magnitude of the velocity vector is equal to the speed.

 

The acceleration vector can be constructed by adding two components:

·         the component of acceleration tangent to the particle’s path is equal  

·         The component of acceleration perpendicular to the path (towards the center of curvature) is equal to .

Mathematically  

 

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  as shown in the figure  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  your mission is to calculate the 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.

 

Calcluation:

 

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

 

The position vector is , so we can calculate the radius of curvature from the formula

 

Note that x acts as the parameter  for this problem, and , so

 

and the acceleration is

 

We are interested in the magnitude of the acceleration… 

 

We see from this that the car has the biggest acceleration when . The maximum acceleration follows as

 

The formula for the speed limit is therefore  

 

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

 

 

Polar coordinates for particles moving in a plane

 

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 coordinates are related to x,y coordinates through

 

 

Suppose that the position of a particle is specified by its ‘polar coordinates’  relative to a fixed origin, as shown in the figure. Let  be a unit vector pointing in the radial direction, and let  be a unit vector pointing in the tangential direction, i.e

 

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

 

Deriving these results takes some tedious algebra, but it’s conceptually simple  here’s what we do:

1.      Write down the position vector in terms of  in a fixed (i,j,k) coordinate system

2.      Take the time derivatives to find acceleration and velocity in the (i,j,k) coordinate system

3.      Convert the results to the  coordinate system.  To do this remember that the component of v parallel to  can be found using a dot product: .  Similarly  

 

Here are the details with Mupad taking care of the tedious algebra.

 

 

These are the answers stated.

 

Example The robotic manipulator shown in the figure rotates with constant angular speed  about the k axis.   Find a formula for the maximum allowable (constant) rate of extension  if the acceleration of the gripper may not exceed g.

 

We can simply write down the acceleration vector, using polar coordinates.  We identify  and r=L, so that

 

 

Other examples using polar coordinates can be found in sections below.

 

 

3.1.5 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  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  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  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  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  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.

 

 

 

3.1.6 Newton’s laws of motion for a particle

 

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)

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

 

 

 

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.

 

 

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 .   The forces acting on the frame must obey  and , (because the frame has no mass). 

 

 

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 we can identify a convenient origin somewhere that we regard as `fixed’.   In addition, to write down an acceleration vector, we need to be able to choose a set of fixed directions in space.

 

For engineering calculations, this 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.

 

But 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 addresses this problem  and in doing so explains many small but noticeable discrepancies between the predictions of Newton’s laws and experiment. 

 

It would be fun to cover the general theory of relativity in this course  but regrettably the mathematics needed to solve any realistic problem is horrendous.  As engineers, we always have to solve realistic problems, and we usually can’t afford to spend a long time doing complicated calculations, so we use the simplest theory that will allow us to make the correct design decisions.  Newton’s laws are fine for us…

 

 

 

3.2 Calculating forces required to cause prescribed motion of a particle

 

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  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  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  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  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  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.   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  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

 

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

 

 

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

 

 

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

 

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 .

4.      EOM. Newton’s laws of motion yield

 

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

 

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

 

where  is the friction coefficient.  Substituting for T and N from (5) shows that for no slip

 

To do the trick, therefore, the acceleration must exceed .  For a friction coefficient of order 0.1, this gives an acceleration of order .   There is a special trick to pulling the tablecloth with a large acceleration  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  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  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  and  - 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  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 .

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  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 , where a is unknown. 

 

4.      EOM: Because this problem includes a massless frame, we must use two equations of motion (  and  ).  It is essential to take moments about the particle (i.e. the rider’s COM).

 

 gives  

 

 It’s very simple to do the moment calculation by hand, but for those of you who find such calculations unbearable here’s a Mupad script to do it. The script simply writes out the position vectors of points A and B relative to the center of mass as 3D vectors, writes down the reactions at A and B as 3D vectors, and calculates the resultant moment (we don’t bother including the weight, because it acts at the origin and so exerts zero moment)

The two nonzero components of  and the one nonzero component of  give us three scalar equations

 

We have four unknowns  the reaction components  and the acceleration a so we need another equation.   The missing equation is the friction law

 

 

5.      Solution: Here’s the solution with Mupad.  It’s easy to get the same answer by hand as well.

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

 

 

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  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  and  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  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   and , and we can treat them as though they were in static equilibrium. 

 

2.      FBD  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

 

 

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

 

We now have three equations for three unknowns (the unknowns are  and T).

 

5.      Solution: As paid up members of ALE (the Academy of Lazy Engineers) we use Mupad to solve the equations

So the tension in the cable is

 

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 .  The bar rotates at constant angular speed .  Find the critical angular velocity that will just make the block start to slip when .  Which way does the block slide?

 

The general approach to this problem is the same as for the Magic trick example  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  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  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

 

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

 

4.      EOM: The equation of motion is

 

5.      Solution: The i and j components of the equation of motion can be solved for N and T  Mupad  makes this painless

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

 

For the block to slip with  

 

so the critical angular velocity is .  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  around a circular arc with radius r.   Its acceleration vector has magnitude  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

 

(iii) Newton’s laws then give

 

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

 

This is the same solution as before.   The short-cut makes the calculation slightly more straightforward.   This is the main purpose of using normal-tangential components.

 

 

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’  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  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 .  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  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

 

 

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

 

4.      EOM: The equation of motion is

 

5.      Solution: The i and j components of the equation of motion can be solved for N and T  Mupad makes this painless

normal reaction force is therefore

 

We are looking for the point where this can first become zero or negative.  Note that   at the point where  =0.  The smallest value of N therefore occurs at this point, and has magnitude

 

The critical speed where N=0 follows as

 

Changing the angle  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

 

 

2.      Kinematics  The acceleration vector is now

 

3.      EOM: The equation of motion is

 

4.      Solution: The  components of the equation of motion can be solved for N and T  again, we can use Mupad for this

 

The normal reaction force is therefore

 

We are looking for the point where this can first become zero or negative.  Note that   at the point where  =0.  The smallest value of N therefore occurs at this point, and has magnitude

 

The critical speed where N=0 follows as

 

Changing the angle  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  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  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  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  The aircraft is idealized as a particle  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 , (iii) a drag force , acting perpendicular to the direction of motion, and (iv) a lift force , acting perpendicular to the plane of the wings.

 

The resultant force is

 

(you may find the components of the lift force difficult to visualize  to see where these come from, note that the lift force can be projected onto components along OR and the k direction as .  Then note that  .) 

 

3.      Kinematics  

a.       The aircraft moves at constant speed around a circle, so the angle , where  is the (constant) angular speed of the line OP.   Since the aircraft completes a turn in two minutes, we know that   rad/sec

b.      The position vector of the plane is

 

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

 

c.       The magnitude of the velocity is , so if the aircraft flies at speed V, the radius of the turn must be  

d.      Differentiating the velocity gives the acceleration

 

 

4.      EOM: The equation of motion is

 

 

5.      Solution: The i j and k components of the equation of motion give three equations that can be solved for ,  and .  We assume that the drag force is known, since this is a function of the aircraft’s speed.

This solution is correct, but you need a PhD to understand what it means (other symbolic manipulation programs like Maple and Mathematica give a comprehensible solution).  Fortunately, I happen to have a PhD…  The solution can be simplified to

 

As we will see below, if you choose to solve this problem in normal-tangential coordinates, you don’t need a PhD to

 

We can calculate values of ,  and the load factor  for a few aircraft

a.       Cessna 150  V=70knots (36 m/s) :  R=690m,  

b.      Boeing 747: V=200 knots (102 m/s)  R=1950m,  

c.       F111  V=300 knots (154 m/s)  R=2950m,  

 

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

 

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

 

 

(iii) Newton’s law  

 

The n, t and k components of this equation give three equations that can be solved for ,  and .  This time it is easy to solve the equations by hand…

 

 

This example again shows why normal-tangential coordinates are useful  describing forces, and solving the resulting force-acceleration relations are much simpler than working with a fixed coordinate system.

 

 

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  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  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  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  conservation of energy, or conservation of momentum  to calculate quantities of interest.  These short-cuts will be discussed in the next chapter.

 

 

3.2.2 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  with initial velocity vector  .  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   

 

2. Write down the position vector in terms of these variables:  

 

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

 

4. Draw a free body diagram.  The only force acting on the particle is gravity  the free body diagram is shown in the figure.  The force vector follows as .

 

 

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

 

The vector equation actually represents three separate differential equations of motion

 

 

6. Eliminate reactions  this is not needed in this example.

 

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

 

 

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

 

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

 

The equation of motion can be re-written in terms of  as

 

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

 

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

 

Again, we can separate variables and integrate

 

so the position and velocity vectors are

 

 

Here’s how to integrate the equations of motion using Mupad

 

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

 

2.      The components of the position of the particle at time t=0 are   

3.      The trajectory of the particle follows as

 

4.      When the particle hits the ground, its position vector is .  This must be on the trajectory, so

 

where  is the time of impact.  

5.      The two components of this vector equation gives us two equations for the two unknowns , which can be solved

The RootOf in MAPLE is scary  any time that MAPLE gives you a scary result, look in the help and see if there is a `convert’ function that might make it less scary. 

 

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

1.      Height of Kremlin palace  71m

2.      Throwing velocity  maybe 25mph?  (pretty pathetic, I know - you can probably do better, especially if you are on the baseball team). 

3.      Throwing angle  45 degrees.

Substituting numbers gives 36m (118ft).

If you want to go wild, you can maximize D with respect to , 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 , 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  and velocity  

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

 

Its velocity is

 

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   is a maximum, then  ).  Therefore, at the instant of maximum height  

5.      This shows that the instant of max height occurs at time  

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

 

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  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 .  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  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  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  and , 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  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

 

 

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

 

 

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

 

The i and j components give two scalar equations of motion

 

6. Eliminate reactions  this is not needed in this example.

 

7. Identify initial conditionsThe initial conditions were given in this problem  at time t=0, we know that  and  

 

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

 

We can re-write this in terms of

 

This gives

 

We can separate variables and integrate

 

Don’t worry if the last line looks mysterious  writing the solution in this form just makes the algebra a bit simpler.  We can now integrate the velocity to find x

 

The integral on the left can be evaluated using the substitution

 

so that

 

Here’s the MAPLE solution

Note that it’s important to include the assume() statements, otherwise Mupad gives the solution in the form of an exponential of a complex number.   The solution in this other form is also correct, but is difficult to visualize.

 

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 .  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 .  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  

4.      The frequency of oscillation (the number of cycles per second) is  (note f=1/T).  Frequency is also sometimes quoted as angular frequency, which is related to f  by .  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 , you can calculate the (angular) vibration frequency as  

 

 

 

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 .  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  the polar coordinates  work for this problem

 

2.      Write down the position vector and differentiate to find acceleration  we don’t need to do this  we can just write down the standard result for polar coordinates

 

 

3.      Draw a free body diagram  shown in the figure  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

 

 

 

5.      Eliminate reactions

 

F=ma gives two equations for N and T.   A third one comes from the friction law  

 

The third solution can be rearranged into an equation of motion for r

 

 

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…   But if not, or you are lazy, you can use MAPLE to solve it for you.

This can be simplified slightly by hand:

 

 

 

Example 4: Motion of a pendulum (R-rated version)

 

A pendulum is a ubiquitous engineering system. You are, of course, familiar with how a pendulum can be used to measure time.  But it’s used for a variety of other scientific applications.  For example, Professor Crisco’s lab uses pendulum to measure properties of human joints, see

Crisco Joseph J; Fujita Lindsey; Spenciner David B, ‘The dynamic flexion/extension properties of the lumbar spine in vitro using a novel pendulum system.’ Journal of biomechanics 2007;40(12):2767-73

 

In this example, we will work through the basic problem of deriving and solving the equations of motion for a pendulum, neglecting air resistance.

 

1. Introduce variables to describe motion:   The angle    shown in the figure is a convenient variable.

 

 

2. Write down the position vector as a function of the variables We introduce a Cartesian coordinate system with origin at O, as shown in the picture.

 

Elementary geometry shows that  

 

Note that we have assumed that the cable remains straight  this will be true as long as the internal force in the cable is tensile.  If calculations predict that the internal force is compressive, this assumption is wrong.  But there is no way to check the assumption at this point so we simply proceed, and check the answer at the end

 

3. Differentiate the position vector to find the acceleration: The computer makes this painless. 

 

 

4. The free body diagram is shown in the figure.   The force exerted by the cable on the particle is introduced as an unknown reaction force.  The force vector is

 

 

5. Newton’s laws of motion can be expressed as

 

Equating the i and j components gives two equations for the two unknowns

 

 

6. Eliminate the reaction forces.  In this problem, it is helpful to eliminate the unknown reaction force R.  You can do this on the computer if you like, but in this case it is simpler to do this by hand.  You can simply multiply the first equation by  and the second equation by  and then add them.   This yields

 

 

7. Identify initial conditions. Some calculations are necessary to determine the initial conditions in this problem.  We are given that  at time t=0, and the horizontal velocity is  at time t=0, but to solve the equation of motion, we need the value of .  We can find the relationship we need by differentiating the position vector to find the velocity

 

Setting  and  at t=0 shows that

 

so  

 

8. Solve the equations of motion  This equation of motion is too difficult for MAPLE but actually the solution does exist and is very well known  this is a classic problem in mathematical physics.  With initial conditions  the solution is

 

The first solution describes swinging motion of the pendulum, while the second solution describes the motion that occurs if you push the pendulum so hard that it whirls around on the pivot. The equations may look scary, but you can simply use MAPLE to calculate and plot them.

  1. In the first equation,
     is a special function called the `sin amplitude.’  You can think of it as a sort of trig function for adults
     in fact for  k=0,
     and we recover the PG version.   You can compute it in Mupad using
    jacobiSN(x,k)
  2. Similarly,
     is a function called the `Amplitude.’  You can calculate it in Mupad using
    jacobiAM(x,k).  In Mupad, the am function has range
    , so the solution predicts that as the pendulum whirls around the pivot, the angle
     increases from 0 to
    , then jumps to
    , increases to
     again, and so on. 

 

You might have solved the pendulum problem already in an elementary physics course, and might remember a different solution.  This is because you probably only derived an approximate solution, by assuming that the angle  remains small.   This occurs when the initial velocity satisfies , in which case the solution can be approximated by

 

 

 

 

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  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  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 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   

 

2. Write down the position vector in terms of these variables:  

 

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

 

 

 

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

 

Gravity  as always we have .

 

 

Air resistance. 

 

The magnitude of the air drag force is given by , where

·           is the air density,

·          is the drag coefficient,

·         D is the projectile’s diameter, and

·          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.

 

 

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  is a unit vector parallel to the particle’s velocity.  The drag force vector is therefore

 

 

The total force vector is therefore

 

 

 

5. Write down Newton’s laws of motion.

 

It is helpful to simplify the equation by defining a specific drag coefficient , so that

 

The vector equation actually represents three separate differential equations of motion

 

 

6. Eliminate reactions  this is not needed in this example.

 

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

 

 

8. Solve the equations of motion.  We can’t use the magic ‘dsolve’ command in MAPLE to solve this equation  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 .  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 , where

 

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

 

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

 

 

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(@eom,[0,stop_time],initial_w);

 

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

 

function dwdt = eom(t,w)

% 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

 

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  and velocity vector .  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:  

 

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

 

 

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

The magnitude of the force is , where

·          is the gravitational constant, and

·          is the distance between the planet and the satellite

 

The direction of the force is always towards the origin:   is therefore a unit vector parallel to the direction of the force.   The total force acting on the satellite is therefore

 

 

5. Write down Newton’s laws of motion.

 

The vector equation represents two separate differential equations of motion

 

 

6. Eliminate reactions  this is not needed in this example.

 

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

 

 

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 , where

 

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

 

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

 

 

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(@odefunc,[0,time],w0);

 

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

 

function dwdt = odefunc(t,w)

   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

 

 

end

 

 

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

 

Do we believe this result?   It is a bit surprising  the satellite seems to be spiraling in towards the planet.   Most satellites don’t do this  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  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

 

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

 

(If you don’t know these formulas, don’t panic  we will discuss energy and angular momentum in the next part of the course)

 

We can have MATLAB plot E/m and , 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  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  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  must increase linearly with time, i.e.  where  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

 

We know that  from the initial conditions, and  is constant. This tells us that

 

Finally, we can substitute this into Newton’s law

 

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

 

So, if we choose initial values of  satisfying this equation, the orbit will be circular.  In fact, our original choice,  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(@odefunc,[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  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  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  (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 .  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  it certainly is not a `small’ object.   However, because the floors move vertically without rotation, the rigid body equations of motion simply reduce to   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)