Chapter 4

Conservation laws for systems of particles

 

In this chapter, we shall introduce the following general concepts:

  1. The power, or rate of work done by a force
  2. The total work done by a force
  3. The kinetic energy of a particle
  4. The power-kinetic energy and work-kinetic energy relations for a single particle
  5. The concepts of a conservative force and a conservative system
  6. The power-total energy and work-total energy relations for a conservative system
  7. Energy conservation for a conservative system
  8. The linear impulse of a force
  9. The linear momentum of a particle (or system of particles)
  10. The linear impulse - linear momentum relations for a single particle
  11. Linear impulse-momentum relations for a system of particles
  12. Conservation of linear momentum for a system
  13. Analyzing collisions between particles using linear momentum
  14. The angular impulse of a force
  15. The angular momentum of a particle
  16. The angular impulse MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  angular momentum relation for a single particle

 

We will also illustrate how these concepts can be used in engineering calculations.  As you will see, to applying these principles to engineering calculations you will need two things: (i) a thorough understanding of the principles themselves; and (ii) Physical insight into how engineering systems behave, so you can see how to apply the theory to practice.   The first is easy.  The second is hard, but practice will help.

 

 

4.1 Work, Power, Potential Energy and Kinetic Energy relations for particles

 

The concepts of work, power and energy are among the most powerful ideas in the physical sciences.  Their most important application is in the field of thermodynamics, which describes the exchange of energy between interacting systems.  In addition, concepts of energy carry over to relativistic systems and quantum mechanics, where the classical versions of Newton’s laws themselves no longer apply.

 

In this section, we develop the basic definitions of mechanical work and energy, and show how they can be used to analyze motion of dynamical systems.  Future courses will expand on these concepts further.

 

 

4.1.1 Definition of the power and work done by a force

 

Suppose that a force F acts on a particle that moves with speed v.

 

By definition:

* The Power developed by the force, (or the rate of work done by the force) is P=Fv MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 da9iaahAeacqGHflY1caWH2baaaa@3BE9@ .  

 

If both force and velocity are expressed in Cartesian components, then

P= F x v x + F y v y + F z v z MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 da9iaadAeadaWgaaWcbaGaamiEaaqabaGccaWG2bWaaSbaaSqaaiaa dIhaaeqaaOGaey4kaSIaamOramaaBaaaleaacaWG5baabeaakiaadA hadaWgaaWcbaGaamyEaaqabaGccqGHRaWkcaWGgbWaaSbaaSqaaiaa dQhaaeqaaOGaamODamaaBaaaleaacaWG6baabeaaaaa@4615@

 

Power has units of Nm/s, or `Watts’ in SI units.

 

 

* The work done by the force during a time interval t 0 t t 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaakiabgsMiJkaadshacqGHKjYOcaWG0bWaaSba aSqaaiaaigdaaeqaaaaa@3F29@  is

W= t 0 t 1 Fv dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadEfacqGH9a qpdaWdXbqaaiaahAeacqGHflY1caWH2baaleaacaWG0bWaaSbaaWqa aiaaicdaaeqaaaWcbaGaamiDamaaBaaameaacaaIXaaabeaaa0Gaey 4kIipakiaadsgacaWG0baaaa@4454@

The work done by the force can also be calculated by integrating the force vector along the path traveled by the force, as

W= r 0 r 1 Fdr MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadEfacqGH9a qpdaWdXbqaaiaahAeacqGHflY1caWGKbGaaCOCaaWcbaGaaCOCamaa BaaameaacaaIWaaabeaaaSqaaiaahkhadaWgaaadbaGaaGymaaqaba aaniabgUIiYdaaaa@4351@

where r 0 , r 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCamaaBa aaleaacaaIWaaabeaakiaacYcacaWHYbWaaSbaaSqaaiaaigdaaeqa aaaa@3A73@  are the initial and final positions of the force.

 

Work has units of Nm in SI units, or `Joules’

 

A moving force can do work on a particle, or on any moving object.  For example, if a force acts to stretch a spring, it is said to do work on the spring. 

 

 

 

4.1.2 Simple examples of power and work calculations

 

 

Example 1: An aircraft with mass 45000 kg flying at 200 knots (102m/s) climbs at 1000ft/min.  Calculate the rate of work done on the aircraft by gravity.

 

The gravitational force is mgj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Iaam yBaiaadEgacaWHQbaaaa@3ABB@ , and the velocity vector of the aircraft is v= v x i+ v y j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9iaadAhadaWgaaWcbaGaamiEaaqabaGccaWHPbGaey4kaSIaamOD amaaBaaaleaacaWG5baabeaakiaahQgaaaa@4026@ .  The rate of work done on the aircraft is therefore

P=Fv=mg v y MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 da9iaahAeacqGHflY1caWH2bGaeyypa0JaeyOeI0IaamyBaiaadEga caWG2bWaaSbaaSqaaiaadMhaaeqaaaaa@42E6@

Substituting numbers gives

 

 

Example 2: Calculate a formula for the work required to stretch a spring with stiffness k and unstretched length L 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamitamaaBa aaleaacaaIWaaabeaaaaa@38B4@  from length l 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiBamaaBa aaleaacaaIWaaabeaaaaa@38D4@  to length l 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiBamaaBa aaleaacaaIXaaabeaaaaa@38D5@ .

 

The figure shows a spring that held fixed at A and is stretched in the horizontal direction by a force F= F x i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaadAeadaWgaaWcbaGaamiEaaqabaGccaWHPbaaaa@3BC2@  acting at B.  At some instant the spring has length l 0 <x< l 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiBamaaBa aaleaacaaIWaaabeaakiabgYda8iaadIhacqGH8aapcaWGSbWaaSba aSqaaiaaigdaaeqaaaaa@3DBB@ .  The spring force law states that the force acting on the spring at B is related to the length of the spring x by

F= F x i=k(x L 0 )i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaadAeadaWgaaWcbaGaamiEaaqabaGccaWHPbGaeyypa0Jaam4A aiaacIcacaWG4bGaeyOeI0IaamitamaaBaaaleaacaaIWaaabeaaki aacMcacaWHPbaaaa@43AE@

The position vector of the force is r=xi MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadIhacaWHPbaaaa@3AED@ , and therefore the work done is

W= r 0 r 1 Fdr= l 0 l 1 k(x L 0 )idxi = k 2 { ( l 1 L 0 ) 2 ( l 0 L 0 ) 2 } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadEfacqGH9a qpdaWdXbqaaiaahAeacqGHflY1caWGKbGaaCOCaiabg2da9aWcbaGa aCOCamaaBaaameaacaaIWaaabeaaaSqaaiaahkhadaWgaaadbaGaaG ymaaqabaaaniabgUIiYdGcdaWdXbqaaiaadUgacaGGOaGaamiEaiab gkHiTiaadYeadaWgaaWcbaGaaGimaaqabaGccaGGPaGaaCyAaiabgw SixlaadsgacaWG4bGaaCyAaaWcbaGaamiBamaaBaaameaacaaIWaaa beaaaSqaaiaadYgadaWgaaadbaGaaGymaaqabaaaniabgUIiYdGccq GH9aqpdaWcaaqaaiaadUgaaeaacaaIYaaaamaacmaabaGaaiikaiaa dYgadaWgaaWcbaGaaGymaaqabaGccqGHsislcaWGmbWaaSbaaSqaai aaicdaaeqaaOGaaiykamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaa cIcacaWGSbWaaSbaaSqaaiaaicdaaeqaaOGaeyOeI0IaamitamaaBa aaleaacaaIWaaabeaakiaacMcadaahaaWcbeqaaiaaikdaaaaakiaa wUhacaGL9baaaaa@6A2F@

 

 

Example 3: Calculate the work done by gravity on a satellite that is launched from the surface of the earth to an altitude of 250km (a typical low earth orbit).

 

Assumptions

  1. The earth’s radius is 6378.145km
  2. The mass of a typical satellite is 4135kg  - see , e.g. http://www.astronautix.com/craft/hs601.htm
  3. The Gravitational parameter μ=GM=3.986012× 10 5 km 3 s 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH8oqBcqGH9aqpcaWGhbGaamytai abg2da9iaaiodacaGGUaGaaGyoaiaaiIdacaaI2aGaaGimaiaaigda caaIYaGaaGPaVlaaykW7caaMc8UaaGPaVlabgEna0kaaykW7caaMc8 UaaGymaiaaicdadaahaaWcbeqaaiaaiwdaaaGccaaMc8UaaGPaVlaa ykW7caqGRbGaaeyBamaaCaaaleqabaGaaG4maaaakiaabohadaahaa WcbeqaaiabgkHiTiaaigdaaaaaaa@560C@   (G= gravitational constant; M=mass of earth)
  4. We will assume that the satellite is launched along a straight line path parallel to the i direction, starting the earths surface and extending to the altitude of the orbit.  It turns out that the work done is independent of the path, but this is not obvious without more elaborate and sophisticated calculations.

 

Calculation:

  1. The gravitational force on the satellite is

F= GMm x 2 i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcqGHsisldaWcaaqaaiaadEeacaWGnbGaamyBaaqaaiaadIhadaah aaWcbeqaaiaaikdaaaaaaOGaaCyAaaaa@3E7D@

  1. The work done follows as

F= R R+h GMm x 2 idxi =GMm( 1 R+h 1 R ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcqGHsisldaWdXbqaamaalaaabaGaam4raiaad2eacaWGTbaabaGa amiEamaaCaaaleqabaGaaGOmaaaaaaGccaWHPbGaeyyXICTaamizai aadIhacaWHPbaaleaacaWGsbaabaGaamOuaiabgUcaRiaadIgaa0Ga ey4kIipakiabg2da9iaadEeacaWGnbGaamyBamaabmaabaWaaSaaae aacaaIXaaabaGaamOuaiabgUcaRiaadIgaaaGaeyOeI0YaaSaaaeaa caaIXaaabaGaamOuaaaaaiaawIcacaGLPaaaaaa@54AE@

  1. Substituting numbers gives 9.7× 10 6 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaaI5aGaaiOlaiaaiEdacqGHxdaTca aIXaGaaGimamaaCaaaleqabaGaaGOnaaaaaaa@3974@  J (be careful with units MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  if you work with kilometers the work done is in N-km instead of SI units Nm)

 

 

Example 4: A Ferrari Testarossa skids to a stop over a distance of 250ft.  Calculate the total work done on the car by the friction forces acting on its wheels.

 

Assumptions:

  1. A Ferrari Testarossa has mass 1506kg (see http://www.ultimatecarpage.com/car/1889/Ferrari-Testarossa.html)
  2. The coefficient of friction between wheels and road is of order 0.8
  3. We assume the brakes are locked so all wheels skid, and air resistance is neglected

 

Calculation The figure shows a free body diagram. The equation of motion for the car is

( T F + T R )i+( N R + N F mg)j=m a x i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaacIcacaWGub WaaSbaaSqaaiaadAeaaeqaaOGaey4kaSIaamivamaaBaaaleaacaWG sbaabeaakiaacMcacaWHPbGaey4kaSIaaiikaiaad6eadaWgaaWcba GaamOuaaqabaGccqGHRaWkcaWGobWaaSbaaSqaaiaadAeaaeqaaOGa eyOeI0IaamyBaiaadEgacaGGPaGaaCOAaiabg2da9iaad2gacaWGHb WaaSbaaSqaaiaadIhaaeqaaOGaaCyAaaaa@4CB8@

  1. The vertical component of the equation of motion yields N R + N F =mg MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaad6eadaWgaa WcbaGaamOuaaqabaGccqGHRaWkcaWGobWaaSbaaSqaaiaadAeaaeqa aOGaeyypa0JaamyBaiaadEgaaaa@3DB3@
  2. The friction law shows that T R =μ N R T F =μ N F T R + T F =μ( N R + N F )=μmg MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsfadaWgaa WcbaGaamOuaaqabaGccqGH9aqpcqaH8oqBcaWGobWaaSbaaSqaaiaa dkfaaeqaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caWGubWaaSbaaSqaaiaadAeaaeqaaOGaeyypa0JaeqiV d0MaamOtamaaBaaaleaacaWGgbaabeaakiabgkDiElaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaadsfadaWgaaWcbaGaamOuaaqa baGccqGHRaWkcaWGubWaaSbaaSqaaiaadAeaaeqaaOGaeyypa0Jaeq iVd02aaeWaaeaacaWGobWaaSbaaSqaaiaadkfaaeqaaOGaey4kaSIa amOtamaaBaaaleaacaWGgbaabeaaaOGaayjkaiaawMcaaiabg2da9i abeY7aTjaad2gacaWGNbaaaa@6D33@
  3. The position vectors of the car’s front and rear wheels are r F =xi r R =(x+L)i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhadaWgaa WcbaGaamOraaqabaGccqGH9aqpcaWG4bGaaCyAaiaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaahkhadaWgaaWcbaGaamOuaaqabaGccqGH9aqpcaGG OaGaamiEaiabgUcaRiaadYeacaGGPaGaaCyAaaaa@55B7@ The work done follows as.  We suppose that the rear wheel starts at some point x 0 i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhadaWgaa WcbaGaaGimaaqabaGccaWHPbaaaa@3918@  when the brakes are applied and skids a total distance d.

W= x 0 x 0 d F R d r R + x 0 +L x 0 d+L F F d r F = x 0 x 0 d T R i(dxi) + x 0 +L x 0 d+L T F i(dxi) =( T R + T F )d MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadEfacqGH9a qpdaWdXbqaaiaahAeadaWgaaWcbaGaamOuaaqabaGccqGHflY1caWG KbGaaCOCamaaBaaaleaacaWGsbaabeaakiabgUcaRmaapehabaGaaC OramaaBaaaleaacaWGgbaabeaakiabgwSixlaadsgacaWHYbWaaSba aSqaaiaadAeaaeqaaOGaeyypa0daleaacaWG4bWaaSbaaWqaaiaaic daaeqaaSGaey4kaSIaamitaaqaaiaadIhadaWgaaadbaGaaGimaaqa baWccqGHsislcaWGKbGaey4kaSIaamitaaqdcqGHRiI8aaWcbaGaam iEamaaBaaameaacaaIWaaabeaaaSqaaiaadIhadaWgaaadbaGaaGim aaqabaWccqGHsislcaWGKbaaniabgUIiYdGcdaWdXbqaaiaadsfada WgaaWcbaGaamOuaaqabaGccaWHPbGaeyyXICTaaiikaiaadsgacaWG 4bGaaCyAaiaacMcaaSqaaiaadIhadaWgaaadbaGaaGimaaqabaaale aacaWG4bWaaSbaaWqaaiaaicdaaeqaaSGaeyOeI0IaamizaaqdcqGH RiI8aOGaey4kaSYaa8qCaeaacaWGubWaaSbaaSqaaiaadAeaaeqaaO GaaCyAaiabgwSixlaacIcacaWGKbGaamiEaiaahMgacaGGPaaaleaa caWG4bWaaSbaaWqaaiaaicdaaeqaaSGaey4kaSIaamitaaqaaiaadI hadaWgaaadbaGaaGimaaqabaWccqGHsislcaWGKbGaey4kaSIaamit aaqdcqGHRiI8aOGaeyypa0JaeyOeI0IaaiikaiaadsfadaWgaaWcba GaamOuaaqabaGccqGHRaWkcaWGubWaaSbaaSqaaiaadAeaaeqaaOGa aiykaiaadsgaaaa@8BCA@

  1. The work done follows as W=μmg MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadEfacqGH9a qpcqGHsislcqaH8oqBcaWGTbGaam4zaaaa@3C9C@ Substituting numbers gives W=9× 10 5 J MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadEfacqGH9a qpcaaI5aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiaaiwdaaaGc caWGkbaaaa@3E2F@ .

 

 

Example 5: The figure shows a box that is pushed up a slope by a force P.  The box moves with speed v. Find a formula for the rate of work done by each of the forces acting on the box.

 

The figure shows a free body diagram. The force vectors are

1.      Applied force Pcosαi+Psinαj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadcfaciGGJb Gaai4BaiaacohacqaHXoqycaWHPbGaey4kaSIaamiuaiGacohacaGG PbGaaiOBaiabeg7aHjaahQgaaaa@4393@

2.      Friction TcosαiTsinαj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTiaads faciGGJbGaai4BaiaacohacqaHXoqycaWHPbGaeyOeI0IaamivaiGa cohacaGGPbGaaiOBaiabeg7aHjaahQgaaaa@4493@

3.      Normal reaction Nsinαi+Tcosαj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTiaad6 eaciGGZbGaaiyAaiaac6gacqaHXoqycaWHPbGaey4kaSIaamivaiGa cogacaGGVbGaai4Caiabeg7aHjaahQgaaaa@4482@

4.      Weight mgj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTiaad2 gacaWGNbGaaCOAaaaa@39F7@

The velocity vector is vcosαi+vsinαj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAhaciGGJb Gaai4BaiaacohacqaHXoqycaWHPbGaey4kaSIaamODaiGacohacaGG PbGaaiOBaiabeg7aHjaahQgaaaa@43DF@

Evaluating the dot products Fv MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGHfl Y1caWH2baaaa@3A51@  for each formula, and recalling that cos 2 α+ sin 2 α=1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiGacogacaGGVb Gaai4CamaaCaaaleqabaGaaGOmaaaakiabeg7aHjabgUcaRiGacoha caGGPbGaaiOBamaaCaaaleqabaGaaGOmaaaakiabeg7aHjabg2da9i aaigdaaaa@43AB@  gives

1.      Applied force Pv MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadcfacaWG2b aaaa@3809@

2.      Friction Tv MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTiaads facaWG2baaaa@38FA@

3.      Normal reaction 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaaicdaaaa@36F3@

4.      Weight mgsinα MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTiaad2 gacaWGNbGaci4CaiaacMgacaGGUbGaeqySdegaaa@3D7B@

 

Force

N

Draw

(cm)

0

0

40

10

90

20

140

30

180

40

220

50

270

60

 

Example 6: The table lists the experimentally measured force-v-draw data for a long-bow.  Calculate the total work done to draw the bow.

 

In this case we don’t have a function that specifies the force as a function of position; instead, we have a table of numerical values.   We have to approximate the integral

W= r 0 r 1 Fdr MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadEfacqGH9a qpdaWdXbqaaiaahAeacqGHflY1caWGKbGaaCOCaaWcbaGaaCOCamaa BaaameaacaaIWaaabeaaaSqaaiaahkhadaWgaaadbaGaaGymaaqaba aaniabgUIiYdaaaa@4351@

numerically.  To understand how to do this, remember that integrating a function can be visualized as computing the area under a curve of the function, as illustrated in the figure. 

 

We can estimate the integral by dividing the area into a series of trapezoids, as shown.  Recall that the area of a trapezoid is (base x average height), so the total area of the function is

W=Δ x 1 ( f 1 + f 2 2 )+Δ x 2 ( f 2 + f 3 2 )+Δ x 3 ( f 3 + f 4 2 )+Δ x 4 ( f 4 + f 5 2 ) =0.1( 40+0 2 )+0.1( 20+10 2 )+.... MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaam4vai abg2da9iabfs5aejaadIhadaWgaaWcbaGaaGymaaqabaGcdaqadaqa amaalaaabaGaamOzamaaBaaaleaacaaIXaaabeaakiabgUcaRiaadA gadaWgaaWcbaGaaGOmaaqabaaakeaacaaIYaaaaaGaayjkaiaawMca aiabgUcaRiabfs5aejaadIhadaWgaaWcbaGaaGOmaaqabaGcdaqada qaamaalaaabaGaamOzamaaBaaaleaacaaIYaaabeaakiabgUcaRiaa dAgadaWgaaWcbaGaaG4maaqabaaakeaacaaIYaaaaaGaayjkaiaawM caaiabgUcaRiabfs5aejaadIhadaWgaaWcbaGaaG4maaqabaGcdaqa daqaamaalaaabaGaamOzamaaBaaaleaacaaIZaaabeaakiabgUcaRi aadAgadaWgaaWcbaGaaGinaaqabaaakeaacaaIYaaaaaGaayjkaiaa wMcaaiabgUcaRiabfs5aejaadIhadaWgaaWcbaGaaGinaaqabaGcda qadaqaamaalaaabaGaamOzamaaBaaaleaacaaI0aaabeaakiabgUca RiaadAgadaWgaaWcbaGaaGynaaqabaaakeaacaaIYaaaaaGaayjkai aawMcaaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7cqGH9aqpcaaIWaGaaiOlaiaaigdadaqadaqaamaalaaabaGaaG inaiaaicdacqGHRaWkcaaIWaaabaGaaGOmaaaaaiaawIcacaGLPaaa cqGHRaWkcaaIWaGaaiOlaiaaigdadaqadaqaamaalaaabaGaaGOmai aaicdacqGHRaWkcaaIXaGaaGimaaqaaiaaikdaaaaacaGLOaGaayzk aaGaey4kaSIaaiOlaiaac6cacaGGUaGaaiOlaaaaaa@8422@

 

You could easily do this calculation by hand MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  but for lazy people like me MATLAB has a convenient function called `trapz’ that does this calculation automatically.  Here’s how to use it

 draw = [0,10,20,30,40,50,60]*0.01;

 force = [0,40,90,140,180,220,270];

 trapz(draw,force)

ans =  80.5000

So the solution is 80.5J

 

 

4.1.3 Definition of the potential energy of a conservative force

 

Preamble: Textbooks nearly always define the `potential energy of a force.’  Strictly speaking, we cannot define a potential energy of a single force MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  instead, we need to define the potential energy of a pair of forces. A force can’t exist by itself MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  there must always be an equal and opposite reaction force acting on a second body.  In all of the discussion to be presented in this section, we implicitly assume that the reaction force is acting on a second body, which is fixed at the origin.  This simplifies calculations, and makes the discussion presented here look like those given in textbooks, but you should remember that the potential energy of a force pair is always a function of the relative positions of the two forces.

 

 

With that proviso, consider a force F acting on a particle at some position r in space. Recall that the work done by a force that moves from position vector r 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhadaWgaa WcbaGaaGimaaqabaaaaa@381A@  to position vector r 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhadaWgaa WcbaGaaGymaaqabaaaaa@381B@  is

W= r 0 r 1 Fdr MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadEfacqGH9a qpdaWdXbqaaiaahAeacqGHflY1caWGKbGaaCOCaaWcbaGaaCOCamaa BaaameaacaaIWaaabeaaaSqaaiaahkhadaWgaaadbaGaaGymaaqaba aaniabgUIiYdaaaa@4351@

In general, the work done by the force depends on the path between r 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhadaWgaa WcbaGaaGimaaqabaaaaa@381A@  to r 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhadaWgaa WcbaGaaGymaaqabaaaaa@381B@ .   For some special forces, however, the work done is independent of the path.  Such forces are said to be conservative.

 

For a force to be conservative:

* The force must be a function only of its position MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  i.e. it can’t depend on the velocity of the force, for example.

 *  The force vector must satisfy  curl(F)=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiGacogacaGG1b GaaiOCaiaacYgacaGGOaGaaCOraiaacMcacqGH9aqpcaWHWaaaaa@3DE8@

 

Examples of conservative forces include gravity, electrostatic forces, and the forces exerted by a spring.  Examples of non-conservative (or should that be liberal?) forces include friction, air resistance, and aerodynamic lift forces.

 

The potential energy of a conservative force is defined as the negative of the work done by the force in moving from some arbitrary initial position r 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhadaWgaa WcbaGaaGimaaqabaaaaa@381A@  to a new position r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhaaaa@3734@ , i.e.

V(r)= r 0 r Fdr +constant MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfacaGGOa GaaCOCaiaacMcacqGH9aqpcqGHsisldaWdXbqaaiaahAeacqGHflY1 caWGKbGaaCOCaaWcbaGaaCOCamaaBaaameaacaaIWaaabeaaaSqaai aahkhaa0Gaey4kIipakiabgUcaRiaabogacaqGVbGaaeOBaiaaboha caqG0bGaaeyyaiaab6gacaqG0baaaa@4E17@

The constant is arbitrary, and the negative sign is introduced by convention (it makes sure that systems try to minimize their potential energy).  If there is a point where the force is zero, it is usual to put r 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhadaWgaa WcbaGaaGimaaqabaaaaa@381A@  at this point, and take the constant to be zero.

 

Note that

  1. The potential energy is a scalar valued function
  2. The potential energy is a function only of the position of the force.  If we choose to describe position in terms of Cartesian components r=xi+yj+zk MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhacqGH9a qpcaWG4bGaaCyAaiabgUcaRiaadMhacaWHQbGaey4kaSIaamOEaiaa hUgaaaa@3FD1@ , then V(r)=V(x,y,z) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfacaGGOa GaaCOCaiaacMcacqGH9aqpcaWGwbGaaiikaiaadIhacaGGSaGaamyE aiaacYcacaWG6bGaaiykaaaa@40FC@ .
  3. The relationship between potential energy and force can also be expressed in differential form (which is often more useful for actual calculations) as

F=grad(V) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcqGHsislciGGNbGaaiOCaiaacggacaGGKbGaciikaiaadAfacaGG Paaaaa@3EE1@

If we choose to work with Cartesian components, then

F x i+ F y j+ F z k=( V x i+ V y j+ V z k ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAeadaWgaa WcbaGaamiEaaqabaGccaWHPbGaey4kaSIaamOramaaBaaaleaacaWG 5baabeaakiaahQgacqGHRaWkcaWGgbWaaSbaaSqaaiaadQhaaeqaaO GaaC4Aaiabg2da9iabgkHiTmaabmaabaWaaSaaaeaacqGHciITcaWG wbaabaGaeyOaIyRaamiEaaaacaWHPbGaey4kaSYaaSaaaeaacqGHci ITcaWGwbaabaGaeyOaIyRaamyEaaaacaWHQbGaey4kaSYaaSaaaeaa cqGHciITcaWGwbaabaGaeyOaIyRaamOEaaaacaWHRbaacaGLOaGaay zkaaaaaa@570B@

 

Occasionally, you might have to calculate a potential energy function by integrating forces MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  for example,  if you are interested in running a molecular dynamic simulation of a collection of atoms in a material, you will need to describe the interatomic forces in some convenient way.   The interatomic forces can be estimated by doing quantum-mechanical calculations, and the results can be approximated by a suitable potential energy function.  Here are a few examples showing how you can integrate forces to calculate potential energy

 

 

Example 1: Potential energy of forces exerted by a spring.  A free body diagram showing the forces exerted by a spring connecting two objects is shown in the figure.

  1. The force exerted by a spring is

F=k(x L 0 )i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iabgkHiTiaadUgacaGGOaGaamiEaiabgkHiTiaadYeadaWgaaWc baGaaGimaaqabaGccaGGPaGaaCyAaaaa@40A5@

  1. The position vector of the force is r=xi MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9iaadIhacaWHPbaaaa@3AED@
  2. The potential energy follows as

V(r)= r 0 r Fdr +constant= L 0 L k(x L 0 )idxi = 1 2 k (L L 0 ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfacaGGOa GaaCOCaiaacMcacqGH9aqpcqGHsisldaWdXbqaaiaahAeacqGHflY1 caWGKbGaaCOCaaWcbaGaaCOCamaaBaaameaacaaIWaaabeaaaSqaai aahkhaa0Gaey4kIipakiabgUcaRiaabogacaqGVbGaaeOBaiaaboha caqG0bGaaeyyaiaab6gacaqG0bGaaGPaVlaaykW7caaMc8Uaaeypam aapehabaGaam4AaiaacIcacaWG4bGaeyOeI0IaamitamaaBaaaleaa caaIWaaabeaakiaacMcacaWHPbGaeyyXICTaamizaiaadIhacaWHPb aaleaacaWGmbWaaSbaaWqaaiaaicdaaeqaaaWcbaGaamitaaqdcqGH RiI8aOGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaacaWGRbGaai ikaiaadYeacqGHsislcaWGmbWaaSbaaSqaaiaaicdaaeqaaOGaaiyk amaaCaaaleqabaGaaGOmaaaaaaa@6DC5@

where we have taken the constant to be zero.

 

 

 

 

Example 2: Potential energy of electrostatic forces exerted by charged particles.

The figure shows two charged particles a distance x apart.  To calculate the potential energy of the force acting on particle 2, we place particle 1 at the origin, and note that the force acting on particle 2 is

F= Q 1 Q 2 4πε x 2 i MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iabgkHiTmaalaaabaGaamyuamaaBaaaleaacaaIXaaabeaakiaa dgfadaWgaaWcbaGaaGOmaaqabaaakeaacaaI0aGaeqiWdaNaeqyTdu MaamiEamaaCaaaleqabaGaaGOmaaaaaaGccaWHPbaaaa@435C@

where   Q 1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyuamaaBa aaleaacaaIXaaabeaaaaa@37B4@  and Q 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyuamaaBa aaleaacaaIYaaabeaaaaa@37B5@  are the charges on the two particles, and ε MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyTdugaaa@379E@  is a fundamental physical constant known as the Permittivity of the medium surrounding the particles.  Since the force is zero when the particles are infinitely far apart, we take r 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhadaWgaa WcbaGaaGimaaqabaaaaa@381A@  at infinity.  The potential energy follows as

V(r)= x Q 1 Q 2 4πε x 2 i dxi= Q 1 Q 2 4πεx MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiaacI cacaWHYbGaaiykaiabg2da9iabgkHiTmaapehabaWaaSaaaeaacaWG rbWaaSbaaSqaaiaaigdaaeqaaOGaamyuamaaBaaaleaacaaIYaaabe aaaOqaaiaaisdacqaHapaCcqaH1oqzcaWG4bWaaWbaaSqabeaacaaI YaaaaaaakiaahMgaaSqaaiabg6HiLcqaaiaadIhaa0Gaey4kIipaki abgwSixlaadsgacaWG4bGaaCyAaiabg2da9maalaaabaGaamyuamaa BaaaleaacaaIXaaabeaakiaadgfadaWgaaWcbaGaaGOmaaqabaaake aacaaI0aGaeqiWdaNaeqyTduMaamiEaaaaaaa@5983@

 

 

 

 

 

 

Table of potential energy relations

 

In practice, however, we rarely need to do the integrals to calculate the potential energy of a force, because there are very few different kinds of force.  For most engineering calculations the potential energy formulas listed in the table below are sufficient.

 

 

Type of force

Force vector

Potential energy

 

Gravity acting on a particle near earths surface

F=mgj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcqGHsislcaWGTbGaam4zaiaahQgaaaa@3BCC@

V=mgy MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfacqGH9a qpcaWGTbGaam4zaiaadMhaaaa@3AF6@

Gravitational force exerted on mass m by mass M at the origin

F= GMm r 3 r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcqGHsisldaWcaaqaaiaadEeacaWGnbGaamyBaaqaaiaadkhadaah aaWcbeqaaiaaiodaaaaaaOGaaCOCaaaa@3E81@

V= GMm r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfacqGH9a qpcqGHsisldaWcaaqaaiaadEeacaWGnbGaamyBaaqaaiaadkhaaaaa aa@3C9E@

Force exerted by a spring with stiffness k and unstretched length L 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamitamaaBa aaleaacaaIWaaabeaaaaa@38B5@

F=k(r L 0 ) r r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcqGHsislcaWGRbGaaiikaiaadkhacqGHsislcaWGmbWaaSbaaSqa aiaaicdaaeqaaOGaaiykamaalaaabaGaaCOCaaqaaiaadkhaaaaaaa@40EB@

V= 1 2 k ( r L 0 ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfacqGH9a qpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaadUgadaqadaqaaiaadkha cqGHsislcaWGmbWaaSbaaSqaaiaaicdaaeqaaaGccaGLOaGaayzkaa WaaWbaaSqabeaacaaIYaaaaaaa@40A8@

Force acting between two charged particles

F= Q 1 Q 2 4πε r 3 r MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9maalaaabaGaamyuamaaBaaaleaacaaIXaaabeaakiaadgfadaWg aaWcbaGaaGOmaaqabaaakeaacaaI0aGaeqiWdaNaeqyTduMaamOCam aaCaaaleqabaGaaG4maaaaaaGccaWHYbaaaa@4273@

V= Q 1 Q 2 4πεr MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 da9maalaaabaGaamyuamaaBaaaleaacaaIXaaabeaakiaadgfadaWg aaWcbaGaaGOmaaqabaaakeaacaaI0aGaeqiWdaNaeqyTduMaamOCaa aaaaa@4090@

Force exerted by one molecule of a noble gas (e.g. He, Ar, etc) on another (Lennard Jones potential). a is the equilibrium spacing between molecules, and E is the energy of the bond.

F=12 E a [ 2 ( a r ) 13 ( a r ) 7 ] r r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iabgkHiTiaaigdacaaIYaWaaSaaaeaacaWGfbaabaGaamyyaaaa daWadaqaaiaaikdadaqadaqaamaalaaabaGaamyyaaqaaiaadkhaaa aacaGLOaGaayzkaaWaaWbaaSqabeaacaaIXaGaaG4maaaakiabgkHi TmaabmaabaWaaSaaaeaacaWGHbaabaGaamOCaaaaaiaawIcacaGLPa aadaahaaWcbeqaaiaaiEdaaaaakiaawUfacaGLDbaacaaMc8UaaGPa VpaalaaabaGaaCOCaaqaaiaadkhaaaaaaa@4F3C@

E[ ( a r ) 12 2 ( a r ) 6 ] MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyramaadm aabaWaaeWaaeaadaWcaaqaaiaadggaaeaacaWGYbaaaaGaayjkaiaa wMcaamaaCaaaleqabaGaaGymaiaaikdaaaGccqGHsislcaaIYaWaae WaaeaadaWcaaqaaiaadggaaeaacaWGYbaaaaGaayjkaiaawMcaamaa CaaaleqabaGaaGOnaaaaaOGaay5waiaaw2faaiaaykW7caaMc8oaaa@4809@

 

 

 

 

 

 

4.1.4 Definition of the Kinetic Energy of a particle

 

Consider a particle with mass m which moves with velocity v= v x i+ v y j+ v z k MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9iaadAhadaWgaaWcbaGaamiEaaqabaGccaWHPbGaey4kaSIaamOD amaaBaaaleaacaWG5baabeaakiaahQgacqGHRaWkcaWG2bWaaSbaaS qaaiaadQhaaeqaaOGaaC4Aaaaa@442D@ .  By definition, its kinetic energy is

T= 1 2 m | v | 2 = 1 2 m( v x 2 + v y 2 + v z 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 da9maalaaabaGaaGymaaqaaiaaikdaaaGaamyBamaaemaabaGaaCOD aaGaay5bSlaawIa7amaaCaaaleqabaGaaGOmaaaakiabg2da9maala aabaGaaGymaaqaaiaaikdaaaGaamyBamaabmaabaGaamODamaaDaaa leaacaWG4baabaGaaGOmaaaakiabgUcaRiaadAhadaqhaaWcbaGaam yEaaqaaiaaikdaaaGccqGHRaWkcaWG2bWaa0baaSqaaiaadQhaaeaa caaIYaaaaaGccaGLOaGaayzkaaaaaa@4FF9@

 

 

 

 

 

4.1.5 Power-Work-kinetic energy relations for a single particle

 

Consider a particle with mass m that moves under the action of a force F. Suppose that

  1. At some time t 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaaaaa@38DD@  the particle has some initial position r 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCamaaBa aaleaacaaIWaaabeaaaaa@38DF@ , velocity v 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaBa aaleaacaaIWaaabeaaaaa@38E3@  and kinetic energy T 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBa aaleaacaaIWaaabeaaaaa@38BD@
  2. At some later time t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaaaa@37F7@  the particle has a new position r, velocity v MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaaaa@37FD@  and kinetic energy T MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaaaa@37D7@ .
  3. Let P=Fv MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 da9iaahAeacqGHflY1caWH2baaaa@3CF1@  denote the rate of work done by the force
  4. Let W= t 0 t Pdt = r 0 r 1 Fdr MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4vaiabg2 da9maapehabaGaamiuaiaadsgacaWG0baaleaacaWG0bWaaSbaaWqa aiaaicdaaeqaaaWcbaGaamiDaaqdcqGHRiI8aOGaeyypa0Zaa8qCae aacaWHgbGaeyyXICTaamizaiaahkhaaSqaaiaahkhadaWgaaadbaGa aGimaaqabaaaleaacaWHYbWaaSbaaWqaaiaaigdaaeqaaaqdcqGHRi I8aaaa@4D2A@  be the total work done by the force

The Power-kinetic energy relation for the particle states that the rate of work done by F is equal to the rate of change of kinetic energy of the particle, i.e.

P= dT dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 da9maalaaabaGaamizaiaadsfaaeaacaWGKbGaamiDaaaaaaa@3C8C@

 

 

Proof: This is just another way of writing Newton’s law for the particle: to see this, note that we can take the dot product of both sides of F=ma with the particle velocity

Fv=mav=m dv dt v= d dt { 1 2 m(vv) } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabgw SixlaahAhacqGH9aqpcaWGTbGaaCyyaiabgwSixlaahAhacqGH9aqp caWGTbWaaSaaaeaacaWGKbGaaCODaaqaaiaadsgacaWG0baaaiabgw SixlaahAhacqGH9aqpdaWcaaqaaiaadsgaaeaacaWGKbGaamiDaaaa daGadaqaamaalaaabaGaaGymaaqaaiaaikdaaaGaamyBaiaacIcaca WH2bGaeyyXICTaaCODaiaacMcaaiaawUhacaGL9baaaaa@5887@

 

To see the last step, do the derivative using the Chain rule and note that ab=ba MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabgw SixlaahkgacqGH9aqpcaWHIbGaeyyXICTaaCyyaaaa@4042@ .

 

The Work-kinetic energy relation for a particle says that the total work done by the force F on the particle is equal to the change in the kinetic energy of the particle.

W=T T 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4vaiabg2 da9iaadsfacqGHsislcaWGubWaaSbaaSqaaiaaicdaaeqaaaaa@3C64@

This follows by integrating the power-kinetic energy relation with respect to time.

 

 

 

 

4.1.6 Examples of simple calculations using work-power-kinetic energy relations

 

There are two main applications of the work-power-kinetic energy relations.  You can use them to calculate the distance over which a force must act in order to produce a given change in velocity.  You can also use them to estimate the energy required to make a particle move in a particular way, or the amount of energy that can be extracted from a collection of moving particles (e.g. using a wind turbine)

 

 

 

 

 

Example 1: The longest single-span escalator in the Western hemisphere is located at the Washington Metro station in Montgomery County, Maryland.  Some technical specifications for the escalator span can be found on Wikipedia (we have not checked this data!).   Additional information concerning escalator standards can be found here.

 

Calculate the kinetic energy of a single 80kg rider standing on the escalator

The KE is m v 2 /2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiaadA hadaahaaWcbeqaaiaaikdaaaGccaGGVaGaaGOmaaaa@3B4D@ ; we have that the speed is 27m per minute, so KE is 8.1J.

 

Calculate the change in potential energy of a single 80kg rider who travels the entire length of the escalator span.

The change in PE is mgh, where h = 35m , so PE is 27468J

 

Assuming the escalator operates at its theoretical capacity of 9000 passengers per hour, estimate the power required to operate the escalator.

 

The power is the number of passengers per second multiplied by the energy change per passenger.   This gives P=68.7 kW

 

 

 

Example 2: Estimate the minimum distance required for a 14 wheeler that travels at the RI speed-limit to brake to a standstill.  Is the distance to stop any different for a Toyota Echo?

 

This problem can be solved by noting that, since we know the initial and final speed of the vehicle, we can calculate the change in kinetic energy as the vehicle stops.  The change in kinetic energy must equal the work done by the forces acting on the vehicle MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  which depends on the distance slid.   Here are the details of the calculation.

 

Assumptions:

  1. We assume that all the wheels are locked and skid over the ground (this will stop the vehicle in the shortest possible distance)
  2. The contacts are assumed to have friction coefficient μ=0.8 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiVd0Maey ypa0JaaGimaiaac6cacaaI4aaaaa@3BE7@
  3. The vehicle is idealized as a particle.
  4. Air resistance will be neglected.

 

Calculation:

  1. The figure shows a free body diagram.
  2. The equation of motion for the vehicle is

( T A + T B + T C + T D )i+( N A + N B + N C + N D mg )j=m a x i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Yaae WaaeaacaWGubWaaSbaaSqaaiaadgeaaeqaaOGaey4kaSIaamivamaa BaaaleaacaWGcbaabeaakiabgUcaRiaadsfadaWgaaWcbaGaam4qaa qabaGccqGHRaWkcaWGubWaaSbaaSqaaiaadseaaeqaaaGccaGLOaGa ayzkaaGaaCyAaiabgUcaRmaabmaabaGaamOtamaaBaaaleaacaWGbb aabeaakiabgUcaRiaad6eadaWgaaWcbaGaamOqaaqabaGccqGHRaWk caWGobWaaSbaaSqaaiaadoeaaeqaaOGaey4kaSIaamOtamaaBaaale aacaWGebaabeaakiabgkHiTiaad2gacaWGNbaacaGLOaGaayzkaaGa aCOAaiabg2da9iaad2gacaWGHbWaaSbaaSqaaiaadIhaaeqaaOGaaC yAaaaa@5979@

The vertical component of the equation shows that N A + N B + N C + N D =mg MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtamaaBa aaleaacaWGbbaabeaakiabgUcaRiaad6eadaWgaaWcbaGaamOqaaqa baGccqGHRaWkcaWGobWaaSbaaSqaaiaadoeaaeqaaOGaey4kaSIaam OtamaaBaaaleaacaWGebaabeaakiabg2da9iaad2gacaWGNbaaaa@43C9@ .

  1. The friction force follows as ( T A + T B + T C + T D )=μ( N A + N B + N C + N D )=μmg MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca WGubWaaSbaaSqaaiaadgeaaeqaaOGaey4kaSIaamivamaaBaaaleaa caWGcbaabeaakiabgUcaRiaadsfadaWgaaWcbaGaam4qaaqabaGccq GHRaWkcaWGubWaaSbaaSqaaiaadseaaeqaaaGccaGLOaGaayzkaaGa eyypa0JaeqiVd02aaeWaaeaacaWGobWaaSbaaSqaaiaadgeaaeqaaO Gaey4kaSIaamOtamaaBaaaleaacaWGcbaabeaakiabgUcaRiaad6ea daWgaaWcbaGaam4qaaqabaGccqGHRaWkcaWGobWaaSbaaSqaaiaads eaaeqaaaGccaGLOaGaayzkaaGaeyypa0JaeqiVd0MaamyBaiaadEga aaa@554D@
  2. If the vehicle skids for a distance d, the total work done by the forces acting on the vehicle is

W= 0 d ( T A + T B + T C + T D )idxi =( T A + T B + T C + T D )d=μmgd MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4vaiabg2 da9maapehabaGaeyOeI0IaaiikaiaadsfadaWgaaWcbaGaamyqaaqa baGccqGHRaWkcaWGubWaaSbaaSqaaiaadkeaaeqaaOGaey4kaSIaam ivamaaBaaaleaacaWGdbaabeaakiabgUcaRiaadsfadaWgaaWcbaGa amiraaqabaGccaGGPaGaaCyAaiabgwSixlaadsgacaWG4bGaaCyAaa WcbaGaaGimaaqaaiaadsgaa0Gaey4kIipakiabg2da9iabgkHiTiaa cIcacaWGubWaaSbaaSqaaiaadgeaaeqaaOGaey4kaSIaamivamaaBa aaleaacaWGcbaabeaakiabgUcaRiaadsfadaWgaaWcbaGaam4qaaqa baGccqGHRaWkcaWGubWaaSbaaSqaaiaadseaaeqaaOGaaiykaiaads gacqGH9aqpcqGHsislcqaH8oqBcaWGTbGaam4zaiaadsgaaaa@63F4@

  1. The work-energy relation states that the total work done on the particle is equal to its change in kinetic energy.  When the brakes are applied the vehicle is traveling at the speed limit, with speed V; at the end of the skid its speed is zero.  The change in kinetic energy is therefore ΔT=0m V 2 /2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam ivaiabg2da9iaaicdacqGHsislcaWGTbGaamOvamaaCaaaleqabaGa aGOmaaaakiaac+cacaaIYaaaaa@4018@ The work-energy relation shows that

mgd=m V 2 /2d= V 2 /μg MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Iaam yBaiaadEgacaWGKbGaeyypa0JaeyOeI0IaamyBaiaadAfadaahaaWc beqaaiaaikdaaaGccaGGVaGaaGOmaiabgkDiElaadsgacqGH9aqpca WGwbWaaWbaaSqabeaacaaIYaaaaOGaai4laiabeY7aTjaadEgaaaa@4A42@

Substituting numbers gives

 

This simple calculation suggests that the braking distance for a vehicle depends only on its speed and the friction coefficient between wheels and tires.  This is unlikely to vary much from one vehicle to another.  In practice there may be more variation between vehicles than this estimate suggests, partly because factors like air resistance and aerodynamic lift forces will influence the results, and also because vehicles usually don’t skid during an emergency stop (if they do, the driver loses control) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  the nature of the braking system therefore also may change the prediction.    

 

 

Example 3: Compare the power consumption of a Ford Excursion to that of a Chevy Cobalt during stop-start driving in a traffic jam.

 

During stop-start driving, the vehicle must be repeatedly accelerated to some (low) velocity; and then braked to a stop. Power is expended to accelerate the vehicle; this power is dissipated as heat in the brakes during braking.  To calculate the energy consumption, we must estimate the energy required to accelerate the vehicle to its maximum speed, and estimate the frequency of this event.

 

 

Calculation/Assumptions:

  1. We assume that the speed in a traffic jam is low enough that air resistance can be neglected.
  2. The energy to accelerate to speed V is m V 2 /2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiaadA fadaahaaWcbeqaaiaaikdaaaGccaGGVaGaaGOmaaaa@3B2D@ .
  3. We assume that the vehicle accelerates and brakes with constant acceleration MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  if so, its average speed is V/2.
  4. If the vehicle travels a distance d between stops, the time between two stops is 2d/V.
  5. The average power is therefore mV/4d MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiaadA facaGGVaGaaGinaiaadsgaaaa@3B25@ .
  6. Taking V=15mph (7m/s) and d=200ft (61m) are reasonable values MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  the power is therefore 0.03m, with m in kg.  A Ford Excursion weighs 9200 lb (4170 kg), requiring 125 Watts (about that of a light bulb) to keep moving.  A Chevy Cobalt weighs 2681lb (1216kg) and requires only 36 Watts MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  a very substantial energy saving.

 

Reducing vehicle weight is the most effective way of improving fuel efficiency during slow driving, and also reduces manufacturing costs and material requirements.  Another, more costly, approach is to use a system that can recover the energy during braking MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  this is the main reason that hybrid vehicles like the Prius have better fuel economy than conventional vehicles.

 

 

 

 

 

 

Example 4: Estimate the power that can be generated by a wind turbine.

 

The figure shows a wind turbine.  The turbine blades deflect the air flowing past them: this changes the air speed and so exerts a force on the blades.  If the blades move, the force exerted by the air on the blades does work MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  this work is the power generated by the turbine.  The rate of work done by the air on the blades must equal the change in kinetic energy of the air as it flows past the blades.   Consequently, we can estimate the power generated by the turbine by calculating the change in kinetic energy of the air flowing through it.

 

To do this properly needs a very sophisticated analysis of the air flow around the turbine.  However, we can get a rather crude estimate of the power by assuming that the turbine is able to extract all the energy from the air that flows through the circular area swept by the blades.

 

Calculation: Let V denote the wind speed, and let ρ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdihaaa@38BE@  denote the density of the air.

1.      In a time t, a cylindrical region of air with radius R and height Vt passes through the fan.

2.      The cylindrical region has mass   π R 2 ρVt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiWdaNaam OuamaaCaaaleqabaGaaGOmaaaakiabeg8aYjaadAfacaWG0baaaa@3E19@

3.      The kinetic energy of the cylindrical region of air is T= 1 2 ( π R 2 ρVt ) V 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 da9maalaaabaGaaGymaaqaaiaaikdaaaWaaeWaaeaacqaHapaCcaWG sbWaaWbaaSqabeaacaaIYaaaaOGaeqyWdiNaamOvaiaadshaaiaawI cacaGLPaaacaWGwbWaaWbaaSqabeaacaaIYaaaaaaa@44CC@

4.      The rate of flow of kinetic energy through the fan is therefore dT dt = π 2 R 2 ρ V 3 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbGaamivaaqaaiaadsgacaWG0baaaiabg2da9maalaaabaGaeqiW dahabaGaaGOmaaaacaWGsbWaaWbaaSqabeaacaaIYaaaaOGaeqyWdi NaamOvamaaCaaaleqabaGaaG4maaaaaaa@4390@

5.      If all this energy could be used to do work on the fan blades the power generated would be P= dT dt = π 2 R 2 ρ V 3 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 da9maalaaabaGaamizaiaadsfaaeaacaWGKbGaamiDaaaacqGH9aqp daWcaaqaaiabec8aWbqaaiaaikdaaaGaamOuamaaCaaaleqabaGaaG Omaaaakiabeg8aYjaadAfadaahaaWcbeqaaiaaiodaaaaaaa@456B@

 

Representative numbers are (i) Air density 1.2 kg/ m 3 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4AaiaadE gacaGGVaGaamyBamaaCaaaleqabaGaaG4maaaaaaa@3B69@ ; (ii) air speed 25mph (11 m/sec); (iii) Radius 30m

This gives 1.8MW.   For comparison, a nuclear power plant generates about 500-1000 MW. 

 

A more sophisticated calculation (which will be covered in EN810) shows that in practice the maximum possible amount of energy that can be extracted from the air is about 60% of this estimate.  On average, a typical household uses about a kW of energy; so a single turbine could provide enough power for about 5-10 houses.

 

 

 

4.1.7 Energy relations for a conservative system of particles.

 

The figure shows a `system of particles’ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  this is just a collection of objects that we might be interested in, which can be idealized as particles. Each particle in the system can experience forces applied by:

 

 * Other particles in the system (e.g. due to gravity, electric charges on the particles, or because the particles are physically connected through springs, or because the particles collide).  We call these internal forces acting in the system.  We will denote the internal force exerted by the ith particle on the jth particle by  R ij MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOuamaaBa aaleaacaWGPbGaamOAaaqabaaaaa@39E1@ .  Note that, because every action has an equal and opposite reaction, the force exerted on the jth particle by the ith particle must be equal and opposite, to R ij MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOuamaaBa aaleaacaWGPbGaamOAaaqabaaaaa@39E1@ , i.e. R ij = R ji MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOuamaaBa aaleaacaWGPbGaamOAaaqabaGccqGH9aqpcqGHsislcaWHsbWaaSba aSqaaiaadQgacaWGPbaabeaaaaa@3EC2@ .

 

* Forces exerted on the particles by the outside world (e.g. by externally applied gravitational or electromagnetic fields, or because the particles are connected to the outside world through mechanical linkages or springs).  We call these external forces acting on the system, and we will denote the external force on the i th particle by F i ext (t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOramaaDa aaleaacaWGPbaabaGaamyzaiaadIhacaWG0baaaOGaaiikaiaadsha caGGPaaaaa@3D1C@

 

We define the rate of external work (or external power) done on the system as

P ext = forces F i ext (t) v i (t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaCa aaleqabaGaamyzaiaadIhacaWG0baaaOGaeyypa0ZaaabCaeaacaWH gbWaa0baaSqaaiaadMgaaeaacaWGLbGaamiEaiaadshaaaGccaGGOa GaamiDaiaacMcacqGHflY1caWH2bWaaSbaaSqaaiaadMgaaeqaaOGa aiikaiaadshacaGGPaaaleaacaWGMbGaam4BaiaadkhacaWGJbGaam yzaiaadohaaeaaa0GaeyyeIuoaaaa@50AF@

We define the total external work done on the system during a time interval t 0 t t 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaakiabgsMiJkaadshacqGHKjYOcaWG0bWaaSba aSqaaiaaigdaaeqaaaaa@3E22@  as the sum of the work done by the external forces. 

Δ W ext = forces t 0 t 1 F i ext (t) v(t)dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam 4vamaaCaaaleqabaGaamyzaiaadIhacaWG0baaaOGaeyypa0ZaaabC aeaadaWdXbqaaiaahAeadaqhaaWcbaGaamyAaaqaaiaadwgacaWG4b GaamiDaaaakiaacIcacaWG0bGaaiykaaWcbaGaamiDamaaBaaameaa caaIWaaabeaaaSqaaiaadshadaWgaaadbaGaaGymaaqabaaaniabgU IiYdaaleaacaWGMbGaam4BaiaadkhacaWGJbGaamyzaiaadohaaeaa a0GaeyyeIuoakiabgwSixlaahAhacaGGOaGaamiDaiaacMcacaWGKb GaamiDaaaa@5919@

The total work done can also include a contribution from external moments acting on the system, but we will worry about this when we analyze rigid body motion..

 

The system of particles is conservative if all the internal forces in the system are conservative.  This means that the particles must interact through conservative forces such as gravity, springs, electrostatic forces, and so on.   The particles can also be connected by rigid links, or touch one another, but contacts between particles must be frictionless.

 

If this is the case, we can define the total potential energy of the system V TOT MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGwbWaaWbaaS qabeaacaWGubGaam4taiaadsfaaaaaaa@3904@  as the sum of potential energies of all the internal forces.    We usually compute the total potential energy by summing up all the terms from the table in Sect 4.1.3.   Mathematically, the potential energy depends in some complicated way on the distances between all the particles, and the resultant force on the ith particle is related to the total potential energy by

ji R ij = V TOT r i MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaaeqbqaaiaahk fadaWgaaWcbaGaamyAaiaadQgaaeqaaaqaaiaadQgacqGHGjsUcaWG PbaabeqdcqGHris5aOGaeyypa0JaeyOeI0YaaSaaaeaacqGHciITca WGwbWaaWbaaSqabeaacaWGubGaam4taiaadsfaaaaakeaacqGHciIT caWHYbWaaSbaaSqaaiaadMgaaeqaaaaaaaa@489C@

 

We also define the total kinetic energy T TOT MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaCa aaleqabaGaamivaiaad+eacaWGubaaaaaa@3A8A@  of the system as the sum of kinetic energies of all the particles.

 

 

The work-energy relation for the system of particles can then be stated as follows.  Suppose that

  1. At some time t 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaaaaa@38DD@  the system has and kinetic energy T 0 TOT MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaDa aaleaacaaIWaaabaGaamivaiaad+eacaWGubaaaaaa@3B43@
  2. At some later time t 1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIXaaabeaaaaa@38DE@  the system has kinetic energy T 1 TOT MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaDa aaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaaaa@3B45@ .
  3. Let V 0 TOT MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaDa aaleaacaaIWaaabaGaamivaiaad+eacaWGubaaaaaa@3B46@  denote the potential energy of the force at time t 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaaaaa@38DD@
  4. Let V 1 TOT MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaDa aaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaaaa@3B47@  denote the potential energy of the force at time t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaaaa@37F7@
  5. Let Δ W ext MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam 4vamaaCaaaleqabaGaamyzaiaadIhacaWG0baaaaaa@3C4D@  denote the total work done on the system between t 0 t t 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaakiabgsMiJkaadshacqGHKjYOcaWG0bWaaSba aSqaaiaaigdaaeqaaaaa@3E22@

 

 

Power Energy Relation: This law states that the rate of external work done on the system is equal to the rate of change of total energy of the system

P ext = d dt ( T 1 TOT + V 1 TOT ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaCa aaleqabaGaamyzaiaadIhacaWG0baaaOGaeyypa0ZaaSaaaeaacaWG KbaabaGaamizaiaadshaaaWaaeWaaeaacaWGubWaa0baaSqaaiaaig daaeaacaWGubGaam4taiaadsfaaaGccqGHRaWkcaWGwbWaa0baaSqa aiaaigdaaeaacaWGubGaam4taiaadsfaaaaakiaawIcacaGLPaaaaa a@49D9@

 

Work Energy Relation: This law states that the external work done on the system is equal to the change in total kinetic and potential energy of the system.

Δ W ext = T 1 TOT + V 1 TOT ( T 0 TOT + V 0 TOT ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam 4vamaaCaaaleqabaGaamyzaiaadIhacaWG0baaaOGaeyypa0Jaamiv amaaDaaaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaOGaey4kaS IaamOvamaaDaaaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaOGa eyOeI0YaaeWaaeaacaWGubWaa0baaSqaaiaaicdaaeaacaWGubGaam 4taiaadsfaaaGccqGHRaWkcaWGwbWaa0baaSqaaiaaicdaaeaacaWG ubGaam4taiaadsfaaaaakiaawIcacaGLPaaaaaa@52DC@

 

Energy conservation law For the special case where no external forces act on the system, the total energy of the system is constant

Δ W ext =0 T 1 TOT + V 1 TOT =( T 0 TOT + V 0 TOT ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam 4vamaaCaaaleqabaGaamyzaiaadIhacaWG0baaaOGaeyypa0JaaGim aiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaeyO0H4TaamivamaaDa aaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaOGaey4kaSIaamOv amaaDaaaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaOGaeyypa0 ZaaeWaaeaacaWGubWaa0baaSqaaiaaicdaaeaacaWGubGaam4taiaa dsfaaaGccqGHRaWkcaWGwbWaa0baaSqaaiaaicdaaeaacaWGubGaam 4taiaadsfaaaaakiaawIcacaGLPaaaaaa@5DC3@

 

It is worth making one final remark before we turn to applications of these laws.  We often invoke the principle of conservation of energy when analyzing the motion of an object that is subjected to the earth’s gravitational field.   For example, the first problem we solve in the next section involves the motion of a projectile launched from the earth’s surface.  We usually glibly say that `the sum of the potential and kinetic energies of the particle are constant’ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  and if you’ve done physics courses you’ve probably used this kind of thinking.  It is not really correct, although it leads to a more or less correct solution.

 

Properly, we should consider the earth and the projectile together as a conservative system.  This means we must include the kinetic energy of the earth in the calculation, which changes by a small, but finite, amount due to gravitational interaction with the projectile.   Fortunately, the principle of conservation of linear momentum (to be covered later) can be used to show that the change in kinetic energy of the earth is negligibly small compared to that of the particle. 

 

 

Proof of the energy relation

 

Recall the power-kinetic energy relation for a single particle

Fv= dT dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHgbGaeyyXIC TaaCODaiabg2da9maalaaabaGaamizaiaadsfaaeaacaWGKbGaamiD aaaaaaa@3E48@

The total force on one particle consists of the external force, plus the sum of all the forces exerted on the particle by other particles.  The power-energy relation for the i th particle therefore becomes

( F i ext + ji R ij ) v i = d T i dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaGGOaGaaCOram aaDaaaleaacaWGPbaabaGaamyzaiaadIhacaWG0baaaOGaey4kaSYa aabuaeaacaWHsbWaaSbaaSqaaiaadMgacaWGQbaabeaaaeaacaWGQb GaeyiyIKRaamyAaaqab0GaeyyeIuoakiaacMcacqGHflY1caWH2bWa aSbaaSqaaiaadMgaaeqaaOGaeyypa0ZaaSaaaeaacaWGKbGaamivam aaBaaaleaacaWGPbaabeaaaOqaaiaadsgacaWG0baaaaaa@4F7A@

Since the internal forces are conservative, we can write

ji R ij = V TOT r i MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaaeqbqaaiaahk fadaWgaaWcbaGaamyAaiaadQgaaeqaaaqaaiaadQgacqGHGjsUcaWG PbaabeqdcqGHris5aOGaeyypa0JaeyOeI0YaaSaaaeaacqGHciITca WGwbWaaWbaaSqabeaacaWGubGaam4taiaadsfaaaaakeaacqGHciIT caWHYbWaaSbaaSqaaiaadMgaaeqaaaaaaaa@489C@

Furthermore, v i = r i /dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWH2bWaaSbaaS qaaiaadMgaaeqaaOGaeyypa0JaeyOaIyRaaCOCamaaBaaaleaacaWG Pbaabeaakiaac+cacaWGKbGaamiDaaaa@3EB9@  , so that

ji R ij v i = V TOT r i d r i dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaaeqbqaaiaahk fadaWgaaWcbaGaamyAaiaadQgaaeqaaOGaeyyXICTaaCODamaaBaaa leaacaWGPbaabeaaaeaacaWGQbGaeyiyIKRaamyAaaqab0GaeyyeIu oakiabg2da9iabgkHiTmaalaaabaGaeyOaIyRaamOvamaaCaaaleqa baGaamivaiaad+eacaWGubaaaaGcbaGaeyOaIyRaaCOCamaaBaaale aacaWGPbaabeaaaaGcdaWcaaqaaiaadsgacaWHYbWaaSbaaSqaaiaa dMgaaeqaaaGcbaGaamizaiaadshaaaaaaa@520D@

We can now sum this over all the particles

i F i ext v i + i V TOT r i d r i dt = i d T i dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaaeqbqaaaWcba GaamyAaaqab0GaeyyeIuoakiaahAeadaqhaaWcbaGaamyAaaqaaiaa dwgacaWG4bGaamiDaaaakiabgwSixlaahAhadaWgaaWcbaGaamyAaa qabaGccqGHRaWkdaaeqbqaaiabgkHiTmaalaaabaGaeyOaIyRaamOv amaaCaaaleqabaGaamivaiaad+eacaWGubaaaaGcbaGaeyOaIyRaaC OCamaaBaaaleaacaWGPbaabeaaaaGcdaWcaaqaaiaadsgacaWHYbWa aSbaaSqaaiaadMgaaeqaaaGcbaGaamizaiaadshaaaaaleaacaWGPb aabeqdcqGHris5aOGaeyypa0ZaaabuaeaadaWcaaqaaiaadsgacaWG ubWaaSbaaSqaaiaadMgaaeqaaaGcbaGaamizaiaadshaaaaaleaaca WGPbaabeqdcqGHris5aaaa@5D38@

We know that V TOT MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGwbWaaWbaaS qabeaacaWGubGaam4taiaadsfaaaaaaa@3904@  depends only on the positions of the particles, so the second term on the left hand side is a total differential

i V TOT r i d r i dt = d V TOT dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaaeqbqaaiabgk HiTmaalaaabaGaeyOaIyRaamOvamaaCaaaleqabaGaamivaiaad+ea caWGubaaaaGcbaGaeyOaIyRaaCOCamaaBaaaleaacaWGPbaabeaaaa GcdaWcaaqaaiaadsgacaWHYbWaaSbaaSqaaiaadMgaaeqaaaGcbaGa amizaiaadshaaaaaleaacaWGPbaabeqdcqGHris5aOGaeyypa0Jaey OeI0YaaSaaaeaacaWGKbGaamOvamaaCaaaleqabaGaamivaiaad+ea caWGubaaaaGcbaGaamizaiaadshaaaaaaa@4F71@

(by the chain rule).   We also recognize the first term as the total external power, and the right hand side as the time derivative of the total KE, so we see that

P ext d V TOT dt = d T TOT dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGqbWaaWbaaS qabeaacaWGLbGaamiEaiaadshaaaGccqGHsisldaWcaaqaaiaadsga caWGwbWaaWbaaSqabeaacaWGubGaam4taiaadsfaaaaakeaacaWGKb GaamiDaaaacqGH9aqpdaWcaaqaaiaadsgacaWGubWaaWbaaSqabeaa caWGubGaam4taiaadsfaaaaakeaacaWGKbGaamiDaaaaaaa@4839@

Rearranging this gives the power-KE relation for the system, and integrating it with respect to time gives the work-energy relation.

 

 

 

4.1.8 Examples of calculations using kinetic and potential energy in conservative systems

 

The kinetic-potential energy relations can be used to quickly calculate relationships between the velocity and position of an object.  Several examples are provided below.

Example 1: (Boring FE exam question) A projectile with mass m is launched from the ground with velocity V 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa aaleaacaaIWaaabeaaaaa@38BF@  at angle α MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySdegaaa@389D@ .  Calculate an expression for the maximum height reached by the projectile.

 

If air resistance can be neglected, we can regard the earth and the projectile together as a conservative system. We neglect the change in the earth’s kinetic energy. In addition, since the gravitational force acting on the particle is vertical, the particle’s horizontal component of velocity must be constant.   

 

Calculation:

  1. Just after launch, the velocity of the particle is v= V 0 cosαi+ V 0 sinαj MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9iaadAfadaWgaaWcbaGaaGimaaqabaGcciGGJbGaai4Baiaacoha cqaHXoqycaWHPbGaey4kaSIaamOvamaaBaaaleaacaaIWaaabeaaki GacohacaGGPbGaaiOBaiabeg7aHjaahQgaaaa@4849@
  2. The kinetic energy of the particle just after launch is m V 0 2 /2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiaadA fadaqhaaWcbaGaaGimaaqaaiaaikdaaaGccaGGVaGaaGOmaaaa@3BE7@ . Its potential energy is zero.
  3. At the peak of the trajectory the vertical velocity is zero. Since the horizontal velocity remains constant, the velocity vector at the peak of the trajectory is v= V 0 cosαi MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9iaadAfadaWgaaWcbaGaaGimaaqabaGcciGGJbGaai4Baiaacoha cqaHXoqycaWHPbaaaa@4032@ . The kinetic energy at this point is therefore m V 0 2 cos 2 α/2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiaadA fadaqhaaWcbaGaaGimaaqaaiaaikdaaaGcciGGJbGaai4Baiaacoha daahaaWcbeqaaiaaikdaaaGccqaHXoqycaGGVaGaaGOmaaaa@414C@
  4. Energy is conserved, so

m V 0 2 /2=mgh+m V 0 2 cos 2 α/2 h= V 0 2 (1 cos 2 α)/2= V 0 2 sin 2 α MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGTb GaamOvamaaDaaaleaacaaIWaaabaGaaGOmaaaakiaac+cacaaIYaGa eyypa0JaamyBaiaadEgacaWGObGaey4kaSIaamyBaiaadAfadaqhaa WcbaGaaGimaaqaaiaaikdaaaGcciGGJbGaai4BaiaacohadaahaaWc beqaaiaaikdaaaGccqaHXoqycaGGVaGaaGOmaaqaaiabgkDiElaadI gacqGH9aqpcaWGwbWaa0baaSqaaiaaicdaaeaacaaIYaaaaOGaaiik aiaaigdacqGHsislciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaik daaaGccqaHXoqycaGGPaGaai4laiaaikdacqGH9aqpcaWGwbWaa0ba aSqaaiaaicdaaeaacaaIYaaaaOGaci4CaiaacMgacaGGUbWaaWbaaS qabeaacaaIYaaaaOGaeqySdegaaaa@6494@

 

 

Example 2: You are asked to design the packaging for a sensitive instrument.  The packaging will be made from an elastic foam, which behaves like a spring. The specifications restrict the maximum acceleration of the instrument to 15g.  Estimate the thickness of the packaging that you must use.

 

This problem can be solved by noting that (i) the max acceleration occurs when the packaging (spring) is fully compressed and so exerts the maximum force on the instrument; (ii) The velocity of the instrument must be zero at this instant, (because the height is a minimum, and the velocity is the derivative of the height); and (iii) The system is conservative, and has zero kinetic energy when the package is dropped, and zero kinetic energy when the spring is fully compressed.

 

 

Assumptions:

  1. The package is dropped from a height of 1.5m
  2. The effects of air resistance during the fall are neglected
  3. The foam is idealized as a linear spring, which can be fully compressed.

 

Calculations: Let h denote the drop height; let d denote the foam thickness.

  1. The potential energy of the system just before the package is dropped is mgh
  2. The potential energy of the system at the instant when the foam is compressed to its maximum extent is 1 2 k d 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca aIXaaabaGaaGOmaaaacaWGRbGaamizamaaCaaaleqabaGaaGOmaaaa aaa@3B47@
  3. The total energy of the system is constant, so

1 2 k d 2 =mgh MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca aIXaaabaGaaGOmaaaacaWGRbGaamizamaaCaaaleqabaGaaGOmaaaa kiabg2da9iaad2gacaWGNbGaamiAaaaa@3F22@

  1. The figure shows a free body diagram for the instrument at the instant of maximum foam compression.  The resultant force acting on the instrument is (kdmg)j MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadU gacaWGKbGaeyOeI0IaamyBaiaadEgacaGGPaGaaCOAaaaa@3DEE@ , so its acceleration follows as a=(kd/mg)j MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyyaiabg2 da9iaacIcacaWGRbGaamizaiaac+cacaWGTbGaeyOeI0Iaam4zaiaa cMcacaWHQbaaaa@4091@ .  The acceleration must not exceed 15g, so

kd/m16g MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Aaiaads gacaGGVaGaamyBaiabgsMiJkaaigdacaaI2aGaam4zaaaa@3E98@

  1. Dividing (3) by (4) shows that

dh/8 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiabgw MiZkaadIgacaGGVaGaaGioaaaa@3C0F@

The thickness of the protective foam must therefore exceed 18.8cm.

 

Example 3: The Charpy Impact Test is a way to measure the work of fracture of a material (i.e. the work per unit area required to separate a material into two pieces).  An example (from www.qualitest-inc.com/qpi.htm) is shown in the picture.  You can see one in Prince Lab if you are curious.

 

It consists of a pendulum, which swings down from a prescribed initial angle to strike a specimen.  The pendulum fractures the specimen, and then continues to swing to a new, smaller angle on the other side of the vertical.   The scale on the pendulum allows the initial and final angles to be measured.  The goal of this example is to deduce a relationship between the angles and the work of fracture of the specimen.

 

The figure shows the pendulum before and after it hits the specimen.  

  1. The potential energy of the mass before it is released is V 1 =mglcos α 1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa aaleaacaaIXaaabeaakiabg2da9iabgkHiTiaad2gacaWGNbGaamiB aiGacogacaGGVbGaai4Caiabeg7aHnaaBaaaleaacaaIXaaabeaaaa a@42E5@ .  Its kinetic energy is zero.
  2. The potential energy of the mass when it comes to rest after striking the specimen is V 2 =mglcos α 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa aaleaacaaIYaaabeaakiabg2da9iabgkHiTiaad2gacaWGNbGaamiB aiGacogacaGGVbGaai4Caiabeg7aHnaaBaaaleaacaaIYaaabeaaaa a@42E7@ . The kinetic energy is again zero.

 

The work of fracture is equal to the change in potential energy -

W F = V 1 V 2 =mgl( cos α 2 cos α 1 ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4vamaaBa aaleaacaWGgbaabeaakiabg2da9iaadAfadaWgaaWcbaGaaGymaaqa baGccqGHsislcaWGwbWaaSbaaSqaaiaaikdaaeqaaOGaeyypa0Jaam yBaiaadEgacaWGSbWaaeWaaeaaciGGJbGaai4BaiaacohacqaHXoqy daWgaaWcbaGaaGOmaaqabaGccqGHsislciGGJbGaai4Baiaacohacq aHXoqydaWgaaWcbaGaaGymaaqabaaakiaawIcacaGLPaaaaaa@4F79@

 

 

 

 

 

 

 

Example 4: Estimate the maximum distance that a long-bow can fire an arrow.

 

We can do this calculation by idealizing the bow as a spring, and estimating the maximum force that a person could apply to draw the bow.   The energy stored in the bow can then be estimated, and energy conservation can be used to estimate the resulting velocity of the arrow.  

 

Assumptions

  1. The long-bow will be idealized as a linear spring
  2. The maximum draw force is likely to be around 60lbf (270N)
  3. The draw length is about 2ft (0.6m)
  4. Arrows come with various masses MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  typical range is between 250-600 grains (16-38 grams)
  5. We will neglect the mass of the bow (this is not a very realistic assumption)

 

Calculation: The calculation needs two steps: (i) we start by calculating the velocity of the arrow just after it is fired. This will be done using the energy conservation law; and (ii) we then calculate the distance traveled by the arrow using the projectile trajectory equations derived in the preceding chapter.

  1. Just before the arrow is released, the spring is stretched to its maximum length, and the arrow is stationary.  The total energy of the system is T 0 + V 0 = 1 2 k L 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBa aaleaacaaIWaaabeaakiabgUcaRiaadAfadaWgaaWcbaGaaGimaaqa baGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaadUgacaWGmb WaaWbaaSqabeaacaaIYaaaaaaa@40AB@ , where L is the draw length and k is the stiffness of the bow.
  2. We can estimate values for the spring stiffness using the draw force: we have that F D =kL MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa aaleaacaWGebaabeaakiabg2da9iaadUgacaWGmbaaaa@3B8F@ , so k= F D /L MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Aaiabg2 da9iaadAeadaWgaaWcbaGaamiraaqabaGccaGGVaGaamitaaaa@3C42@ Thus T 0 + V 0 = 1 2 L F D MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBa aaleaacaaIWaaabeaakiabgUcaRiaadAfadaWgaaWcbaGaaGimaaqa baGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaadYeacaWGgb WaaSbaaSqaaiaadseaaeqaaaaa@4092@ .
  3. Just after the arrow is fired, the spring returns to its un-stretched length, and the arrow has velocity V. The total energy of the system is T 1 + V 1 = 1 2 m V 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBa aaleaacaaIXaaabeaakiabgUcaRiaadAfadaWgaaWcbaGaaGymaaqa baGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaad2gacaWGwb WaaWbaaSqabeaacaaIYaaaaaaa@40B9@ , where m is the mass of the arrow
  4. The system is conservative, therefore T 0 + V 0 = T 1 + V 1 1 2 L F D = 1 2 m V 2 V= L F D /m MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBa aaleaacaaIWaaabeaakiabgUcaRiaadAfadaWgaaWcbaGaaGimaaqa baGccqGH9aqpcaWGubWaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaam OvamaaBaaaleaacaaIXaaabeaakiabgkDiEpaalaaabaGaaGymaaqa aiaaikdaaaGaamitaiaadAeadaWgaaWcbaGaamiraaqabaGccqGH9a qpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaad2gacaWGwbWaaWbaaSqa beaacaaIYaaaaOGaeyO0H4TaamOvaiabg2da9maakaaabaGaamitai aadAeadaWgaaWcbaGaamiraaqabaGccaGGVaGaamyBaaWcbeaaaaa@5557@
  5. We suppose that the arrow is launched from the origin at an angle θ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUdehaaa@38B4@  to the horizontal. The horizontal and vertical components of velocity are V x =Vcosθ V y =Vsinθ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa aaleaacaWG4baabeaakiabg2da9iaadAfaciGGJbGaai4Baiaacoha cqaH4oqCcaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaadAfadaWgaaWcbaGaamyEaaqabaGccqGH9aqpcaWGwbGa ci4CaiaacMgacaGGUbGaeqiUdehaaa@544C@ . The position vector of the arrow can be calculated using the method outlined in Section 3.2.2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  the result is

r=( Vtcosθ )i+( Vtsinθ 1 2 g t 2 )j MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabg2 da9maabmaabaGaamOvaiaadshaciGGJbGaai4BaiaacohacqaH4oqC aiaawIcacaGLPaaacaWHPbGaey4kaSYaaeWaaeaacaWGwbGaamiDai GacohacaGGPbGaaiOBaiabeI7aXjabgkHiTmaalaaabaGaaGymaaqa aiaaikdaaaGaam4zaiaadshadaahaaWcbeqaaiaaikdaaaaakiaawI cacaGLPaaacaWHQbaaaa@50E2@

We can calculate the distance traveled by noting that its position vector when it lands is di.  This gives  

( Vtcosθ )i+( Vtsinθ 1 2 g t 2 )j=di MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca WGwbGaamiDaiGacogacaGGVbGaai4CaiabeI7aXbGaayjkaiaawMca aiaahMgacqGHRaWkdaqadaqaaiaadAfacaWG0bGaci4CaiaacMgaca GGUbGaeqiUdeNaeyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaaaacaWG NbGaamiDamaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaiaahQ gacqGH9aqpcaWGKbGaaCyAaaaa@51C2@

where t is the time of flight.  The i and j components of this equation can be solved for t and d, with the result

t= 2Vsinθ g d= 2Vsinθ gcosθ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 da9maalaaabaGaaGOmaiaadAfaciGGZbGaaiyAaiaac6gacqaH4oqC aeaacaWGNbaaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8Uaamizaiabg2da9maalaaabaGaaGOmaiaadA faciGGZbGaaiyAaiaac6gacqaH4oqCaeaacaWGNbGaci4yaiaac+ga caGGZbGaeqiUdehaaaaa@67F1@

The arrow travels furthest when fired at an angle that maximizes sinθ/cosθ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4CaiaacM gacaGGUbGaeqiUdeNaai4laiGacogacaGGVbGaai4CaiabeI7aXbaa @40C8@  - i.e. 45 degrees.  The distance follows as

d= 2 V 2 g = 2L F D mg MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiabg2 da9maalaaabaGaaGOmaiaadAfadaahaaWcbeqaaiaaikdaaaaakeaa caWGNbaaaiabg2da9maalaaabaGaaGOmaiaadYeacaWGgbWaaSbaaS qaaiaadseaaeqaaaGcbaGaamyBaiaadEgaaaaaaa@42BE@

 

  1. Substituting numbers gives 2064m for a 250 grain arrow MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  over a mile!  Of course air resistance will reduce this value, and in practice the kinetic energy associated with the motion of the bow and bowstring (neglected here) will reduce the distance.

 

 

 

Example 5: Find a formula for the escape velocity of a space vehicle as a function of altitude above the earths surface.

 

The term ‘Escape velocity’ means that the space vehicle has a large enough velocity to completely escape the earth’s gravitational field MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  i.e. the space vehicle will never stop after being launched.

 

Assumptions

  1. The space vehicle is initially in orbit at an altitude h above the earth’s surface
  2. The earth’s radius is 6378.145km
  3. While in orbit, a rocket is burned on the vehicle to increase its speed to v (the escape velocity), placing it on a hyperbolic trajectory that will eventually escape the earth’s gravitational field.
  4. The Gravitational parameter μ=GM=3.986012× 10 5 km 3 s 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH8oqBcqGH9aqpcaWGhbGaamytai abg2da9iaaiodacaGGUaGaaGyoaiaaiIdacaaI2aGaaGimaiaaigda caaIYaGaaGPaVlaaykW7caaMc8UaaGPaVlabgEna0kaaykW7caaMc8 UaaGymaiaaicdadaahaaWcbeqaaiaaiwdaaaGccaaMc8UaaGPaVlaa ykW7caqGRbGaaeyBamaaCaaaleqabaGaaG4maaaakiaabohadaahaa WcbeqaaiabgkHiTiaaigdaaaaaaa@560C@   (G= gravitational constant; M=mass of earth)
  5.  

Calculation

  1. Just after the rocket is burned, the potential energy of the system is V=GMm/(R+h) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGwbGaeyypa0JaeyOeI0Iaam4rai aad2eacaWGTbGaai4laiaacIcacaWGsbGaey4kaSIaamiAaiaacMca aaa@3CD5@ , while its kinetic energy is T=m v 2 /2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGubGaeyypa0JaamyBaiaadAhada ahaaWcbeqaaiaaikdaaaGccaGGVaGaaGOmaaaa@38F3@
  2. When it escapes the earth’s gravitational field (at an infinite height above the earth’s surface) the potential energy is zero.  At the critical escape velocity, the velocity of the spacecraft at this point drops to zero.  The total energy at escape is therefore zero.
  3. This is a conservative system, so

T+V=m v 2 /2GMm/(R+h)=0 v= 2GM/(R+h) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiaadsfacqGHRaWkcaWGwbGaey ypa0JaamyBaiaadAhadaahaaWcbeqaaiaaikdaaaGccaGGVaGaaGOm aiabgkHiTiaadEeacaWGnbGaamyBaiaac+cacaGGOaGaamOuaiabgU caRiaadIgacaGGPaGaeyypa0JaaGimaaqaaiabgkDiElaadAhacqGH 9aqpdaGcaaqaaiaaikdacaWGhbGaamytaiaac+cacaGGOaGaamOuai abgUcaRiaadIgacaGGPaaaleqaaaaaaa@502B@

  1. A typical low earth orbit has altitude of 250km.  For this altitude the escape velocity is 10.9km/sec.

 

 

 

 

 

 

4.2 Linear impulse-momentum relations

 

 

 

4.2.1 Definition of the linear impulse of a force

In most dynamic problems, particles are subjected to forces that vary with time.  We can write this mathematically by saying that the force is a vector valued function of time F(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiaacI cacaWG0bGaaiykaaaa@3917@ .  If we express the force as components in a fixed basis { i,j,k } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaaca WHPbGaaiilaiaahQgacaGGSaGaaC4AaaGaay5Eaiaaw2haaaaa@3C60@ , then

F(t)= F x (t)i+ F y (t)j+ F z (t)k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiaacI cacaWG0bGaaiykaiabg2da9iaadAeadaWgaaWcbaGaamiEaaqabaGc caGGOaGaamiDaiaacMcacaWHPbGaey4kaSIaamOramaaBaaaleaaca WG5baabeaakiaacIcacaWG0bGaaiykaiaahQgacqGHRaWkcaWGgbWa aSbaaSqaaiaadQhaaeqaaOGaaiikaiaadshacaGGPaGaaC4Aaaaa@4BAD@

where each component of the force is a function of time.

 

The Linear Impulse exerted by a force during a time interval t 0 t t 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaakiabgsMiJkaadshacqGHKjYOcaWG0bWaaSba aSqaaiaaigdaaeqaaaaa@3E22@  is defined as

= t 0 t 1 F(t)dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyeHeSaey ypa0Zaa8qCaeaacaWHgbGaaiikaiaadshacaGGPaGaamizaiaadsha aSqaaiaadshadaWgaaadbaGaaGimaaqabaaaleaacaWG0bWaaSbaaW qaaiaaigdaaeqaaaqdcqGHRiI8aaaa@43B3@

The linear impulse is a vector, and can be expressed as components in a basis

= x i+ y j+ z k x = t 0 t 1 F x (t)dt y = t 0 t 1 F y (t)dt z = t 0 t 1 F z (t)dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHre sWcqGH9aqpcqGHresWdaWgaaWcbaGaamiEaaqabaGccaWHPbGaey4k aSIaeyyeHe8aaSbaaSqaaiaadMhaaeqaaOGaaCOAaiabgUcaRiabgg ribpaaBaaaleaacaWG6baabeaakiaahUgaaeaacqGHresWdaWgaaWc baGaamiEaaqabaGccqGH9aqpdaWdXbqaaiaadAeadaWgaaWcbaGaam iEaaqabaGccaGGOaGaamiDaiaacMcacaWGKbGaamiDaaWcbaGaamiD amaaBaaameaacaaIWaaabeaaaSqaaiaadshadaWgaaadbaGaaGymaa qabaaaniabgUIiYdGccaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaeyyeHe8aaSbaaSqaaiaadM haaeqaaOGaeyypa0Zaa8qCaeaacaWGgbWaaSbaaSqaaiaadMhaaeqa aOGaaiikaiaadshacaGGPaGaamizaiaadshaaSqaaiaadshadaWgaa adbaGaaGimaaqabaaaleaacaWG0bWaaSbaaWqaaiaaigdaaeqaaaqd cqGHRiI8aOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7cqGHresWdaWgaaWcbaGaamOE aaqabaGccqGH9aqpdaWdXbqaaiaadAeadaWgaaWcbaGaamOEaaqaba GccaGGOaGaamiDaiaacMcacaWGKbGaamiDaaWcbaGaamiDamaaBaaa meaacaaIWaaabeaaaSqaaiaadshadaWgaaadbaGaaGymaaqabaaani abgUIiYdaaaaa@9616@

 

If you know the force as a function of time, you can calculate its impulse using simple calculus.  For example:

1.      For a constant force, with vector value F 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOramaaBa aaleaacaaIWaaabeaaaaa@37AB@ , the impulse is = F 0 Δt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyeHeSaey ypa0JaaCOramaaBaaaleaacaaIWaaabeaakiabfs5aejaadshaaaa@3C99@

2.      For a harmonic force of the form F(t)= F 0 sinωt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiaacI cacaWG0bGaaiykaiabg2da9iaahAeadaWgaaWcbaGaaGimaaqabaGc ciGGZbGaaiyAaiaac6gacqaHjpWDcaWG0baaaa@417A@  the impulse is

= F 0 [ cos(ω( t 0 +Δt))cosω t 0 ]/ω MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyeHeSaey ypa0JaeyOeI0IaaCOramaaBaaaleaacaaIWaaabeaakmaadmaabaGa ci4yaiaac+gacaGGZbGaaiikaiabeM8a3jaacIcacaWG0bWaaSbaaS qaaiaaicdaaeqaaOGaey4kaSIaeuiLdqKaamiDaiaacMcacaGGPaGa eyOeI0Iaci4yaiaac+gacaGGZbGaeqyYdCNaamiDamaaBaaaleaaca aIWaaabeaaaOGaay5waiaaw2faaiaac+cacqaHjpWDaaa@538B@

 

It is rather rare in practice to have to calculate the impulse of a force from its time variation.

 

 

4.2.2 Definition of the linear momentum of a particle

 

The linear momentum of a particle is simply the product of its mass and velocity

p=mv MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiCaiabg2 da9iaad2gacaWH2baaaa@39E6@

The linear momentum is a vector MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  its direction is parallel to the velocity of the particle.

 

 

4.2.3 Impulse-momentum relations for a single particle

 

* Consider a particle that is subjected to a force F(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiaacI cacaWG0bGaaiykaaaa@3917@  for a time interval t 0 t t 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaakiabgsMiJkaadshacqGHKjYOcaWG0bWaaSba aSqaaiaaigdaaeqaaaaa@3E22@ .

* Let MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyeHemaaa@3775@  denote the impulse exerted by F on the particle

* Let p 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiCamaaBa aaleaacaaIWaaabeaaaaa@37D5@  denote the linear momentum of the particle at time t 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG0bWaaSbaaS qaaiaaicdaaeqaaaaa@3755@  .

* Let p 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiCamaaBa aaleaacaaIXaaabeaaaaa@37D6@  denote the linear momentum of the particle at time t 1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG0bWaaSbaaS qaaiaaigdaaeqaaaaa@3756@  .

 

The impulse-momentum equation can be expressed either in differential or integral form. 

1.      In differential form

F(t)= dp dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiaacI cacaWG0bGaaiykaiabg2da9maalaaabaGaamizaiaahchaaeaacaWG KbGaamiDaaaaaaa@3DF1@

This is clearly just a different way of writing Newton’s law F=ma.

 

2.      In integral form

=Δp MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyeHeSaey ypa0JaeuiLdqKaaCiCaaaa@3ADA@

This is the integral of Newton’s law of motion with respect to time.

 

The impulse-momentum relations for a single particle are useful if you need to calculate the change in velocity of an object that is subjected to a prescribed force.

 

 

4.2.4 Examples using impulse-momentum relations for a single particle

 

Example 1: Estimate the time that it takes for a Ferrari Testarossa traveling at the RI speed limit to make an emergency stop. (Like many textbook problems this one is totally unrealistic MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  nobody in a Ferrari is going to travel at the speed limit!)

 

We can do this calculation using the impulse-momentum relation for a single particle.  Assume that the car has mass m, and travels with speed V before the brakes are applied.  Let Δt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam iDaaaa@3855@  denote the time required to stop.

1.      Start by estimating the force acting on the car during an emergency stop.  The figure shows a free body diagram for the car as it brakes to a standstill.   We assume that the driver brakes hard enough to lock the wheels, so that the car skids over the road.  The horizontal friction forces must oppose the sliding, as shown in the picture.  F=ma for the car follows as

( T R + T F )i+( N F + N R mg)j=m a x i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca WGubWaaSbaaSqaaiaadkfaaeqaaOGaey4kaSIaamivamaaBaaaleaa caWGgbaabeaaaOGaayjkaiaawMcaaiaahMgacqGHRaWkcaGGOaGaam OtamaaBaaaleaacaWGgbaabeaakiabgUcaRiaad6eadaWgaaWcbaGa amOuaaqabaGccqGHsislcaWGTbGaam4zaiaacMcacaWHQbGaeyypa0 JaamyBaiaadggadaWgaaWcbaGaamiEaaqabaGccaWHPbaaaa@4CA5@

The vertical component of the equation of motion shows that  N F + N R =mg MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtamaaBa aaleaacaWGgbaabeaakiabgUcaRiaad6eadaWgaaWcbaGaamOuaaqa baGccqGH9aqpcaWGTbGaam4zaaaa@3D70@ .  The friction law then shows that T F + T R =μ( N F + N R )=μmg MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBa aaleaacaWGgbaabeaakiabgUcaRiaadsfadaWgaaWcbaGaamOuaaqa baGccqGH9aqpcqaH8oqBdaqadaqaaiaad6eadaWgaaWcbaGaamOraa qabaGccqGHRaWkcaWGobWaaSbaaSqaaiaadkfaaeqaaaGccaGLOaGa ayzkaaGaeyypa0JaeqiVd0MaamyBaiaadEgaaaa@480D@

2.      The force acting on the car is constant, so the impulse that brings the car to a halt is

=( T R + T F )Δti+( N F + N R mg)Δtj=μmgi MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyeHeSaey ypa0ZaaeWaaeaacaWGubWaaSbaaSqaaiaadkfaaeqaaOGaey4kaSIa amivamaaBaaaleaacaWGgbaabeaaaOGaayjkaiaawMcaaiabfs5aej aadshacaWHPbGaey4kaSIaaiikaiaad6eadaWgaaWcbaGaamOraaqa baGccqGHRaWkcaWGobWaaSbaaSqaaiaadkfaaeqaaOGaeyOeI0Iaam yBaiaadEgacaGGPaGaeuiLdqKaamiDaiaahQgacqGH9aqpcqaH8oqB caWGTbGaam4zaiaahMgaaaa@5471@

3.      The linear momentum of the car before the brakes are applied is p 0 =mVi MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiCamaaBa aaleaacaaIWaaabeaakiabg2da9iabgkHiTiaad2gacaWGwbGaaCyA aaaa@3C91@ .   The linear momentum after the car stops is zero.  Therefore, Δp=mVi MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaaC iCaiabg2da9iaad2gacaWGwbGaaCyAaaaa@3C1A@ .

4.      The linear impulse-momentum relation shows that

=ΔpμmgΔti=mVi Δt=V/(μg) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHre sWcqGH9aqpcqqHuoarcaWHWbGaeyO0H4TaeqiVd0MaamyBaiaadEga cqqHuoarcaWG0bGaaCyAaiabg2da9iaad2gacaWGwbGaaCyAaaqaai abgkDiElabfs5aejaadshacqGH9aqpcaWGwbGaai4laiaacIcacqaH 8oqBcaWGNbGaaiykaaaaaa@5333@

5.      We can take the friction coefficient as μ0.8 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiVd0Maey isISRaaGimaiaac6cacaaI4aaaaa@3B8B@ , and 65mph is 29m/s. We take the gravitational acceleration g=9.81 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zaiabg2 da9iaaiMdacaGGUaGaaGioaiaaigdaaaa@3ADA@ . The time follows as Δt3.7s MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam iDaiabgIKi7kaaiodacaGGUaGaaG4naiaabohaaaa@3D2C@ .   Note that a TestaRossa can’t stop any faster than a Honda Civic, despite the price difference…

 

 

4.2.5 Impulse-momentum relation for a system of particles

Suppose we are interested in calculating the motion of several particles, sketched in the figure. 

 

Total external impulse on a system of particles: Each particle in the system can experience forces applied by:

 * Other particles in the system (e.g. due to gravity, electric charges on the particles, or because the particles are physically connected through springs, or because the particles collide).  We call these internal forces acting in the system.  We will denote the internal force exerted by the ith particle on the jth particle by  R ij MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOuamaaBa aaleaacaWGPbGaamOAaaqabaaaaa@39E1@ .  Note that, because every action has an equal and opposite reaction, the force exerted on the jth particle by the ith particle must be equal and opposite, to R ij MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOuamaaBa aaleaacaWGPbGaamOAaaqabaaaaa@39E1@ , i.e. R ij = R ji MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOuamaaBa aaleaacaWGPbGaamOAaaqabaGccqGH9aqpcqGHsislcaWHsbWaaSba aSqaaiaadQgacaWGPbaabeaaaaa@3EC2@ .

 

* Forces exerted on the particles by the outside world (e.g. by externally applied gravitational or electromagnetic fields, or because the particles are connected to the outside world through mechanical linkages or springs).  We call these external forces acting on the system, and we will denote the external force on the i th particle by F i ext (t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOramaaDa aaleaacaWGPbaabaGaamyzaiaadIhacaWG0baaaOGaaiikaiaadsha caGGPaaaaa@3D1C@

 

We define the total impulse exerted on the system during a time interval t 0 t t 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaakiabgsMiJkaadshacqGHKjYOcaWG0bWaaSba aSqaaiaaigdaaeqaaaaa@3E22@  as the sum of all the impulses on all the particles.  It’s easy to see that the total impulse due to the internal forces is zero MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  because the ith and jth particles must exert equal and opposite impulses on one another, and when you add them up they cancel out.   So the total impulse on the system is simply

TOT = particles t 0 t 1 F i ext (t) dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyeHe8aaW baaSqabeaacaWGubGaam4taiaadsfaaaGccqGH9aqpdaaeWbqaamaa pehabaGaaCOramaaDaaaleaacaWGPbaabaGaamyzaiaadIhacaWG0b aaaOGaaiikaiaadshacaGGPaaaleaacaWG0bWaaSbaaWqaaiaaicda aeqaaaWcbaGaamiDamaaBaaameaacaaIXaaabeaaa0Gaey4kIipaaS qaaiaadchacaWGHbGaamOCaiaadshacaWGPbGaam4yaiaadYgacaWG LbGaam4CaaqaaaqdcqGHris5aOGaamizaiaadshaaaa@5535@

 

Total linear momentum of a system of particles: The total linear momentum of a system of particles is simply the sum of the momenta of all the particles, i.e.

 

p TOT = particles m i v i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiCamaaCa aaleqabaGaamivaiaad+eacaWGubaaaOGaeyypa0ZaaabuaeaacaWG TbWaaSbaaSqaaiaadMgaaeqaaOGaaCODamaaBaaaleaacaWGPbaabe aaaeaacaWGWbGaamyyaiaadkhacaWG0bGaamyAaiaadogacaWGSbGa amyzaiaadohaaeqaniabggHiLdaaaa@496D@

 

 

 

 

 

 

 

The impulse-momentum equation

 

1.      In differential form

particles F i ext (t)= d p TOT dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaabuaeaaca WHgbWaa0baaSqaaiaadMgaaeaacaWGLbGaamiEaiaadshaaaaabaGa amiCaiaadggacaWGYbGaamiDaiaadMgacaWGJbGaamiBaiaadwgaca WGZbaabeqdcqGHris5aOGaaiikaiaadshacaGGPaGaeyypa0ZaaSaa aeaacaWGKbGaaCiCamaaCaaaleqabaGaamivaiaad+eacaWGubaaaa GcbaGaamizaiaadshaaaaaaa@4F3F@

 

2.      In integral form

TOT = p 1 TOT p 0 TOT MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyeHe8aaW baaSqabeaacaWGubGaam4taiaadsfaaaGccqGH9aqpcaWHWbWaa0ba aSqaaiaaigdaaeaacaWGubGaam4taiaadsfaaaGccqGHsislcaWHWb Waa0baaSqaaiaaicdaaeaacaWGubGaam4taiaadsfaaaaaaa@44FC@

This is the integral of Newton’s law of motion with respect to time.

 

Conservation of momentum: If no external forces act on a system of particles, their total linear momentum is conserved, i.e.  

p 1 TOT = p 0 TOT MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiCamaaDa aaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaOGaeyypa0JaaCiC amaaDaaaleaacaaIWaaabaGaamivaiaad+eacaWGubaaaaaa@3FD3@

 

 

 

 

Deriving the impulse-momentum equation

 

We start with the impulse-momentum relation for a single particle in differential form

F i ext + ji R ij = d p i dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHgbWaa0baaS qaaiaadMgaaeaacaWGLbGaamiEaiaadshaaaGccqGHRaWkdaaeqbqa aiaahkfadaWgaaWcbaGaamyAaiaadQgaaeqaaaqaaiaadQgacqGHGj sUcaWGPbaabeqdcqGHris5aOGaeyypa0ZaaSaaaeaacaWGKbGaaCiC amaaBaaaleaacaWGPbaabeaaaOqaaiaadsgacaWG0baaaaaa@49D4@

(The left hand side is the total force on the ith particle MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  it includes the external force, as well as all the forces exerted by all the other particles).

 

Now sum over all the particles

i ( F i ext + ji R ij ) = i d p i dt MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaaeqbqaamaabm aabaGaaCOramaaDaaaleaacaWGPbaabaGaamyzaiaadIhacaWG0baa aOGaey4kaSYaaabuaeaacaWHsbWaaSbaaSqaaiaadMgacaWGQbaabe aaaeaacaWGQbGaeyiyIKRaamyAaaqab0GaeyyeIuoaaOGaayjkaiaa wMcaaaWcbaGaamyAaaqab0GaeyyeIuoakiabg2da9maaqafabaWaaS aaaeaacaWGKbGaaCiCamaaBaaaleaacaWGPbaabeaaaOqaaiaadsga caWG0baaaaWcbaGaamyAaaqab0GaeyyeIuoaaaa@5189@

 

Notice that since R ij = R ji MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWHsbWaaSbaaS qaaiaadMgacaWGQbaabeaakiabg2da9iabgkHiTiaahkfadaWgaaWc baGaamOAaiaadMgaaeqaaaaa@3D3B@  

i ji R ij =0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaaeqbqaamaaqa fabaGaaCOuamaaBaaaleaacaWGPbGaamOAaaqabaaabaGaamOAaiab gcMi5kaadMgaaeqaniabggHiLdaaleaacaWGPbaabeqdcqGHris5aO Gaeyypa0JaaCimaaaa@42F0@

(To see this just write out the full sum MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  every internal force on one particle is equal and opposite to the force on another, so the total has to cancel)

 

Therefore, evaluating the sums

particles F i ext (t)= d p TOT dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaabuaeaaca WHgbWaa0baaSqaaiaadMgaaeaacaWGLbGaamiEaiaadshaaaaabaGa amiCaiaadggacaWGYbGaamiDaiaadMgacaWGJbGaamiBaiaadwgaca WGZbaabeqdcqGHris5aOGaaiikaiaadshacaGGPaGaeyypa0ZaaSaa aeaacaWGKbGaaCiCamaaCaaaleqabaGaamivaiaad+eacaWGubaaaa GcbaGaamizaiaadshaaaaaaa@4F3F@

We can integrate this expression with respect to time to get the integral version of the theorem.

 

 

 

 

4.2.6 Examples of applications of momentum conservation for systems of particles

 

The impulse-momentum equations for systems of particles are particularly useful for (i) analyzing the recoil of a gun; and (ii) analyzing rocket and jet propulsion systems.  In both these applications, the internal forces acting between the gun on the projectile, or the motor and propellant, are much larger than any external forces, so the total momentum of the system is conserved. 

Example 1: Estimate the recoil velocity of a rifle (youtube abounds with recoil demonstrations MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  see. e.g. http://www.youtube.com/watch?v=F4juEIK_zRM for samples.  Be warned, however MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  a lot of the videos are tasteless and/or sexist… )

 

The recoil velocity can be estimated by noting that the total momentum of bullet and rifle together must be conserved.  If we can estimate the mass of rifle and bullet, and the bullet’s velocity, the recoil velocity can be computed from the momentum conservation equation.

 

Assumptions:

1.      The mass of a typical 0.22 (i.e. 0.22” diameter) caliber rifle bullet is about 7× 10 4 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaG4naiabgE na0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI0aaaaaaa@3D22@  kg (idealizing the bullet as a sphere, with density 7860 kg/ m 3 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4AaiaadE gacaGGVaGaamyBamaaCaaaleqabaGaaG4maaaaaaa@3B68@  )

2.      The muzzle velocity of a 0.22 is about 1000 ft/sec (305 m/s)

3.      A typical rifle weighs between 5 and 10 lb (2.5-5 kg)

 

Calculation:

1.      Let m b MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaWGIbaabeaaaaa@3902@  denote the bullet mass, and let m R MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaWGsbaabeaaaaa@38F2@  denote the mass of the rifle.

2.      The rifle and bullet are idealized as two particles.  Before firing, both are at rest.  After firing, the bullet has velocity v b =vi MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaBa aaleaacaWGIbaabeaakiabg2da9iaadAhacaWHPbaaaa@3C0C@ ; the rifle has velocity v R =Vi MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaBa aaleaacaWGsbaabeaakiabg2da9iaadAfacaWHPbaaaa@3BDC@

3.      External forces acting on the system can be neglected, so

p tot ( t 0 +Δt)= p tot ( t 0 ) m b v b + m R v R =0 m b vi+ m R Vi=0V= m b v/ m R MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWHWb WaaSbaaSqaaiaadshacaWGVbGaamiDaaqabaGccaGGOaGaamiDamaa BaaaleaacaaIWaaabeaakiabgUcaRiabfs5aejaadshacaGGPaGaey ypa0JaaCiCamaaBaaaleaacaWG0bGaam4BaiaadshaaeqaaOGaaiik aiaadshadaWgaaWcbaGaaGimaaqabaGccaGGPaGaeyO0H4TaamyBam aaBaaaleaacaWGIbaabeaakiaahAhadaWgaaWcbaGaamOyaaqabaGc cqGHRaWkcaWGTbWaaSbaaSqaaiaadkfaaeqaaOGaaCODamaaBaaale aacaWGsbaabeaakiabg2da9iaahcdaaeaacqGHshI3caWGTbWaaSba aSqaaiaadkgaaeqaaOGaamODaiaahMgacqGHRaWkcaWGTbWaaSbaaS qaaiaadkfaaeqaaOGaamOvaiaahMgacqGH9aqpcaWHWaGaeyO0H4Ta amOvaiabg2da9iabgkHiTiaad2gadaWgaaWcbaGaamOyaaqabaGcca WG2bGaai4laiaad2gadaWgaaWcbaGaamOuaaqabaaaaaa@6DC9@ .

4.      Substituting numbers gives |V| between 0.04 and 0.08 m/s (about 0.14 ft/sec)

 

Example 2: Derive a formula that can be used to estimate the mass of a handgun required to keep its recoil within acceptable limits.  

 

The preceding example shows that the firearm will recoil with a velocity that depends on the ratio of the mass of the bullet to the firearm.  The firearm must be brought to rest by the person holding it.

 

Assumptions:

1.      We will idealize a person’s hand holding the gun as a spring, with stiffness k, fixed at one end.  The ‘end-point stiffness’ of a human hand has been extensively studied MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  see, e.g. Shadmehr et al J. Neuroscience, 13 (1) 45 (1993).  Typical values of stiffness during quasi-static deflections are of order 0.2 N/mm. During dynamic loading stiffnesses are likely to be larger than this.

2.      We idealize the handgun and bullet as particles, with mass m b MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaWGIbaabeaaaaa@37FB@  and m g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaWGNbaabeaaaaa@3800@ , respectively.

 

Calculation.

1.      The preceding problem shows that the firearm will recoil with velocity V= m b v/ m g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 da9iabgkHiTiaad2gadaWgaaWcbaGaamOyaaqabaGccaWG2bGaai4l aiaad2gadaWgaaWcbaGaam4zaaqabaaaaa@3E8B@

2.      Energy conservation can be used to calculate the recoil distance.  We consider the firearm and the hand holding it a system.  At time t=0 it has zero potential energy; and has kinetic energy T= 1 2 m g V 2 = 1 2 ( m b v ) 2 / m g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 da9maalaaabaGaaGymaaqaaiaaikdaaaGaamyBamaaBaaaleaacaWG NbaabeaakiaadAfadaahaaWcbeqaaiaaikdaaaGccqGH9aqpdaWcaa qaaiaaigdaaeaacaaIYaaaamaabmaabaGaamyBamaaBaaaleaacaWG IbaabeaakiaadAhaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaa GccaGGVaGaamyBamaaBaaaleaacaWGNbaabeaaaaa@480E@ .  At the end of the recoil, the gun is at rest, and the spring is fully compressed MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  the kinetic energy is zero, and potential energy is 1 2 k d 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca aIXaaabaGaaGOmaaaacaWGRbGaamizamaaCaaaleqabaGaaGOmaaaa aaa@3A3F@ .  Energy conservation gives k d 2 = ( m b v ) 2 / m g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Aaiaads gadaahaaWcbeqaaiaaikdaaaGccqGH9aqpdaqadaqaaiaad2gadaWg aaWcbaGaamOyaaqabaGccaWG2baacaGLOaGaayzkaaWaaWbaaSqabe aacaaIYaaaaOGaai4laiaad2gadaWgaaWcbaGaam4zaaqabaaaaa@420B@

3.      The required mass follows as m g = ( m b v ) 2 /k d 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaWGNbaabeaakiabg2da9maabmaabaGaamyBamaaBaaaleaa caWGIbaabeaakiaadAhaaiaawIcacaGLPaaadaahaaWcbeqaaiaaik daaaGccaGGVaGaam4AaiaadsgadaahaaWcbeqaaiaaikdaaaaaaa@420B@

 

 

Example 3 Rocket propulsion equations. Rocket motors and jet engines exploit the momentum conservation law in order to produce motion.  They work by expelling mass from a vehicle at very high velocity, in a direction opposite to the motion of the vehicle.   The momentum of the expelled mass must be balanced by an equal and opposite change in the momentum of the vehicle; so the velocity of the vehicle increases.

 

Analyzing a rocket engine is quite complicated, because the propellant carried by the engine is usually a  very significant fraction of the total mass of the vehicle.  Consequently, it is important to account for the fact that the mass decreases as the propellant is used.

 

 

Assumptions:

1.      The figure shows a rocket motor attached to a rocket with mass M.

2.      The rocket motor contains an initial mass m 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaaIWaaabeaaaaa@38D5@  of propellant and expels propellant at rate dm dt = μ ˙ 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbGaamyBaaqaaiaadsgacaWG0baaaiabg2da9iqbeY7aTzaacaWa aSbaaSqaaiaaicdaaeqaaaaa@3E75@  (kg/sec) with a velocity v 0 = v 0 i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaBa aaleaacaaIWaaabeaakiabg2da9iabgkHiTiaadAhadaWgaaWcbaGa aGimaaqabaGccaWHPbaaaa@3DBC@  relative to the rocket.

3.      We assume straight line motion, and assume that no external forces act on the rocket or motor.

 

Calculations:

The figure shows the rocket at an instant of time t, and then a very short time interval dt later.

1.      At time t, the rocket moves at speed v, and the system has momentum p=(M+m)vi MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiCaiabg2 da9iaacIcacaWGnbGaey4kaSIaamyBaiaacMcacaWG2bGaaCyAaaaa @3EE8@ , where m is the motor’s mass.

2.       During the time interval dt a mass dm= μ ˙ dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaad2 gacqGH9aqpcuaH8oqBgaGaaiaadsgacaWG0baaaa@3D7F@  is expelled from the motor. The velocity of the expelled mass is v=(v v 0 )i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9iaacIcacaWG2bGaeyOeI0IaamODamaaBaaaleaacaaIWaaabeaa kiaacMcacaWHPbaaaa@3F20@ .

3.      At time t+dt the mass of the motor has decreased to  m μ ˙ dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabgk HiTiqbeY7aTzaacaGaamizaiaadshaaaa@3C7D@ .

4.      At time t+dt, the rocket has velocity v=(v+dv)i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabg2 da9iaacIcacaWG2bGaey4kaSIaamizaiaadAhacaGGPaGaaCyAaaaa @3F0E@ .

5.       The total momentum of the system at time t+dt is therefore

p=(M+m μ 0 dt)(v+dv)i+ μ 0 dt(v v 0 )i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiCaiabg2 da9iaacIcacaWGnbGaey4kaSIaamyBaiabgkHiTiabeY7aTnaaBaaa leaacaaIWaaabeaakiaadsgacaWG0bGaaiykaiaacIcacaWG2bGaey 4kaSIaamizaiaadAhacaGGPaGaaCyAaiabgUcaRiabeY7aTnaaBaaa leaacaaIWaaabeaakiaadsgacaWG0bGaaiikaiaadAhacqGHsislca WG2bWaaSbaaSqaaiaaicdaaeqaaOGaaiykaiaahMgaaaa@5404@

6.      Momentum must be conserved, so

(M+m)vi=(M+m μ 0 dt)(v+dv)i+ μ 0 dt(v v 0 )i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaad2 eacqGHRaWkcaWGTbGaaiykaiaadAhacaWHPbGaeyypa0Jaaiikaiaa d2eacqGHRaWkcaWGTbGaeyOeI0IaeqiVd02aaSbaaSqaaiaaicdaae qaaOGaamizaiaadshacaGGPaGaaiikaiaadAhacqGHRaWkcaWGKbGa amODaiaacMcacaWHPbGaey4kaSIaeqiVd02aaSbaaSqaaiaaicdaae qaaOGaamizaiaadshacaGGOaGaamODaiabgkHiTiaadAhadaWgaaWc baGaaGimaaqabaGccaGGPaGaaCyAaaaa@58F7@

7.      Multiplying this out and simplifying shows that

(M+m)dvi μ ˙ dt v 0 i=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaad2 eacqGHRaWkcaWGTbGaaiykaiaadsgacaWG2bGaaCyAaiabgkHiTiqb eY7aTzaacaGaamizaiaadshacaWG2bWaaSbaaSqaaiaaicdaaeqaaO GaaCyAaiabg2da9iaahcdaaaa@46FC@

where the term μ 0 dtdv MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiVd02aaS baaSqaaiaaicdaaeqaaOGaamizaiaadshacaWGKbGaamODaaaa@3D69@  has been neglected.

8.      Finally, we see that

(M+m) dv dt = μ ˙ v 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaad2 eacqGHRaWkcaWGTbGaaiykamaalaaabaGaamizaiaadAhaaeaacaWG KbGaamiDaaaacqGH9aqpcuaH8oqBgaGaaiaadAhadaWgaaWcbaGaaG imaaqabaaaaa@4378@

This result is called the `rocket equation.’   

 

Specific Impulse of a rocket motor: The performance of a rocket engine is usually specified by its `specific impulse.’  Confusingly, two different definitions of specific impulse are commonly used:

I sp = v 0 I ¯ sp = v 0 g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGjb WaaSbaaSqaaiaadohacaWGWbaabeaakiabg2da9iaadAhadaWgaaWc baGaaGimaaqabaaakeaaceWGjbGbaebadaWgaaWcbaGaam4Caiaadc haaeqaaOGaeyypa0ZaaSaaaeaacaWG2bWaaSbaaSqaaiaaicdaaeqa aaGcbaGaam4zaaaaaaaa@43DC@

The first definition quantifies the impulse exerted by the motor per unit mass of propellant; the second is the impulse per unit weight of propellant.  You can usually tell which definition is being used from the units MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  the first definition has units of m/s; the second has units of s.  In terms of the specific impulse, the rocket equation is

(M+m) dv dt = μ ˙ I sp = μ ˙ g I ¯ sp MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaad2 eacqGHRaWkcaWGTbGaaiykamaalaaabaGaamizaiaadAhaaeaacaWG KbGaamiDaaaacqGH9aqpcuaH8oqBgaGaaiaadMeadaWgaaWcbaGaam 4CaiaadchaaeqaaOGaeyypa0JafqiVd0MbaiaacaWGNbGabmysayaa raWaaSbaaSqaaiaadohacaWGWbaabeaaaaa@4B38@

 

 

Integrated form of the rocket equation: If the motor expels propellant at constant rate, the equation of motion can be integrated.  Assume that

1.      The rocket is at the origin at time t=0;

2.      The rocket has speed v 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaaIWaaabeaaaaa@38DE@  at time t=0

3.      The motor has mass m 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaaIWaaabeaaaaa@38D5@  at time t=0; this means that at time t it has mass m 0 μt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaaIWaaabeaakiabgkHiTiabeY7aTjaadshaaaa@3C7B@

Then the rocket’s speed can be calculated as a function of time:

(M+ m 0 μt) dv dt = μ ˙ I sp v0 v dv = 0 t μ ˙ I sp dt (M+ m 0 μt) v v 0 = I sp log( M+ m 0 M+ m 0 μt ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaGGOa GaamytaiabgUcaRiaad2gadaWgaaWcbaGaaGimaaqabaGccqGHsisl cqaH8oqBcaWG0bGaaiykamaalaaabaGaamizaiaadAhaaeaacaWGKb GaamiDaaaacqGH9aqpcuaH8oqBgaGaaiaadMeadaWgaaWcbaGaam4C aiaadchaaeqaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaeyO0H49aa8qCaeaacaWGKbGaamODaaWcbaGaamODaiaaicdaaeaa caWG2baaniabgUIiYdGccqGH9aqpdaWdXbqaamaalaaabaGafqiVd0 MbaiaacaWGjbWaaSbaaSqaaiaadohacaWGWbaabeaakiaadsgacaWG 0baabaGaaiikaiaad2eacqGHRaWkcaWGTbWaaSbaaSqaaiaaicdaae qaaOGaeyOeI0IaeqiVd0MaamiDaiaacMcaaaaaleaacaaIWaaabaGa amiDaaqdcqGHRiI8aaGcbaGaeyO0H4TaamODaiabgkHiTiaadAhada WgaaWcbaGaaGimaaqabaGccqGH9aqpcaWGjbWaaSbaaSqaaiaadoha caWGWbaabeaakiGacYgacaGGVbGaai4zamaabmaabaWaaSaaaeaaca WGnbGaey4kaSIaamyBamaaBaaaleaacaaIWaaabeaaaOqaaiaad2ea cqGHRaWkcaWGTbWaaSbaaSqaaiaaicdaaeqaaOGaeyOeI0IaeqiVd0 MaamiDaaaaaiaawIcacaGLPaaaaaaa@8972@

Similarly, the position follows as

v= dx dt = I sp log( M+ m 0 M+ m 0 μt )+ v 0 0 x dx = 0 t ( I sp log( M+ m 0 M+ m 0 μt )+ v 0 ) dt x=( v 0 + I sp )t( M+ m 0 μt ) I sp μ log( M+ m 0 M+ m 0 μt ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsh I3caWG2bGaeyypa0ZaaSaaaeaacaWGKbGaamiEaaqaaiaadsgacaWG 0baaaiabg2da9iaadMeadaWgaaWcbaGaam4CaiaadchaaeqaaOGaci iBaiaac+gacaGGNbWaaeWaaeaadaWcaaqaaiaad2eacqGHRaWkcaWG TbWaaSbaaSqaaiaaicdaaeqaaaGcbaGaamytaiabgUcaRiaad2gada WgaaWcbaGaaGimaaqabaGccqGHsislcqaH8oqBcaWG0baaaaGaayjk aiaawMcaaiabgUcaRiaadAhadaWgaaWcbaGaaGimaaqabaGccqGHsh I3daWdXbqaaiaadsgacaWG4baaleaacaaIWaaabaGaamiEaaqdcqGH RiI8aOGaeyypa0Zaa8qCaeaadaqadaqaaiaadMeadaWgaaWcbaGaam 4CaiaadchaaeqaaOGaciiBaiaac+gacaGGNbWaaeWaaeaadaWcaaqa aiaad2eacqGHRaWkcaWGTbWaaSbaaSqaaiaaicdaaeqaaaGcbaGaam ytaiabgUcaRiaad2gadaWgaaWcbaGaaGimaaqabaGccqGHsislcqaH 8oqBcaWG0baaaaGaayjkaiaawMcaaiabgUcaRiaadAhadaWgaaWcba GaaGimaaqabaaakiaawIcacaGLPaaaaSqaaiaaicdaaeaacaWG0baa niabgUIiYdGccaWGKbGaamiDaaqaaiabgkDiElaadIhacqGH9aqpca GGOaGaamODamaaBaaaleaacaaIWaaabeaakiabgUcaRiaadMeadaWg aaWcbaGaam4CaiaadchaaeqaaOGaaiykaiaadshacqGHsisldaqada qaaiaad2eacqGHRaWkcaWGTbWaaSbaaSqaaiaaicdaaeqaaOGaeyOe I0IaeqiVd0MaamiDaaGaayjkaiaawMcaamaalaaabaGaamysamaaBa aaleaacaWGZbGaamiCaaqabaaakeaacqaH8oqBaaGaciiBaiaac+ga caGGNbWaaeWaaeaadaWcaaqaaiaad2eacqGHRaWkcaWGTbWaaSbaaS qaaiaaicdaaeqaaaGcbaGaamytaiabgUcaRiaad2gadaWgaaWcbaGa aGimaaqabaGccqGHsislcqaH8oqBcaWG0baaaaGaayjkaiaawMcaaa aaaa@A522@

 

These calculations assume that no external forces act on the rocket.  It is quite straightforward to generalize them to account for external forces as well MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  the details are left as an exercise.

 

 

Example 4 Application of rocket propulsion equation: Calculate the maximum payload that can be launched to escape velocity on the Ares I launch vehicle.

 

‘Escape velocity’ means that after the motor burns out, the space vehicle can escape the earth’s gravitational field MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  see example 5 in Section 4.1.6. 

 

Assumptions

1.      The specifications for the Ares I are at http://www.braeunig.us/space/specs/ares.htm  Relevant variables are listed in the table below.

 

 

Total mass (kg)

Specific impulse (s)

Propellant mass (kg)

Stage I

586344

268.8

504516

Stage II

183952

452.1

163530

2.      As an approximation, we will neglect the motion of the rocket during the burn, and will neglect aerodynamic forces.

3.      We will assume that the first stage is jettisoned before burning the second stage.

4.      Note that the change in velocity due to burning a stage can be expressed as

v v 0 = I ¯ sp glog( M 0 M 0 Δm ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODaiabgk HiTiaadAhadaWgaaWcbaGaaGimaaqabaGccqGH9aqpceWGjbGbaeba daWgaaWcbaGaam4CaiaadchaaeqaaOGaam4zaiGacYgacaGGVbGaai 4zamaabmaabaWaaSaaaeaacaWGnbWaaSbaaSqaaiaaicdaaeqaaaGc baGaamytamaaBaaaleaacaaIWaaabeaakiabgkHiTiabfs5aejaad2 gaaaaacaGLOaGaayzkaaaaaa@4AFD@

where M 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytamaaBa aaleaacaaIWaaabeaaaaa@38B5@  is the total mass before the burn, and Δm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam yBaaaa@3955@  is the mass of propellant burned.

  1. The earth’s radius is 6378.145km
  2. The Gravitational parameter μ=GM=3.986012× 10 5 km 3 s 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH8oqBcqGH9aqpcaWGhbGaamytai abg2da9iaaiodacaGGUaGaaGyoaiaaiIdacaaI2aGaaGimaiaaigda caaIYaGaaGPaVlaaykW7caaMc8UaaGPaVlabgEna0kaaykW7caaMc8 UaaGymaiaaicdadaahaaWcbeqaaiaaiwdaaaGccaaMc8UaaGPaVlaa ykW7caqGRbGaaeyBamaaCaaaleqabaGaaG4maaaakiaabohadaahaa WcbeqaaiabgkHiTiaaigdaaaaaaa@560C@   (G= gravitational constant; M=mass of earth)
  3. Escape velocity is from the earths surface is v e = 2GM/R MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG2bWaaSbaaSqaaiaadwgaaeqaaO Gaeyypa0ZaaOaaaeaacaaIYaGaam4raiaad2eacaGGVaGaamOuaaWc beaaaaa@39E5@ , where R is the earth’s radius.

 

 

 

Calculation:

1.      Let m denote the payload mass; let m 1 , m 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaaIXaaabeaakiaacYcacaWGTbWaaSbaaSqaaiaaikdaaeqa aaaa@3B6A@  denote the total masses of stages I and II, let m 01 , m 02 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaaIWaGaaGymaaqabaGccaGGSaGaamyBamaaBaaaleaacaaI WaGaaGOmaaqabaaaaa@3CDE@  denote the propellant masses of stages I and II; and let I ¯ sp1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmysayaara WaaSbaaSqaaiaadohacaWGWbGaaGymaaqabaaaaa@3AB7@ , I ¯ sp2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmysayaara WaaSbaaSqaaiaadohacaWGWbGaaGOmaaqabaaaaa@3AB8@  denote the specific impulses of the two stages.

2.      The rocket is at rest before burning the first stage; and its total mass is m+ m 1 + m 2 + m 01 + m 02 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabgU caRiaad2gadaWgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGTbWaaSba aSqaaiaaikdaaeqaaOGaey4kaSIaamyBamaaBaaaleaacaaIWaGaaG ymaaqabaGccqGHRaWkcaWGTbWaaSbaaSqaaiaaicdacaaIYaaabeaa aaa@446F@ .  After burn, the mass is m+ m 1 + m 2 + m 02 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabgU caRiaad2gadaWgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGTbWaaSba aSqaaiaaikdaaeqaaOGaey4kaSIaamyBamaaBaaaleaacaaIWaGaaG Omaaqabaaaaa@40F0@ .  The velocity after burning the first stage is therefore

v 1 = I ¯ sp1 glog( m+ m 1 + m 2 + m 01 + m 02 m+ m 1 + m 2 + m 02 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaaIXaaabeaakiabg2da9iqadMeagaqeamaaBaaaleaacaWG ZbGaamiCaiaaigdaaeqaaOGaam4zaiGacYgacaGGVbGaai4zamaabm aabaWaaSaaaeaacaWGTbGaey4kaSIaamyBamaaBaaaleaacaaIXaaa beaakiabgUcaRiaad2gadaWgaaWcbaGaaGOmaaqabaGccqGHRaWkca WGTbWaaSbaaSqaaiaaicdacaaIXaaabeaakiabgUcaRiaad2gadaWg aaWcbaGaaGimaiaaikdaaeqaaaGcbaGaamyBaiabgUcaRiaad2gada WgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGTbWaaSbaaSqaaiaaikda aeqaaOGaey4kaSIaamyBamaaBaaaleaacaaIWaGaaGOmaaqabaaaaa GccaGLOaGaayzkaaaaaa@5A81@

3.      The first stage is then jettisoned MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  the mass before starting the second burn is m+ m 2 + m 02 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabgU caRiaad2gadaWgaaWcbaGaaGOmaaqabaGccqGHRaWkcaWGTbWaaSba aSqaaiaaicdacaaIYaaabeaaaaa@3E2B@ , and after the second burn it is m+ m 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabgU caRiaad2gadaWgaaWcbaGaaGOmaaqabaaaaa@3AAB@ .  The velocity after the second burn is therefore

v 2 = I ¯ sp1 glog( m+ m 1 + m 2 + m 01 + m 02 m+ m 1 + m 2 + m 02 )+ I ¯ sp2 glog( m+ m 2 + m 02 m+ m 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaaIYaaabeaakiabg2da9iqadMeagaqeamaaBaaaleaacaWG ZbGaamiCaiaaigdaaeqaaOGaam4zaiGacYgacaGGVbGaai4zamaabm aabaWaaSaaaeaacaWGTbGaey4kaSIaamyBamaaBaaaleaacaaIXaaa beaakiabgUcaRiaad2gadaWgaaWcbaGaaGOmaaqabaGccqGHRaWkca WGTbWaaSbaaSqaaiaaicdacaaIXaaabeaakiabgUcaRiaad2gadaWg aaWcbaGaaGimaiaaikdaaeqaaaGcbaGaamyBaiabgUcaRiaad2gada WgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGTbWaaSbaaSqaaiaaikda aeqaaOGaey4kaSIaamyBamaaBaaaleaacaaIWaGaaGOmaaqabaaaaa GccaGLOaGaayzkaaGaey4kaSIabmysayaaraWaaSbaaSqaaiaadoha caWGWbGaaGOmaaqabaGccaWGNbGaciiBaiaac+gacaGGNbWaaeWaae aadaWcaaqaaiaad2gacqGHRaWkcaWGTbWaaSbaaSqaaiaaikdaaeqa aOGaey4kaSIaamyBamaaBaaaleaacaaIWaGaaGOmaaqabaaakeaaca WGTbGaey4kaSIaamyBamaaBaaaleaacaaIYaaabeaaaaaakiaawIca caGLPaaaaaa@6F6E@

4.      Substituting numbers into the escape velocity formula gives v e =11.18 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaWGLbaabeaakiabg2da9iaaigdacaaIXaGaaiOlaiaaigda caaI4aaaaa@3DC3@  km/sec.  Substituting numbers for the masses shows that to reach this velocity, the payload mass must satisfy

11.18<2.621log( m+770296 m+265780 )+4.435log( m+183952 m+20422 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiaaig dacaGGUaGaaGymaiaaiIdacqGH8aapcaaIYaGaaiOlaiaaiAdacaaI YaGaaGymaiGacYgacaGGVbGaai4zamaabmaabaWaaSaaaeaacaWGTb Gaey4kaSIaaG4naiaaiEdacaaIWaGaaGOmaiaaiMdacaaI2aaabaGa amyBaiabgUcaRiaaikdacaaI2aGaaGynaiaaiEdacaaI4aGaaGimaa aaaiaawIcacaGLPaaacqGHRaWkcaaI0aGaaiOlaiaaisdacaaIZaGa aGynaiGacYgacaGGVbGaai4zamaabmaabaWaaSaaaeaacaWGTbGaey 4kaSIaaGymaiaaiIdacaaIZaGaaGyoaiaaiwdacaaIYaaabaGaamyB aiabgUcaRiaaikdacaaIWaGaaGinaiaaikdacaaIYaaaaaGaayjkai aawMcaaaaa@6510@

where m is in kg.

5.      We can use Matlab to solve for the critical value of m for equality

clear all
syms m
eq = 11.1798==2.62908*log((m+770296)/(m+265780))+4.435101*log((m+183952)/(m+20422))
solve(eq,m)

 

so the solution is 8300kg MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  a very small mass compared with that of the launch vehicle, but you could pack in a large number of people you would like to launch into outer space nonetheless (the entire faculty of the school of engineering, if you wish).

 

 

 

4.2.7 Analyzing collisions between particles: the restitution coefficient

 

The momentum conservation equations are particularly helpful if you want to analyze collisions between two or more objects.  If the impact occurs over a very short time, the impulse exerted by the contact forces acting at the point of collision is huge compared with the impulse exerted by any other forces.  If we consider the two colliding particles as a system, the external impulse exerted on the system can be taken to be zero, and so the total momentum of the system is conserved.

 

The momentum conservation equation can be used to relate the velocities of the particles before collision to those after collision.  These relations are not enough to be able to determine the velocities completely, however MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  to do this, we also need to be able to quantify the energy lost (or more accurately, dissipated as heat) during the collision.

 

In practice we don’t directly specify the energy loss during a collision MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  instead, the relative velocities are related by a property of the impact called the coefficient of restitution

Restitution coefficient for straight line motion

 

Suppose that two colliding particles A and B move in a straight line parallel to a unit vector i.  Let

v A0 = v x A0 i, v B0 = v x B0 i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaicdaaaGccqGH9aqpcaWG2bWaa0baaSqaaiaa dIhaaeaacaWGbbGaaGimaaaakiaahMgacaGGSaGaaGPaVlaaykW7ca aMc8UaaGPaVlaahAhadaahaaWcbeqaaiaadkeacaaIWaaaaOGaeyyp a0JaamODamaaDaaaleaacaWG4baabaGaamOqaiaaicdaaaGccaWHPb aaaa@4D8E@  denote the velocities of A and B just before the collision

v A1 = v x A1 i, v B1 = v x B1 i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaigdaaaGccqGH9aqpcaWG2bWaa0baaSqaaiaa dIhaaeaacaWGbbGaaGymaaaakiaahMgacaGGSaGaaGPaVlaaykW7ca aMc8UaaGPaVlaahAhadaahaaWcbeqaaiaadkeacaaIXaaaaOGaeyyp a0JaamODamaaDaaaleaacaWG4baabaGaamOqaiaaigdaaaGccaWHPb aaaa@4D92@  denote the velocities of A and B just after the collision.

 

The velocities before and after impact are related by

v B1 v A1 =e( v B0 v A0 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaCa aaleqabaGaamOqaiaaigdaaaGccqGHsislcaWG2bWaaWbaaSqabeaa caWGbbGaaGymaaaakiabg2da9iabgkHiTiaadwgadaqadaqaaiaadA hadaahaaWcbeqaaiaadkeacaaIWaaaaOGaeyOeI0IaamODamaaCaaa leqabaGaamyqaiaaicdaaaaakiaawIcacaGLPaaaaaa@4702@

where e is the restitution coefficient. The negative sign is needed because the particles approach one another before impact, and separate afterwards.

 

 

Restitution coefficient for 3D frictionless collisions

For a more general contact, we define

v A0 v B0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaicdaaaGccaaMc8UaaGPaVlaahAhadaahaaWc beqaaiaadkeacaaIWaaaaaaa@3E6F@  to denote velocities of the particles before collision

v A1 v B1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaigdaaaGccaaMc8UaaGPaVlaahAhadaahaaWc beqaaiaadkeacaaIXaaaaaaa@3E71@  to denote velocities of the particles after collision

 

In addition, we let n be a unit vector normal to the common tangent plane at the point of contact (if the two colliding particles are spheres or disks the vector is parallel to the line joining their centers).

 

The velocities before and after impact are related by two vector equations:

( v B1 v A1 )n=e( v B0 v A0 )n ( v B1 v A1 )[ ( v B1 v A1 )n ]n=( v B0 v A0 )[ ( v B0 v A0 )n ]n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaqada qaaiaahAhadaahaaWcbeqaaiaadkeacaaIXaaaaOGaeyOeI0IaaCOD amaaCaaaleqabaGaamyqaiaaigdaaaaakiaawIcacaGLPaaacqGHfl Y1caWHUbGaeyypa0JaeyOeI0IaamyzamaabmaabaGaaCODamaaCaaa leqabaGaamOqaiaaicdaaaGccqGHsislcaWH2bWaaWbaaSqabeaaca WGbbGaaGimaaaaaOGaayjkaiaawMcaaiabgwSixlaah6gaaeaadaqa daqaaiaahAhadaahaaWcbeqaaiaadkeacaaIXaaaaOGaeyOeI0IaaC ODamaaCaaaleqabaGaamyqaiaaigdaaaaakiaawIcacaGLPaaacqGH sisldaWadaqaamaabmaabaGaaCODamaaCaaaleqabaGaamOqaiaaig daaaGccqGHsislcaWH2bWaaWbaaSqabeaacaWGbbGaaGymaaaaaOGa ayjkaiaawMcaaiabgwSixlaah6gaaiaawUfacaGLDbaacaWHUbGaey ypa0ZaaeWaaeaacaWH2bWaaWbaaSqabeaacaWGcbGaaGimaaaakiab gkHiTiaahAhadaahaaWcbeqaaiaadgeacaaIWaaaaaGccaGLOaGaay zkaaGaeyOeI0YaamWaaeaadaqadaqaaiaahAhadaahaaWcbeqaaiaa dkeacaaIWaaaaOGaeyOeI0IaaCODamaaCaaaleqabaGaamyqaiaaic daaaaakiaawIcacaGLPaaacqGHflY1caWHUbaacaGLBbGaayzxaaGa aCOBaaaaaa@7EEF@

 

To interpret these equations, note that

1.      The first equation states that the component of relative velocity normal to the contact plane is reduced by a factor e (just as for 1D contacts)

2.      The second equation states that the component of relative velocity tangent to the contact plane is unchanged

 

To understand the second equation, note that.

·         During the collision, a very large force acts on both A and B at the contact between them.  Because the contact is frictionless, the direction of the force must be parallel to n.   Also, the force on A must be equal and opposite to the force on B.

·         There is no force acting on either A or B parallel to t.   This means that momentum must be conserved in the t direction for both A and B individually.  We can write this mathematically as

v B1 [ v B1 n ]n= v B0 [ v B0 n ]n v A1 [ v A1 n ]n= v A0 [ v A0 n ]n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWH2b WaaWbaaSqabeaacaWGcbGaaGymaaaakiabgkHiTmaadmaabaGaaCOD amaaCaaaleqabaGaamOqaiaaigdaaaGccqGHflY1caWHUbaacaGLBb GaayzxaaGaaCOBaiabg2da9iaahAhadaahaaWcbeqaaiaadkeacaaI WaaaaOGaeyOeI0YaamWaaeaacaWH2bWaaWbaaSqabeaacaWGcbGaaG imaaaakiabgwSixlaah6gaaiaawUfacaGLDbaacaWHUbaabaGaaCOD amaaCaaaleqabaGaamyqaiaaigdaaaGccqGHsisldaWadaqaaiaahA hadaahaaWcbeqaaiaadgeacaaIXaaaaOGaeyyXICTaaCOBaaGaay5w aiaaw2faaiaah6gacqGH9aqpcaWH2bWaaWbaaSqabeaacaWGbbGaaG imaaaakiabgkHiTmaadmaabaGaaCODamaaCaaaleqabaGaamyqaiaa icdaaaGccqGHflY1caWHUbaacaGLBbGaayzxaaGaaCOBaaaaaa@6B24@

(this looks funny, but we have just subtracted the normal component of velocity from the total velocity.   This leaves just the tangential component).    Subtracting the second equation from the first shows that

( v B1 v A1 )[ ( v B1 v A1 )n ]n=( v B0 v A0 )[ ( v B0 v A0 )n ]n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca WH2bWaaWbaaSqabeaacaWGcbGaaGymaaaakiabgkHiTiaahAhadaah aaWcbeqaaiaadgeacaaIXaaaaaGccaGLOaGaayzkaaGaeyOeI0Yaam WaaeaadaqadaqaaiaahAhadaahaaWcbeqaaiaadkeacaaIXaaaaOGa eyOeI0IaaCODamaaCaaaleqabaGaamyqaiaaigdaaaaakiaawIcaca GLPaaacqGHflY1caWHUbaacaGLBbGaayzxaaGaaCOBaiabg2da9maa bmaabaGaaCODamaaCaaaleqabaGaamOqaiaaicdaaaGccqGHsislca WH2bWaaWbaaSqabeaacaWGbbGaaGimaaaaaOGaayjkaiaawMcaaiab gkHiTmaadmaabaWaaeWaaeaacaWH2bWaaWbaaSqabeaacaWGcbGaaG imaaaakiabgkHiTiaahAhadaahaaWcbeqaaiaadgeacaaIWaaaaaGc caGLOaGaayzkaaGaeyyXICTaaCOBaaGaay5waiaaw2faaiaah6gaaa a@65C1@

 

 

Combined 3D restitution formula

 

The two equations for the normal and tangential behavior can be combined (just add them) into a single vector equation relating velocities before and after impact MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  this form is more compact, and often more useful, but more difficult to visualize physically

( v B1 v A1 )=( v B0 v A0 )(1+e)[ ( v B0 v A0 )n ]n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca WH2bWaaWbaaSqabeaacaWGcbGaaGymaaaakiabgkHiTiaahAhadaah aaWcbeqaaiaadgeacaaIXaaaaaGccaGLOaGaayzkaaGaeyypa0Zaae WaaeaacaWH2bWaaWbaaSqabeaacaWGcbGaaGimaaaakiabgkHiTiaa hAhadaahaaWcbeqaaiaadgeacaaIWaaaaaGccaGLOaGaayzkaaGaey OeI0IaaiikaiaaigdacqGHRaWkcaWGLbGaaiykamaadmaabaWaaeWa aeaacaWH2bWaaWbaaSqabeaacaWGcbGaaGimaaaakiabgkHiTiaahA hadaahaaWcbeqaaiaadgeacaaIWaaaaaGccaGLOaGaayzkaaGaeyyX ICTaaCOBaaGaay5waiaaw2faaiaah6gaaaa@599E@

 

Values of restitution coefficient

 

The restitution coefficient almost always lies in the range 0e1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGimaiabgs MiJkaadwgacqGHKjYOcaaIXaaaaa@3BBF@ .  It can only be less than zero if one object can penetrate and pass through the other (e.g. a bullet); and can only be greater than 1 if the collision generates energy somehow (e.g. releasing a preloaded spring, or causing an explosion).

 

If e=0, the two colliding objects stick together; if e=1 the collision is perfectly elastic, with no energy loss. 

 

The restitution coefficient is strongly sensitive to the material properties of the two colliding objects, and also varies weakly with their geometry and the velocity of impact.  The two latter effects are usually ignored. 

 

Collision between a particle and a fixed rigid surface. The collision formulas can be applied to impact between a rigid fixed surface by taking the surface to be particle B, and noting that the velocity of particle B is then zero both before and after impact. 

 

For straight line motion, v A1 =e v A0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaCa aaleqabaGaamyqaiaaigdaaaGccqGH9aqpcqGHsislcaWGLbGaamOD amaaCaaaleqabaGaamyqaiaaicdaaaaaaa@3E2E@

 

For collision with an angled wall v A1 = v A0 (1+e)[ v A0 n ]n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaigdaaaGccqGH9aqpcaWH2bWaaWbaaSqabeaa caWGbbGaaGimaaaakiabgkHiTiaacIcacaaIXaGaey4kaSIaamyzai aacMcadaWadaqaaiaahAhadaahaaWcbeqaaiaadgeacaaIWaaaaOGa eyyXICTaaCOBaaGaay5waiaaw2faaiaah6gaaaa@4A16@ , where

n is a unit vector perpendicular to the wall.

 

 

 

 

 

 

4.2.8 Examples of collision problems

 

Example 1 Suppose that a moving car hits a stationary (parked) vehicle from behind.  Derive a formula that will enable an accident investigator to determine the velocity of the moving car from the length of the skid marks left on the road.

 

Assumptions:

1.      We will assume both cars move in a straight line

2.      The moving and stationary cars will be assumed to have masses m 1 , m 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaaIXaaabeaakiaacYcacaWGTbWaaSbaaSqaaiaaikdaaeqa aaaa@3B6A@ , respectively

3.      We will assume the cars stick together after the collision (i.e. the restitution coefficient is zero)

4.      We will assume that only the parked car has brakes applied after the collision

 

This calculation takes two steps: first, we will use work-energy to relate the distance slid by the cars after impact to their velocity just after the impact occurs. Then, we will use momentum and the restitution formula to work out the velocity of the moving car just before impact. 

 

Calculation: Let V denote the velocity of the moving car just before impact; let v 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaaIXaaabeaaaaa@38DF@  denote the velocity of the two (connected) cars just after impact, and let d denote the distance slid.

1.      The figure shows a free body diagram for each of the two cars during sliding after the collision.  Newton’s law of motion for each car shows that

Ri+( N 1 + N 2 m 1 g )j= m 1 a x i ( R+ T 3 + T 4 )i+( N 3 + N 4 m 2 g )j= m 2 a x i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsi slcaWGsbGaaCyAaiabgUcaRmaabmaabaGaamOtamaaBaaaleaacaaI XaaabeaakiabgUcaRiaad6eadaWgaaWcbaGaaGOmaaqabaGccqGHsi slcaWGTbWaaSbaaSqaaiaaigdaaeqaaOGaam4zaaGaayjkaiaawMca aiaahQgacqGH9aqpcaWGTbWaaSbaaSqaaiaaigdaaeqaaOGaamyyam aaBaaaleaacaWG4baabeaakiaahMgaaeaadaqadaqaaiaadkfacqGH RaWkcaWGubWaaSbaaSqaaiaaiodaaeqaaOGaey4kaSIaamivamaaBa aaleaacaaI0aaabeaaaOGaayjkaiaawMcaaiaahMgacqGHRaWkdaqa daqaaiaad6eadaWgaaWcbaGaaG4maaqabaGccqGHRaWkcaWGobWaaS baaSqaaiaaisdaaeqaaOGaeyOeI0IaamyBamaaBaaaleaacaaIYaaa beaakiaadEgaaiaawIcacaGLPaaacaWHQbGaeyypa0JaamyBamaaBa aaleaacaaIYaaabeaakiaadggadaWgaaWcbaGaamiEaaqabaGccaWH Pbaaaaa@6561@

2.      The vertical component of the equations of motion give N 1 + N 2 = m 1 g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtamaaBa aaleaacaaIXaaabeaakiabgUcaRiaad6eadaWgaaWcbaGaaGOmaaqa baGccqGH9aqpcaWGTbWaaSbaaSqaaiaaigdaaeqaaOGaam4zaaaa@3F3D@ ; N 3 + N 4 m 2 g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtamaaBa aaleaacaaIZaaabeaakiabgUcaRiaad6eadaWgaaWcbaGaaGinaaqa baGccqGHsislcaWGTbWaaSbaaSqaaiaaikdaaeqaaOGaam4zaaaa@3F29@ .

3.      The parked car has locked wheels and skids over the road; the friction law gives the tangential forces at the contacts as T 3 + T 4 =μ( N 3 + N 4 )=μ m 2 g MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBa aaleaacaaIZaaabeaakiabgUcaRiaadsfadaWgaaWcbaGaaGinaaqa baGccqGH9aqpcqaH8oqBcaGGOaGaamOtamaaBaaaleaacaaIZaaabe aakiabgUcaRiaad6eadaWgaaWcbaGaaGinaaqabaGccaGGPaGaeyyp a0JaeqiVd0MaamyBamaaBaaaleaacaaIYaaabeaakiaadEgaaaa@4988@

4.      We can calculate the velocity of the cars just after impact using the work-kinetic energy relation during skidding.  For this purpose, we consider the two connected cars as a single particle.  The work done on the particle by the friction forces is ( T 3 + T 4 )d=μ m 2 gd MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Yaae WaaeaacaWGubWaaSbaaSqaaiaaiodaaeqaaOGaey4kaSIaamivamaa BaaaleaacaaI0aaabeaaaOGaayjkaiaawMcaaiaadsgacqGH9aqpcq GHsislcqaH8oqBcaWGTbWaaSbaaSqaaiaaikdaaeqaaOGaam4zaiaa dsgaaaa@4639@ .  The work done is equal to the change in kinetic energy of the particle, so

μ m 2 gd=0 1 2 ( m 1 + m 2 ) v 1 2 v 1 = 2μ m 2 gd m 1 + m 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsi slcqaH8oqBcaWGTbWaaSbaaSqaaiaaikdaaeqaaOGaam4zaiaadsga cqGH9aqpcaaIWaGaeyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaaaaca GGOaGaamyBamaaBaaaleaacaaIXaaabeaakiabgUcaRiaad2gadaWg aaWcbaGaaGOmaaqabaGccaGGPaGaamODamaaDaaaleaacaaIXaaaba GaaGOmaaaaaOqaaiaadAhadaWgaaWcbaGaaGymaaqabaGccqGH9aqp daGcaaqaamaalaaabaGaaGOmaiabeY7aTjaad2gadaWgaaWcbaGaaG OmaaqabaGccaWGNbGaamizaaqaaiaad2gadaWgaaWcbaGaaGymaaqa baGccqGHRaWkcaWGTbWaaSbaaSqaaiaaikdaaeqaaaaaaeqaaaaaaa@581B@

5.      Finally, we can use momentum conservation to calculate the velocity just before impact.  The momentum after impact is h 1 =( m 1 + m 2 ) v 1 i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiAamaaBa aaleaacaaIXaaabeaakiabg2da9iaacIcacaWGTbWaaSbaaSqaaiaa igdaaeqaaOGaey4kaSIaamyBamaaBaaaleaacaaIYaaabeaakiaacM cacaWG2bWaaSbaaSqaaiaaigdaaeqaaOGaaCyAaaaa@42C5@ , while before impact h 0 = m 1 Vi MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiAamaaBa aaleaacaaIWaaabeaakiabg2da9iaad2gadaWgaaWcbaGaaGymaaqa baGccaWGwbGaaCyAaaaa@3D94@ .  Equating the two gives

V= ( m 1 + m 2 ) m 1 v 1 = 2μ( m 1 + m 2 ) m 2 gd m 1 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 da9maalaaabaGaaiikaiaad2gadaWgaaWcbaGaaGymaaqabaGccqGH RaWkcaWGTbWaaSbaaSqaaiaaikdaaeqaaOGaaiykaaqaaiaad2gada WgaaWcbaGaaGymaaqabaaaaOGaamODamaaBaaaleaacaaIXaaabeaa kiabg2da9maakaaabaWaaSaaaeaacaaIYaGaeqiVd0Maaiikaiaad2 gadaWgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGTbWaaSbaaSqaaiaa ikdaaeqaaOGaaiykaiaad2gadaWgaaWcbaGaaGOmaaqabaGccaWGNb Gaamizaaqaaiaad2gadaqhaaWcbaGaaGymaaqaaiaaikdaaaaaaaqa baaaaa@52A8@

 

 

 

 

 

 

Example 2: Two frictionless spheres with radius R have initial velocity v A0 v B0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaicdaaaGccaaMc8UaaGPaVlaahAhadaahaaWc beqaaiaadkeacaaIWaaaaaaa@3E6F@ .  At some instant of time, the two particles collide.  At the point of collision, the centers of the spheres have position vectors y A y B MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyEamaaCa aaleqabaGaamyqaaaakiaaykW7caaMc8UaaCyEamaaCaaaleqabaGa amOqaaaaaaa@3D01@ .  The restitution coefficient for the contact is denoted by e. Find a formula for the velocities of the spheres after impact.  Hence, deduce an expression for the change in kinetic energy during the impact.

 

 

This is a straightforward vector algebra exercise.  We have two unknown velocity vectors: v A1 v B1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaigdaaaGccaaMc8UaaGPaVlaahAhadaahaaWc beqaaiaadkeacaaIXaaaaaaa@3E71@ , and two vector equations MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  momentum conservation, and the restitution coefficient formula.

 

Calculation

1.      Note that a unit vector normal to the tangent plane can be calculated from the position vectors of the centers at the impact as n=( y A y B )/| y A y B | MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOBaiabg2 da9maabmaabaGaaCyEamaaCaaaleqabaGaamyqaaaakiabgkHiTiaa ykW7caaMc8UaaCyEamaaCaaaleqabaGaamOqaaaaaOGaayjkaiaawM caaiaac+cadaabdaqaaiaahMhadaahaaWcbeqaaiaadgeaaaGccqGH sislcaaMc8UaaGPaVlaahMhadaahaaWcbeqaaiaadkeaaaaakiaawE a7caGLiWoaaaa@4D55@ .  It doesn’t matter whether you choose to take n to point from A to B or the other way around MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  the formula will work either way.

2.      Momentum conservation requires that

m B v B1 + m A v A1 = m B v B0 + m A v A0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaWGcbaabeaakiaaykW7caaMc8UaaCODamaaCaaaleqabaGa amOqaiaaigdaaaGccqGHRaWkcaWGTbWaaSbaaSqaaiaadgeaaeqaaO GaaCODamaaCaaaleqabaGaamyqaiaaigdaaaGccqGH9aqpcaWGTbWa aSbaaSqaaiaadkeaaeqaaOGaaGPaVlaaykW7caWH2bWaaWbaaSqabe aacaWGcbGaaGimaaaakiabgUcaRiaad2gadaWgaaWcbaGaamyqaaqa baGccaWH2bWaaWbaaSqabeaacaWGbbGaaGimaaaaaaa@5178@

3.      The restitution coefficient formula gives

( v B1 v A1 )=( v B0 v A0 )(1+e)[ ( v B0 v A0 )n ]n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca WH2bWaaWbaaSqabeaacaWGcbGaaGymaaaakiabgkHiTiaahAhadaah aaWcbeqaaiaadgeacaaIXaaaaaGccaGLOaGaayzkaaGaeyypa0Zaae WaaeaacaWH2bWaaWbaaSqabeaacaWGcbGaaGimaaaakiabgkHiTiaa hAhadaahaaWcbeqaaiaadgeacaaIWaaaaaGccaGLOaGaayzkaaGaey OeI0IaaiikaiaaigdacqGHRaWkcaWGLbGaaiykamaadmaabaWaaeWa aeaacaWH2bWaaWbaaSqabeaacaWGcbGaaGimaaaakiabgkHiTiaahA hadaahaaWcbeqaaiaadgeacaaIWaaaaaGccaGLOaGaayzkaaGaeyyX ICTaaCOBaaGaay5waiaaw2faaiaah6gaaaa@599E@

4.      We can solve (2) and (3) for v B1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamOqaiaaigdaaaaaaa@38A4@  by multiplying (3) by m A MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaWGbbaabeaaaaa@37DA@  and adding the equations, which gives

( m B + m A ) v B1 = m B v B0 + m A v A0 + m A [ ( v B0 v A0 )(1+e)[ ( v B0 v A0 )n ]n ] v B1 = v B0 m A m B + m A (1+e)[ ( v B0 v A0 )n ]n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaqada qaaiaad2gadaWgaaWcbaGaamOqaaqabaGccaaMc8Uaey4kaSIaamyB amaaBaaaleaacaWGbbaabeaakiaaykW7aiaawIcacaGLPaaacaWH2b WaaWbaaSqabeaacaWGcbGaaGymaaaakiabg2da9iaad2gadaWgaaWc baGaamOqaaqabaGccaaMc8UaaGPaVlaahAhadaahaaWcbeqaaiaadk eacaaIWaaaaOGaey4kaSIaamyBamaaBaaaleaacaWGbbaabeaakiaa hAhadaahaaWcbeqaaiaadgeacaaIWaaaaOGaey4kaSIaamyBamaaBa aaleaacaWGbbaabeaakmaadmaabaWaaeWaaeaacaWH2bWaaWbaaSqa beaacaWGcbGaaGimaaaakiabgkHiTiaahAhadaahaaWcbeqaaiaadg eacaaIWaaaaaGccaGLOaGaayzkaaGaeyOeI0IaaiikaiaaigdacqGH RaWkcaWGLbGaaiykamaadmaabaWaaeWaaeaacaWH2bWaaWbaaSqabe aacaWGcbGaaGimaaaakiabgkHiTiaahAhadaahaaWcbeqaaiaadgea caaIWaaaaaGccaGLOaGaayzkaaGaeyyXICTaaCOBaaGaay5waiaaw2 faaiaah6gaaiaawUfacaGLDbaaaeaacqGHshI3caWH2bWaaWbaaSqa beaacaWGcbGaaGymaaaakiabg2da9iaaykW7caWH2bWaaWbaaSqabe aacaWGcbGaaGimaaaakiabgkHiTmaalaaabaGaamyBamaaBaaaleaa caWGbbaabeaaaOqaaiaad2gadaWgaaWcbaGaamOqaaqabaGccaaMc8 Uaey4kaSIaamyBamaaBaaaleaacaWGbbaabeaaaaGccaGGOaGaaGym aiabgUcaRiaadwgacaGGPaWaamWaaeaadaqadaqaaiaahAhadaahaa WcbeqaaiaadkeacaaIWaaaaOGaeyOeI0IaaCODamaaCaaaleqabaGa amyqaiaaicdaaaaakiaawIcacaGLPaaacqGHflY1caWHUbaacaGLBb GaayzxaaGaaCOBaaaaaa@9559@

5.      Similarly, we can solve for v A1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaigdaaaaaaa@38A3@  by multiplying (3) by m B MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaWGcbaabeaaaaa@37DB@  and subtracting (3) from (2), with the result

v A1 = v A0 + m B m B + m A (1+e)[ ( v B0 v A0 )n ]n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaigdaaaGccqGH9aqpcaaMc8UaaCODamaaCaaa leqabaGaamyqaiaaicdaaaGccqGHRaWkdaWcaaqaaiaad2gadaWgaa WcbaGaamOqaaqabaaakeaacaWGTbWaaSbaaSqaaiaadkeaaeqaaOGa aGPaVlabgUcaRiaad2gadaWgaaWcbaGaamyqaaqabaaaaOGaaiikai aaigdacqGHRaWkcaWGLbGaaiykamaadmaabaWaaeWaaeaacaWH2bWa aWbaaSqabeaacaWGcbGaaGimaaaakiabgkHiTiaahAhadaahaaWcbe qaaiaadgeacaaIWaaaaaGccaGLOaGaayzkaaGaeyyXICTaaCOBaaGa ay5waiaaw2faaiaah6gaaaa@590C@

6.      The change in kinetic energy during the collision can be calculated as

ΔT= m A 2 v A1 v A1 + m B 2 v B1 v B1 ( m A 2 v A0 v A0 + m B 2 v B0 v B0 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam ivaiabg2da9maalaaabaGaamyBamaaBaaaleaacaWGbbaabeaaaOqa aiaaikdaaaGaaCODamaaCaaaleqabaGaamyqaiaaigdaaaGccqGHfl Y1caWH2bWaaWbaaSqabeaacaWGbbGaaGymaaaakiabgUcaRmaalaaa baGaamyBamaaBaaaleaacaWGcbaabeaaaOqaaiaaikdaaaGaaCODam aaCaaaleqabaGaamOqaiaaigdaaaGccqGHflY1caWH2bWaaWbaaSqa beaacaWGcbGaaGymaaaakiabgkHiTmaabmaabaWaaSaaaeaacaWGTb WaaSbaaSqaaiaadgeaaeqaaaGcbaGaaGOmaaaacaWH2bWaaWbaaSqa beaacaWGbbGaaGimaaaakiabgwSixlaahAhadaahaaWcbeqaaiaadg eacaaIWaaaaOGaey4kaSYaaSaaaeaacaWGTbWaaSbaaSqaaiaadkea aeqaaaGcbaGaaGOmaaaacaWH2bWaaWbaaSqabeaacaWGcbGaaGimaa aakiabgwSixlaahAhadaahaaWcbeqaaiaadkeacaaIWaaaaaGccaGL OaGaayzkaaaaaa@673F@

7.      Substituting the results of (4) and (5) for v B1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamOqaiaaigdaaaaaaa@38A4@  and v A1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaigdaaaaaaa@38A3@  and simplifying the result gives

ΔT= m A m B ( e 2 1 ) 2( m A + m B ) [ ( v B0 v A0 )n ] 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam ivaiabg2da9maalaaabaGaamyBamaaBaaaleaacaWGbbaabeaakiaa d2gadaWgaaWcbaGaamOqaaqabaGcdaqadaqaaiaadwgadaahaaWcbe qaaiaaikdaaaGccqGHsislcaaIXaaacaGLOaGaayzkaaaabaGaaGOm amaabmaabaGaamyBamaaBaaaleaacaWGbbaabeaakiabgUcaRiaad2 gadaWgaaWcbaGaamOqaaqabaaakiaawIcacaGLPaaaaaWaamWaaeaa daqadaqaaiaahAhadaahaaWcbeqaaiaadkeacaaIWaaaaOGaeyOeI0 IaaCODamaaCaaaleqabaGaamyqaiaaicdaaaaakiaawIcacaGLPaaa cqGHflY1caWHUbaacaGLBbGaayzxaaWaaWbaaSqabeaacaaIYaaaaa aa@5840@

 

Note that the energy change is zero if e=1 (perfectly elastic collisions) and is always negative for e<1 (i.e. the kinetic energy after collision is less than that before collision). 

 

 

 

Example 3: This is just a boring example to help illustrate the practical application of the vector formulas in the preceding example.  In the figure shown, disk A has a vertical velocity V at time t=0, while disk B is stationary.   The two disks both have radius R, have the same mass, and the restitution coefficient between them is e. Gravity can be neglected. Calculate the velocity vector of each disk after collision. 

 

Calculation

1.      Before impact, the velocity vectors are v A0 =Vj v B0 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaicdaaaGccqGH9aqpcaWGwbGaaCOAaiaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWH2bWaaWbaaS qabeaacaWGcbGaaGimaaaakiabg2da9iaahcdaaaa@4AC3@

2.      A unit vector parallel to the line joining the two centers is n=(3i+ 7 j)/4 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOBaiabg2 da9iaacIcacaaIZaGaaCyAaiabgUcaRmaakaaabaGaaG4naaWcbeaa kiaahQgacaGGPaGaai4laiaaisdaaaa@3F27@  (to see this, apply Pythagoras theorem to the triangle shown in the figure).

3.      The velocities after impact are

v B1 = v B0 m A m B + m A (1+e)[ ( v B0 v A0 )n ]n v A1 = v A0 + m B m B + m A (1+e)[ ( v B0 v A0 )n ]n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWH2b WaaWbaaSqabeaacaWGcbGaaGymaaaakiabg2da9iaaykW7caWH2bWa aWbaaSqabeaacaWGcbGaaGimaaaakiabgkHiTmaalaaabaGaamyBam aaBaaaleaacaWGbbaabeaaaOqaaiaad2gadaWgaaWcbaGaamOqaaqa baGccaaMc8Uaey4kaSIaamyBamaaBaaaleaacaWGbbaabeaaaaGcca GGOaGaaGymaiabgUcaRiaadwgacaGGPaWaamWaaeaadaqadaqaaiaa hAhadaahaaWcbeqaaiaadkeacaaIWaaaaOGaeyOeI0IaaCODamaaCa aaleqabaGaamyqaiaaicdaaaaakiaawIcacaGLPaaacqGHflY1caWH UbaacaGLBbGaayzxaaGaaCOBaaqaaiaahAhadaahaaWcbeqaaiaadg eacaaIXaaaaOGaeyypa0JaaCODamaaCaaaleqabaGaamyqaiaaicda aaGccqGHRaWkdaWcaaqaaiaad2gadaWgaaWcbaGaamOqaaqabaaake aacaWGTbWaaSbaaSqaaiaadkeaaeqaaOGaaGPaVlabgUcaRiaad2ga daWgaaWcbaGaamyqaaqabaaaaOGaaiikaiaaigdacqGHRaWkcaWGLb GaaiykamaadmaabaWaaeWaaeaacaWH2bWaaWbaaSqabeaacaWGcbGa aGimaaaakiabgkHiTiaahAhadaahaaWcbeqaaiaadgeacaaIWaaaaa GccaGLOaGaayzkaaGaeyyXICTaaCOBaaGaay5waiaaw2faaiaah6ga aaaa@7AAA@

Substituting the vectors gives

v B1 = 1 2 (1+e)[ ( Vj )(3i+ 7 j)/4 ](3i+ 7 j)/4= 1+e 32 V(3 7 i+7j) v A1 =Vj+ 1 2 (1+e)[ ( Vj )(3i+ 7 j)/4 ](3i+ 7 j)/4= 3 7 (1+e) 32 Vi+ 257e 32 Vj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWH2b WaaWbaaSqabeaacaWGcbGaaGymaaaakiabg2da9iaaykW7cqGHsisl daWcaaqaaiaaigdaaeaacaaIYaaaaiaacIcacaaIXaGaey4kaSIaam yzaiaacMcadaWadaqaamaabmaabaGaeyOeI0IaamOvaiaahQgaaiaa wIcacaGLPaaacqGHflY1caGGOaGaaG4maiaahMgacqGHRaWkdaGcaa qaaiaaiEdaaSqabaGccaWHQbGaaiykaiaac+cacaaI0aaacaGLBbGa ayzxaaGaaiikaiaaiodacaWHPbGaey4kaSYaaOaaaeaacaaI3aaale qaaOGaaCOAaiaacMcacaGGVaGaaGinaiabg2da9maalaaabaGaaGym aiabgUcaRiaadwgaaeaacaaIZaGaaGOmaaaacaWGwbGaaiikaiaaio dadaGcaaqaaiaaiEdaaSqabaGccaWHPbGaey4kaSIaaG4naiaahQga caGGPaaabaGaaCODamaaCaaaleqabaGaamyqaiaaigdaaaGccqGH9a qpcaWGwbGaaCOAaiabgUcaRmaalaaabaGaaGymaaqaaiaaikdaaaGa aiikaiaaigdacqGHRaWkcaWGLbGaaiykamaadmaabaWaaeWaaeaacq GHsislcaWGwbGaaCOAaaGaayjkaiaawMcaaiabgwSixlaacIcacaaI ZaGaaCyAaiabgUcaRmaakaaabaGaaG4naaWcbeaakiaahQgacaGGPa Gaai4laiaaisdaaiaawUfacaGLDbaacaGGOaGaaG4maiaahMgacqGH RaWkdaGcaaqaaiaaiEdaaSqabaGccaWHQbGaaiykaiaac+cacaaI0a Gaeyypa0JaeyOeI0YaaSaaaeaacaaIZaWaaOaaaeaacaaI3aaaleqa aOGaaiikaiaaigdacqGHRaWkcaWGLbGaaiykaaqaaiaaiodacaaIYa aaaiaadAfacaWHPbGaey4kaSYaaSaaaeaacaaIYaGaaGynaiabgkHi TiaaiEdacaWGLbaabaGaaG4maiaaikdaaaGaamOvaiaahQgaaaaa@9ACB@

 

Example 4: How to play pool (or snooker, billiards, or your own favorite bar game involving balls, a stick, and a table…).  The figure shows a typical problem faced by a pool player MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  where should the queue ball hit the eight ball to send it into the pocket?

 

This is easily solved MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  the eight ball is stationary before impact, and after impact has a velocity

v B1 = 1 2 (1+e)[ v A0 n ]n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamOqaiaaigdaaaGccqGH9aqpdaWcaaqaaiaaigdaaeaa caaIYaaaaiaacIcacaaIXaGaey4kaSIaamyzaiaacMcadaWadaqaai aahAhadaahaaWcbeqaaiaadgeacaaIWaaaaOGaeyyXICTaaCOBaaGa ay5waiaaw2faaiaah6gaaaa@47FB@

Notice that the velocity is parallel to the unit vector n.  This vector is parallel to a line connecting the centers of the two balls at the instant of impact. So the correct thing to do is to visualize an imaginary ball just touching the eight ball, in line with the pocket, and aim the queue ball at the imaginary ball.   Easy!

 

The real secret to being a successful pool player is not potting the balls MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  that part is easy.  It is controlling where the queue ball goes after impact.   You may have seen experts make a queue ball reverse its direction after an impact (appearing to bounce off the stationary ball); or make the queue ball follow the struck ball after the impact.  According to the simple equations developed here, this is impossible MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  but a pool table is more complicated, because the balls rotate, and are in contact with a table.  By giving the queue ball spin, an expert player can move the queue ball around at will.   To make the ball rebound, it must be struck low down (below the ‘center of percussion’) to give it a reverse spin; to make it follow the struck ball, it should be struck high up, to make it roll towards the ball to be potted.   Giving the ball a sideways spin can make it rebound in a controllable direction laterally as well.  And it is even possible to make a queue ball travel in a curved path with the right spin.

 

Never let it be said that you don’t learn useful skills in engineering classes!

 

 

 

4.3 Angular impulse-momentum relations

 

 

4.3.1 Definition of the angular impulse of a force

The angular impulse of a force is the time integral of the moment exerted by the force. 

 

To make the concept precise, consider a particle that is subjected to a time varying force F(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiaacI cacaWG0bGaaiykaaaa@3917@ , with components in a fixed basis { i,j,k } MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaaca WHPbGaaiilaiaahQgacaGGSaGaaC4AaaGaay5Eaiaaw2haaaaa@3C60@ , then

F(t)= F x (t)i+ F y (t)j+ F z (t)k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiaacI cacaWG0bGaaiykaiabg2da9iaadAeadaWgaaWcbaGaamiEaaqabaGc caGGOaGaamiDaiaacMcacaWHPbGaey4kaSIaamOramaaBaaaleaaca WG5baabeaakiaacIcacaWG0bGaaiykaiaahQgacqGHRaWkcaWGgbWa aSbaaSqaaiaadQhaaeqaaOGaaiikaiaadshacaGGPaGaaC4Aaaaa@4BAD@

Let

r(t)=x(t)i+y(t)j+z(t)k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiaacI cacaWG0bGaaiykaiabg2da9iaadIhacaGGOaGaamiDaiaacMcacaWH PbGaey4kaSIaamyEaiaacIcacaWG0bGaaiykaiaahQgacqGHRaWkca WG6bGaaiikaiaadshacaGGPaGaaC4Aaaaa@48D6@

denote the position vector of the particle, and

M(t)=r(t)×F(t)=( y(t) F z (t)z(t) F y (t) )i+( z(t) F x (t)x(t) F z (t) )j+( x(t) F y (t)y(t) F x (t) )k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytaiaacI cacaWG0bGaaiykaiabg2da9iaahkhacaGGOaGaamiDaiaacMcacqGH xdaTcaWHgbGaaiikaiaadshacaGGPaGaeyypa0ZaaeWaaeaacaWG5b GaaiikaiaadshacaGGPaGaamOramaaBaaaleaacaWG6baabeaakiaa cIcacaWG0bGaaiykaiabgkHiTiaadQhacaGGOaGaamiDaiaacMcaca WGgbWaaSbaaSqaaiaadMhaaeqaaOGaaiikaiaadshacaGGPaaacaGL OaGaayzkaaGaaCyAaiabgUcaRmaabmaabaGaamOEaiaacIcacaWG0b GaaiykaiaadAeadaWgaaWcbaGaamiEaaqabaGccaGGOaGaamiDaiaa cMcacqGHsislcaWG4bGaaiikaiaadshacaGGPaGaamOramaaBaaale aacaWG6baabeaakiaacIcacaWG0bGaaiykaaGaayjkaiaawMcaaiaa hQgacqGHRaWkdaqadaqaaiaadIhacaGGOaGaamiDaiaacMcacaWGgb WaaSbaaSqaaiaadMhaaeqaaOGaaiikaiaadshacaGGPaGaeyOeI0Ia amyEaiaacIcacaWG0bGaaiykaiaadAeadaWgaaWcbaGaamiEaaqaba GccaGGOaGaamiDaiaacMcaaiaawIcacaGLPaaacaWHRbaaaa@7D74@

the moment of the force about the origin.

 

The Angular Impulse exerted by the force about O during a time interval t 0 t t 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaakiabgsMiJkaadshacqGHKjYOcaWG0bWaaSba aSqaaiaaigdaaeqaaaaa@3E22@  is defined as

A= t 0 t 1 M(t)dt = t 0 t 1 r(t)×F(t)dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyqaiabg2 da9maapehabaGaaCytaiaacIcacaWG0bGaaiykaiaadsgacaWG0baa leaacaWG0bWaaSbaaWqaaiaaicdaaeqaaaWcbaGaamiDamaaBaaame aacaaIXaaabeaaa0Gaey4kIipakiabg2da9maapehabaGaaCOCaiaa cIcacaWG0bGaaiykaiabgEna0kaahAeacaGGOaGaamiDaiaacMcaca WGKbGaamiDaaWcbaGaamiDamaaBaaameaacaaIWaaabeaaaSqaaiaa dshadaWgaaadbaGaaGymaaqabaaaniabgUIiYdaaaa@54B1@

The angular impulse is a vector, and can be expressed as components in a basis

A= A x i+ A y j+ A z k A x = t 0 t 1 ( y(t) F z (t)z(t) F y (t) )dt A y = t 0 t 1 ( z(t) F x (t)x(t) F z (t) )dt A z = t 0 t 1 ( x(t) F y (t)y(t) F x (t) )dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWHbb Gaeyypa0JaamyqamaaBaaaleaacaWG4baabeaakiaahMgacqGHRaWk caWGbbWaaSbaaSqaaiaadMhaaeqaaOGaaCOAaiabgUcaRiaadgeada WgaaWcbaGaamOEaaqabaGccaWHRbaabaGaamyqamaaBaaaleaacaWG 4baabeaakiabg2da9maapehabaWaaeWaaeaacaWG5bGaaiikaiaads hacaGGPaGaamOramaaBaaaleaacaWG6baabeaakiaacIcacaWG0bGa aiykaiabgkHiTiaadQhacaGGOaGaamiDaiaacMcacaWGgbWaaSbaaS qaaiaadMhaaeqaaOGaaiikaiaadshacaGGPaaacaGLOaGaayzkaaGa amizaiaadshaaSqaaiaadshadaWgaaadbaGaaGimaaqabaaaleaaca WG0bWaaSbaaWqaaiaaigdaaeqaaaqdcqGHRiI8aOGaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVd qaaiaadgeadaWgaaWcbaGaamyEaaqabaGccqGH9aqpdaWdXbqaamaa bmaabaGaamOEaiaacIcacaWG0bGaaiykaiaadAeadaWgaaWcbaGaam iEaaqabaGccaGGOaGaamiDaiaacMcacqGHsislcaWG4bGaaiikaiaa dshacaGGPaGaamOramaaBaaaleaacaWG6baabeaakiaacIcacaWG0b GaaiykaaGaayjkaiaawMcaaiaadsgacaWG0baaleaacaWG0bWaaSba aWqaaiaaicdaaeqaaaWcbaGaamiDamaaBaaameaacaaIXaaabeaaa0 Gaey4kIipakiaaykW7aeaacaWGbbWaaSbaaSqaaiaadQhaaeqaaOGa eyypa0Zaa8qCaeaadaqadaqaaiaadIhacaGGOaGaamiDaiaacMcaca WGgbWaaSbaaSqaaiaadMhaaeqaaOGaaiikaiaadshacaGGPaGaeyOe I0IaamyEaiaacIcacaWG0bGaaiykaiaadAeadaWgaaWcbaGaamiEaa qabaGccaGGOaGaamiDaiaacMcaaiaawIcacaGLPaaacaWGKbGaamiD aaWcbaGaamiDamaaBaaameaacaaIWaaabeaaaSqaaiaadshadaWgaa adbaGaaGymaaqabaaaniabgUIiYdaaaaa@A9D4@

 

If you know the moment as a function of time, you can calculate its angular impulse using simple calculus.  For example for a constant moment, with vector value M 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytamaaBa aaleaacaaIWaaabeaaaaa@37B2@ , the impulse is A= M 0 ( t 1 t 0 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyqaiabg2 da9iaah2eadaWgaaWcbaGaaGimaaqabaGccaGGOaGaamiDamaaBaaa leaacaaIXaaabeaakiabgkHiTiaadshadaWgaaWcbaGaaGimaaqaba GccaGGPaaaaa@3FA5@

 

 

 

4.3.2 Definition of the angular momentum of a particle

 

The angular momentum of a particle is simply the cross product of the particle’s position vector with its linear momentum

h=r×p=r×mv MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiAaiabg2 da9iaahkhacqGHxdaTcaWHWbGaeyypa0JaaCOCaiabgEna0kaad2ga caWH2baaaa@4201@

The angular momentum is a vector MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  the direction of the vector is perpendicular to its velocity and its position vectors.

 

 

4.3.3 Angular impulse MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieqajugybabaaa aaaaaapeGaa83eGaaa@3723@  Angular Momentum relation for a single particle

 

* Consider a particle that is subjected to a force F(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiaacI cacaWG0bGaaiykaaaa@3917@  for a time interval t 0 t t 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaakiabgsMiJkaadshacqGHKjYOcaWG0bWaaSba aSqaaiaaigdaaeqaaaaa@3E22@ .

* Let r(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiaacI cacaWG0bGaaiykaaaa@3943@  denote the position vector of the particle

* Let M(t)=r(t)×F(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytaiaacI cacaWG0bGaaiykaiabg2da9iaahkhacaGGOaGaamiDaiaacMcacqGH xdaTcaWHgbGaaiikaiaadshacaGGPaaaaa@42A9@  denote the moment of F about the origin

* Let A denote the angular impulse exerted on the particle

* Let h 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiAamaaBa aaleaacaaIWaaabeaaaaa@37CD@  denote the angular momentum at time t 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG0bWaaSbaaS qaaiaaicdaaeqaaaaa@3755@  

* Let h 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiAamaaBa aaleaacaaIXaaabeaaaaa@37CE@  denote the angular momentum at time t 1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG0bWaaSbaaS qaaiaaigdaaeqaaaaa@3756@  

 

 

The momentum conservation equation can be expressed either in differential or integral form. 

1.      In differential form

M= dh dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytaiabg2 da9maalaaabaGaamizaiaahIgaaeaacaWGKbGaamiDaaaaaaa@3B9E@

2.      In integral form

A= h 1 h 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyqaiabg2 da9iaahIgadaWgaaWcbaGaaGymaaqabaGccqGHsislcaWHObWaaSba aSqaaiaaicdaaeqaaaaa@3C6C@

 

 

Proof: Although it’s not obvious, these are just another way of writing Newton’s laws of motion.  To show this, we’ll derive the differential form.  Start with Newton’s law

F=ma=m dv dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iaad2gacaWHHbGaeyypa0JaamyBamaalaaabaGaamizaiaahAha aeaacaWGKbGaamiDaaaaaaa@3F79@

Take the cross product of both sides with r

r×F=r×m dv dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabgE na0kaahAeacqGH9aqpcaWHYbGaey41aqRaamyBamaalaaabaGaamiz aiaahAhaaeaacaWGKbGaamiDaaaaaaa@42BB@

Note that v×mv=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODaiabgE na0kaad2gacaWH2bGaeyypa0JaaCimaaaa@3CBC@  since the cross product of two parallel vectors is zero.  We can add this to the right hand side, which shows that

r×F=r×m dv dt +v×mv=r×m dv dt + dr dt ×mv= d dt ( r×mv )= dh dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabgE na0kaahAeacqGH9aqpcaWHYbGaey41aqRaamyBamaalaaabaGaamiz aiaahAhaaeaacaWGKbGaamiDaaaacqGHRaWkcaWH2bGaey41aqRaam yBaiaahAhacqGH9aqpcaWHYbGaey41aqRaamyBamaalaaabaGaamiz aiaahAhaaeaacaWGKbGaamiDaaaacqGHRaWkdaWcaaqaaiaadsgaca WHYbaabaGaamizaiaadshaaaGaey41aqRaamyBaiaahAhacqGH9aqp daWcaaqaaiaadsgaaeaacaWGKbGaamiDaaaadaqadaqaaiaahkhacq GHxdaTcaWGTbGaaCODaaGaayjkaiaawMcaaiabg2da9maalaaabaGa amizaiaahIgaaeaacaWGKbGaamiDaaaaaaa@6987@

This yields the required relation.

 

Angular momentum conservation: For the special case where the force is parallel to r, the moment of the force acting on the particle is zero ( r×F=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiabgE na0kaahAeacqGH9aqpcaWHWaaaaa@3B96@  ), and angular momentum is constant

d dt ( r×mv )=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbaabaGaamizaiaadshaaaWaaeWaaeaacaWHYbGaey41aqRaamyB aiaahAhaaiaawIcacaGLPaaacqGH9aqpcaWHWaaaaa@411C@

 

4.3.4 Examples using Angular Impulse MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieqajugybabaaa aaaaaapeGaa83eGaaa@3723@  Angular Momentum relations for a single particle

 

The angular impulse-angular momentum equations are particularly helpful when you need to solve problems where particles are subjected to a single force, which acts through a fixed point. They can also be used to analyze rotational motion of a massless frame containing one or more particles.

Example 1: Orbital motion.  A satellite is launched into a geostationary transfer orbit by the ARIANE V launch facility.  At its perigee (the point where the satellite is closest to the earth) the satellite has speed 10.2km/sec and altitude 250km.  At apogee (the point where the satellite is furthest from the earth) the satellite has altitude 35950 km.  Calculate the speed of the satellite at apogee.

 

Assumptions:

1.      We assume that the only force acting on the satellite is the gravitational attraction of the earth

2.      The earth’s radius is 6378.145km

 

Calculation:

1.      Since the gravitational force on the satellite always acts towards the center of the earth, angular momentum about the earth’s center is conserved.

2.      At both perigee and apogee, the velocity vector of the satellite must be perpendicular to its position vector. To see this, note that at the point where the satellite is closest and furthest from the earth, the distance to the earth is a max or min, and so the derivative of the distance of the satellite from the earth must vanish, i.e.

d dt ( | r | )= d dt ( rr )=0 1 2 1 rr ( r dr dt + dr dt r )=0rv=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WGKbaabaGaamizaiaadshaaaWaaeWaaeaadaabdaqaaiaahkhaaiaa wEa7caGLiWoaaiaawIcacaGLPaaacqGH9aqpdaWcaaqaaiaadsgaae aacaWGKbGaamiDaaaadaqadaqaamaakaaabaGaaCOCaiabgwSixlaa hkhaaSqabaaakiaawIcacaGLPaaacqGH9aqpcaWHWaGaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlabgkDiEpaalaaabaGaaGymaaqaaiaaikdaaaWaaSaaaeaacaaIXa aabaWaaOaaaeaacaWHYbGaeyyXICTaaCOCaaWcbeaaaaGcdaqadaqa aiaahkhacqGHflY1daWcaaqaaiaadsgacaWHYbaabaGaamizaiaads haaaGaey4kaSYaaSaaaeaacaWGKbGaaCOCaaqaaiaadsgacaWG0baa aiabgwSixlaahkhaaiaawIcacaGLPaaacqGH9aqpcaaIWaGaeyO0H4 TaaCOCaiabgwSixlaahAhacqGH9aqpcaaIWaaaaa@7D49@

where we have used the chain rule to evaluate the time derivative of rr MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaOaaaeaaca WHYbGaeyyXICTaaCOCaaWcbeaaaaa@3A51@ .  Recall that if the dot product of two vectors vanishes, they are mutually perpendicular. We take a coordinate system with i and j in the plane of the orbit, and k perpendicular to the orbit. 

3.      We take the satellite orbit to lie in the i , j plane with k perpendicular to the orbit.  The angular momentum at apogee and perigee is then

h p = r p ×m v p = r p m v p k h a = r a ×m v a = r a m v a k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWHOb WaaSbaaSqaaiaadchaaeqaaOGaeyypa0JaaCOCamaaBaaaleaacaWG WbaabeaakiabgEna0kaad2gacaWH2bWaaSbaaSqaaiaadchaaeqaaO Gaeyypa0JaamOCamaaBaaaleaacaWGWbaabeaakiaad2gacaWG2bWa aSbaaSqaaiaadchaaeqaaOGaaC4AaaqaaiaahIgadaWgaaWcbaGaam yyaaqabaGccqGH9aqpcaWHYbWaaSbaaSqaaiaadggaaeqaaOGaey41 aqRaamyBaiaahAhadaWgaaWcbaGaamyyaaqabaGccqGH9aqpcaWGYb WaaSbaaSqaaiaadggaaeqaaOGaamyBaiaadAhadaWgaaWcbaGaamyy aaqabaGccaWHRbaaaaa@5910@

where r a , r p MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOCamaaBa aaleaacaWGHbaabeaakiaacYcacaWGYbWaaSbaaSqaaiaadchaaeqa aaaa@3AD1@  are the distance of the satellite from the center of the earth at apogee and perigee, and v a , v p MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaWGHbaabeaakiaacYcacaWG2bWaaSbaaSqaaiaadchaaeqa aaaa@3AD9@  are the corresponding speeds.

4.      Since angular momentum is conserved it follows that

h p = h a r p m v p = r a m v a v a = r p v p / r a MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiAamaaBa aaleaacaWGWbaabeaakiabg2da9iaahIgadaWgaaWcbaGaamyyaaqa baGccqGHshI3caWGYbWaaSbaaSqaaiaadchaaeqaaOGaamyBaiaadA hadaWgaaWcbaGaamiCaaqabaGccqGH9aqpcaWGYbWaaSbaaSqaaiaa dggaaeqaaOGaamyBaiaadAhadaWgaaWcbaGaamyyaaqabaGccaaMc8 UaaGPaVlaaykW7cqGHshI3caWG2bWaaSbaaSqaaiaadggaaeqaaOGa eyypa0JaamOCamaaBaaaleaacaWGWbaabeaakiaadAhadaWgaaWcba GaamiCaaqabaGccaGGVaGaamOCamaaBaaaleaacaWGHbaabeaaaaa@59FD@

5.      Substituting numbers yields 1.6 km/s

 

 

 

 

Example 2: More orbital motion. The orbit for a satellite is normally specified by a set of angles specifying the inclination of the orbit, and by quoting the distance of the satellite from the earths center at apogee and perigee r a , r p MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOCamaaBa aaleaacaWGHbaabeaakiaacYcacaWGYbWaaSbaaSqaaiaadchaaeqa aaaa@3AD1@ .   It is possible to calculate the speed of the satellite at perigee and apogee from this information.

 

Calculation

1.      From the previous example, we know that the distances and velocities are related by

r p m v p = r a m v a MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOCamaaBa aaleaacaWGWbaabeaakiaad2gacaWG2bWaaSbaaSqaaiaadchaaeqa aOGaeyypa0JaamOCamaaBaaaleaacaWGHbaabeaakiaad2gacaWG2b WaaSbaaSqaaiaadggaaeqaaaaa@4148@

2.      The system is conservative, so the total energy of the system is conserved.

3.      The potential energies when the satellite is at perigee and apogee are

V p = GMm r p V a = GMm r a MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa aaleaacaWGWbaabeaakiabg2da9iabgkHiTmaalaaabaGaam4raiaa d2eacaWGTbaabaGaamOCamaaBaaaleaacaWGWbaabeaaaaGccaaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caWGwbWaaSbaaSqaaiaadggaaeqa aOGaeyypa0JaeyOeI0YaaSaaaeaacaWGhbGaamytaiaad2gaaeaaca WGYbWaaSbaaSqaaiaadggaaeqaaaaaaaa@67AB@

4.      The kinetic energies of the satellite at perigee and apogee are

T p = 1 2 m v p 2 T a = 1 2 m v a 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBa aaleaacaWGWbaabeaakiabg2da9maalaaabaGaaGymaaqaaiaaikda aaGaamyBaiaadAhadaqhaaWcbaGaamiCaaqaaiaaikdaaaGccaaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caWGubWaaSbaaSqaaiaadggaaeqa aOGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaacaWGTbGaamODam aaDaaaleaacaWGHbaabaGaaGOmaaaaaaa@6701@

5.      The kinetic energy of the earth can be assumed to be constant.  Energy conservation therefore shows that

1 2 m v p 2 GMm r p = 1 2 m v a 2 GMm r a MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca aIXaaabaGaaGOmaaaacaWGTbGaamODamaaDaaaleaacaWGWbaabaGa aGOmaaaakiabgkHiTmaalaaabaGaam4raiaad2eacaWGTbaabaGaam OCamaaBaaaleaacaWGWbaabeaaaaGccaaMc8Uaeyypa0JaaGPaVpaa laaabaGaaGymaaqaaiaaikdaaaGaamyBaiaadAhadaqhaaWcbaGaam yyaaqaaiaaikdaaaGccqGHsisldaWcaaqaaiaadEeacaWGnbGaamyB aaqaaiaadkhadaWgaaWcbaGaamyyaaqabaaaaaaa@5000@

6.      The results of (1) and (5) give two equations that can be solved for v a , v p MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaWGHbaabeaakiaacYcacaWG2bWaaSbaaSqaaiaadchaaeqa aaaa@3AD9@  in terms of known parameters.  For example, (1) shows that v a = v p r p / r a MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaWGHbaabeaakiabg2da9iaadAhadaWgaaWcbaGaamiCaaqa baGccaWGYbWaaSbaaSqaaiaadchaaeqaaOGaai4laiaadkhadaWgaa WcbaGaamyyaaqabaaaaa@4017@  - this can be substituted into (5) to see that

1 2 m v p 2 GMm r p = 1 2 m v p 2 r p 2 r a 2 GMm r a v p 2 r a 2 r p 2 r a 2 =2GM( 1 r p 1 r a ) v p = 2GM r a r p ( r a + r p ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWcaa qaaiaaigdaaeaacaaIYaaaaiaad2gacaWG2bWaa0baaSqaaiaadcha aeaacaaIYaaaaOGaeyOeI0YaaSaaaeaacaWGhbGaamytaiaad2gaae aacaWGYbWaaSbaaSqaaiaadchaaeqaaaaakiaaykW7cqGH9aqpcaaM c8+aaSaaaeaacaaIXaaabaGaaGOmaaaacaWGTbGaamODamaaDaaale aacaWGWbaabaGaaGOmaaaakmaalaaabaGaamOCamaaDaaaleaacaWG WbaabaGaaGOmaaaaaOqaaiaadkhadaqhaaWcbaGaamyyaaqaaiaaik daaaaaaOGaeyOeI0YaaSaaaeaacaWGhbGaamytaiaad2gaaeaacaWG YbWaaSbaaSqaaiaadggaaeqaaaaaaOqaaiabgkDiElaadAhadaqhaa WcbaGaamiCaaqaaiaaikdaaaGcdaWcaaqaaiaadkhadaqhaaWcbaGa amyyaaqaaiaaikdaaaGccqGHsislcaWGYbWaa0baaSqaaiaadchaae aacaaIYaaaaaGcbaGaamOCamaaDaaaleaacaWGHbaabaGaaGOmaaaa aaGccqGH9aqpcaaIYaGaam4raiaad2eadaqadaqaamaalaaabaGaaG ymaaqaaiaadkhadaWgaaWcbaGaamiCaaqabaaaaOGaeyOeI0YaaSaa aeaacaaIXaaabaGaamOCamaaBaaaleaacaWGHbaabeaaaaaakiaawI cacaGLPaaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlabgkDiElaadAhadaWgaaWcbaGaamiCaaqabaGccqGH9a qpdaGcaaqaamaalaaabaGaaGOmaiaadEeacaWGnbGaamOCamaaBaaa leaacaWGHbaabeaaaOqaaiaadkhadaWgaaWcbaGaamiCaaqabaGcca GGOaGaamOCamaaBaaaleaacaWGHbaabeaakiabgUcaRiaadkhadaWg aaWcbaGaamiCaaqabaGccaGGPaaaaaWcbeaaaaaa@8F47@

Similarly, at apogee

v a = 2GM r p r a ( r a + r p ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa aaleaacaWGHbaabeaakiabg2da9maakaaabaWaaSaaaeaacaaIYaGa am4raiaad2eacaWGYbWaaSbaaSqaaiaadchaaeqaaaGcbaGaamOCam aaBaaaleaacaWGHbaabeaakiaacIcacaWGYbWaaSbaaSqaaiaadgga aeqaaOGaey4kaSIaamOCamaaBaaaleaacaWGWbaabeaakiaacMcaaa aaleqaaaaa@463D@

 

 


4.4 Summary of definitions and equations

 

 

Work, Power, Kinetic Energy

* The Power developed by a force, (or the rate of work done by the force) is

P=Fv MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 da9iaahAeacqGHflY1caWH2baaaa@3BE9@ .

 

* The work done by the force during a time interval t 0 t t 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaakiabgsMiJkaadshacqGHKjYOcaWG0bWaaSba aSqaaiaaigdaaeqaaaaa@3F29@  is

W= t 0 t 1 P dt= t 0 t 1 Fv dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadEfacqGH9a qpdaWdXbqaaiaadcfaaSqaaiaadshadaWgaaadbaGaaGimaaqabaaa leaacaWG0bWaaSbaaWqaaiaaigdaaeqaaaqdcqGHRiI8aOGaamizai aadshacqGH9aqpdaWdXbqaaiaahAeacqGHflY1caWH2baaleaacaWG 0bWaaSbaaWqaaiaaicdaaeqaaaWcbaGaamiDamaaBaaameaacaaIXa aabeaaa0Gaey4kIipakiaadsgacaWG0baaaa@4E50@

The work done by the force can also be calculated by integrating the force vector along the path traveled by the force, as

W= r 0 r 1 Fdr MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadEfacqGH9a qpdaWdXbqaaiaahAeacqGHflY1caWGKbGaaCOCaaWcbaGaaCOCamaa BaaameaacaaIWaaabeaaaSqaaiaahkhadaWgaaadbaGaaGymaaqaba aaniabgUIiYdaaaa@4351@

where r 0 , r 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCamaaBa aaleaacaaIWaaabeaakiaacYcacaWHYbWaaSbaaSqaaiaaigdaaeqa aaaa@3A73@  are the initial and final positions of the force.

 

* The Kinetic Energy of a particle is

T= 1 2 m | v | 2 = 1 2 m( v x 2 + v y 2 + v z 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGubGaeyypa0 ZaaSaaaeaacaaIXaaabaGaaGOmaaaacaWGTbWaaqWaaeaacaWH2baa caGLhWUaayjcSdWaaWbaaSqabeaacaaIYaaaaOGaeyypa0ZaaSaaae aacaaIXaaabaGaaGOmaaaacaWGTbWaaeWaaeaacaWG2bWaa0baaSqa aiaadIhaaeaacaaIYaaaaOGaey4kaSIaamODamaaDaaaleaacaWG5b aabaGaaGOmaaaakiabgUcaRiaadAhadaqhaaWcbaGaamOEaaqaaiaa ikdaaaaakiaawIcacaGLPaaaaaa@4E72@  

 

 

Conservative forces and potential energy

A force (or pair of forces) is conservative if the work done by the force when it moves between any two points is the same for all paths joining the two points

 

The potential energy of a conservative force is defined as the negative of the work done by the force in moving from some arbitrary initial position r 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhadaWgaa WcbaGaaGimaaqabaaaaa@381A@  to a new position r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahkhaaaa@3734@ , i.e.

V(r)= r 0 r Fdr +constant MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfacaGGOa GaaCOCaiaacMcacqGH9aqpcqGHsisldaWdXbqaaiaahAeacqGHflY1 caWGKbGaaCOCaaWcbaGaaCOCamaaBaaameaacaaIWaaabeaaaSqaai aahkhaa0Gaey4kIipakiabgUcaRiaabogacaqGVbGaaeOBaiaaboha caqG0bGaaeyyaiaab6gacaqG0baaaa@4E17@

Alternatively

F=grad(V) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcqGHsislciGGNbGaaiOCaiaacggacaGGKbGaciikaiaadAfacaGG Paaaaa@3EE1@

F x i+ F y j+ F z k=( V x i+ V y j+ V z k ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAeadaWgaa WcbaGaamiEaaqabaGccaWHPbGaey4kaSIaamOramaaBaaaleaacaWG 5baabeaakiaahQgacqGHRaWkcaWGgbWaaSbaaSqaaiaadQhaaeqaaO GaaC4Aaiabg2da9iabgkHiTmaabmaabaWaaSaaaeaacqGHciITcaWG wbaabaGaeyOaIyRaamiEaaaacaWHPbGaey4kaSYaaSaaaeaacqGHci ITcaWGwbaabaGaeyOaIyRaamyEaaaacaWHQbGaey4kaSYaaSaaaeaa cqGHciITcaWGwbaabaGaeyOaIyRaamOEaaaacaWHRbaacaGLOaGaay zkaaaaaa@570B@

 

 

 

 

Table of potential energy relations

 

In practice, however, we rarely need to do the integrals to calculate the potential energy of a force, because there are very few different kinds of force.  For most engineering calculations the potential energy formulas listed in the table below are sufficient.

 

 

Type of force

Force vector

Potential energy

 

Gravity acting on a particle near earths surface

F=mgj MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcqGHsislcaWGTbGaam4zaiaahQgaaaa@3BCC@

V=mgy MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfacqGH9a qpcaWGTbGaam4zaiaadMhaaaa@3AF6@

Gravitational force exerted on mass m by mass M at the origin

F= GMm r 3 r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcqGHsisldaWcaaqaaiaadEeacaWGnbGaamyBaaqaaiaadkhadaah aaWcbeqaaiaaiodaaaaaaOGaaCOCaaaa@3E81@

V= GMm r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfacqGH9a qpcqGHsisldaWcaaqaaiaadEeacaWGnbGaamyBaaqaaiaadkhaaaaa aa@3C9E@

Force exerted by a spring with stiffness k and unstretched length L 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamitamaaBa aaleaacaaIWaaabeaaaaa@38B5@

F=k(r L 0 ) r r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaahAeacqGH9a qpcqGHsislcaWGRbGaaiikaiaadkhacqGHsislcaWGmbWaaSbaaSqa aiaaicdaaeqaaOGaaiykamaalaaabaGaaCOCaaqaaiaadkhaaaaaaa@40EB@

V= 1 2 k ( r L 0 ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfacqGH9a qpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaadUgadaqadaqaaiaadkha cqGHsislcaWGmbWaaSbaaSqaaiaaicdaaeqaaaGccaGLOaGaayzkaa WaaWbaaSqabeaacaaIYaaaaaaa@40A8@

Force acting between two charged particles

F= Q 1 Q 2 4πε r 3 r MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9maalaaabaGaamyuamaaBaaaleaacaaIXaaabeaakiaadgfadaWg aaWcbaGaaGOmaaqabaaakeaacaaI0aGaeqiWdaNaeqyTduMaamOCam aaCaaaleqabaGaaG4maaaaaaGccaWHYbaaaa@4273@

V= Q 1 Q 2 4πεr MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 da9maalaaabaGaamyuamaaBaaaleaacaaIXaaabeaakiaadgfadaWg aaWcbaGaaGOmaaqabaaakeaacaaI0aGaeqiWdaNaeqyTduMaamOCaa aaaaa@4090@

Force exerted by one molecule of a noble gas (e.g. He, Ar, etc) on another (Lennard Jones potential). a is the equilibrium spacing between molecules, and E is the energy of the bond.

F=12 E a [ 2 ( a r ) 13 ( a r ) 7 ] r r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiabg2 da9iabgkHiTiaaigdacaaIYaWaaSaaaeaacaWGfbaabaGaamyyaaaa daWadaqaaiaaikdadaqadaqaamaalaaabaGaamyyaaqaaiaadkhaaa aacaGLOaGaayzkaaWaaWbaaSqabeaacaaIXaGaaG4maaaakiabgkHi TmaabmaabaWaaSaaaeaacaWGHbaabaGaamOCaaaaaiaawIcacaGLPa aadaahaaWcbeqaaiaaiEdaaaaakiaawUfacaGLDbaacaaMc8UaaGPa VpaalaaabaGaaCOCaaqaaiaadkhaaaaaaa@4F3C@

E[ ( a r ) 12 2 ( a r ) 6 ] MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyramaadm aabaWaaeWaaeaadaWcaaqaaiaadggaaeaacaWGYbaaaaGaayjkaiaa wMcaamaaCaaaleqabaGaaGymaiaaikdaaaGccqGHsislcaaIYaWaae WaaeaadaWcaaqaaiaadggaaeaacaWGYbaaaaGaayjkaiaawMcaamaa CaaaleqabaGaaGOnaaaaaOGaay5waiaaw2faaiaaykW7caaMc8oaaa@4809@

 

 

 

 

Power-Work-kinetic energy relations for a single particle

 

The Power-kinetic energy relation for the particle states that the rate of work done by F is equal to the rate of change of kinetic energy of the particle, i.e.

P= dT dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 da9maalaaabaGaamizaiaadsfaaeaacaWGKbGaamiDaaaaaaa@3C8C@

 

 

The Work-kinetic energy relation for a particle says that the total work done by the force F on the particle is equal to the change in the kinetic energy of the particle.

W=T T 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4vaiabg2 da9iaadsfacqGHsislcaWGubWaaSbaaSqaaiaaicdaaeqaaaaa@3C64@

 

Power-Work-kinetic energy relations for a conservative system of particles

 

A system is conservative if a potential energy can be defined for all internal forces in the system

 

 

 

  1. At some time t 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaaaaa@38DD@  the system has and kinetic energy T 0 TOT MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaDa aaleaacaaIWaaabaGaamivaiaad+eacaWGubaaaaaa@3B43@
  2. At some later time t 1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIXaaabeaaaaa@38DE@  the system has kinetic energy T 1 TOT MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaDa aaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaaaa@3B45@ .
  3. Let V 0 TOT MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaDa aaleaacaaIWaaabaGaamivaiaad+eacaWGubaaaaaa@3B46@  denote the potential energy of the internal forces at time t 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaaaaa@38DD@
  4. Let V 1 TOT MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaDa aaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaaaa@3B47@  denote the potential energy of the internal forces at time t MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaaaa@37F7@
  5. Let Δ W ext MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam 4vamaaCaaaleqabaGaamyzaiaadIhacaWG0baaaaaa@3C4D@  denote the total work done on the system between t 0 t t 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaakiabgsMiJkaadshacqGHKjYOcaWG0bWaaSba aSqaaiaaigdaaeqaaaaa@3E22@

 

Power Energy Relation:  P ext = d dt ( T 1 TOT + V 1 TOT ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaCa aaleqabaGaamyzaiaadIhacaWG0baaaOGaeyypa0ZaaSaaaeaacaWG KbaabaGaamizaiaadshaaaWaaeWaaeaacaWGubWaa0baaSqaaiaaig daaeaacaWGubGaam4taiaadsfaaaGccqGHRaWkcaWGwbWaa0baaSqa aiaaigdaaeaacaWGubGaam4taiaadsfaaaaakiaawIcacaGLPaaaaa a@49D9@

Work Energy Relation:    Δ W ext = T 1 TOT + V 1 TOT ( T 0 TOT + V 0 TOT ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam 4vamaaCaaaleqabaGaamyzaiaadIhacaWG0baaaOGaeyypa0Jaamiv amaaDaaaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaOGaey4kaS IaamOvamaaDaaaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaOGa eyOeI0YaaeWaaeaacaWGubWaa0baaSqaaiaaicdaaeaacaWGubGaam 4taiaadsfaaaGccqGHRaWkcaWGwbWaa0baaSqaaiaaicdaaeaacaWG ubGaam4taiaadsfaaaaakiaawIcacaGLPaaaaaa@52DC@

Energy conservation law  Δ W ext =0 T 1 TOT + V 1 TOT =( T 0 TOT + V 0 TOT ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKc9crFfpeea0xh9v8qiW7rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam 4vamaaCaaaleqabaGaamyzaiaadIhacaWG0baaaOGaeyypa0JaaGim aiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaeyO0H4TaamivamaaDa aaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaOGaey4kaSIaamOv amaaDaaaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaOGaeyypa0 ZaaeWaaeaacaWGubWaa0baaSqaaiaaicdaaeaacaWGubGaam4taiaa dsfaaaGccqGHRaWkcaWGwbWaa0baaSqaaiaaicdaaeaacaWGubGaam 4taiaadsfaaaaakiaawIcacaGLPaaaaaa@5DC3@

 

Linear impulse of a force

= t 0 t 1 F(t)dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyeHeSaey ypa0Zaa8qCaeaacaWHgbGaaiikaiaadshacaGGPaGaamizaiaadsha aSqaaiaadshadaWgaaadbaGaaGimaaqabaaaleaacaWG0bWaaSbaaW qaaiaaigdaaeqaaaqdcqGHRiI8aaaa@43B3@

 

Linear momentum of a particle

 

The linear momentum of a particle is simply the product of its mass and velocity

p=mv MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiCaiabg2 da9iaad2gacaWH2baaaa@39E6@

The linear momentum is a vector MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieaajugybabaaa aaaaaapeGaa83eGaaa@3722@  its direction is parallel to the velocity of the particle.

 

 

Impulse-momentum relations for a single particle

 

* Consider a particle that is subjected to a force F(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiaacI cacaWG0bGaaiykaaaa@3917@  for a time interval t 0 t t 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaakiabgsMiJkaadshacqGHKjYOcaWG0bWaaSba aSqaaiaaigdaaeqaaaaa@3E22@ .

* Let MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyeHemaaa@3775@  denote the impulse exerted by F on the particle

* Let p 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiCamaaBa aaleaacaaIWaaabeaaaaa@37D5@  denote the linear momentum of the particle at time t 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG0bWaaSbaaS qaaiaaicdaaeqaaaaa@3755@  .

* Let p 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiCamaaBa aaleaacaaIXaaabeaaaaa@37D6@  denote the linear momentum of the particle at time t 1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG0bWaaSbaaS qaaiaaigdaaeqaaaaa@3756@  .

 

In differential form

F(t)= dp dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiaacI cacaWG0bGaaiykaiabg2da9maalaaabaGaamizaiaahchaaeaacaWG KbGaamiDaaaaaaa@3DF1@

In integral form

= p 1 p 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyeHeSaey ypa0JaaCiCamaaBaaaleaacaaIXaaabeaakiabgkHiTiaahchadaWg aaWcbaGaaGimaaqabaaaaa@3D31@

This is the integral of Newton’s law of motion with respect to time.

 

 

Impulse-momentum relations for a system of particles

 

 

In differential form  particles F i ext (t)= d p TOT dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaabuaeaaca WHgbWaa0baaSqaaiaadMgaaeaacaWGLbGaamiEaiaadshaaaaabaGa amiCaiaadggacaWGYbGaamiDaiaadMgacaWGJbGaamiBaiaadwgaca WGZbaabeqdcqGHris5aOGaaiikaiaadshacaGGPaGaeyypa0ZaaSaa aeaacaWGKbGaaCiCamaaCaaaleqabaGaamivaiaad+eacaWGubaaaa GcbaGaamizaiaadshaaaaaaa@4F3F@

 

In integral form  TOT = p 1 TOT p 0 TOT MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyeHe8aaW baaSqabeaacaWGubGaam4taiaadsfaaaGccqGH9aqpcaWHWbWaa0ba aSqaaiaaigdaaeaacaWGubGaam4taiaadsfaaaGccqGHsislcaWHWb Waa0baaSqaaiaaicdaaeaacaWGubGaam4taiaadsfaaaaaaa@44FC@

 

If no external forces act on a system of particles, their total linear momentum is conserved, i.e.  

p 1 TOT = p 0 TOT MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiCamaaDa aaleaacaaIXaaabaGaamivaiaad+eacaWGubaaaOGaeyypa0JaaCiC amaaDaaaleaacaaIWaaabaGaamivaiaad+eacaWGubaaaaaa@3FD3@

 

Collisions

 

If no external impulse acts on two particles as they collide, their total momentum is conserved.

 

To calculate velocities after impact we define the restitution coefficient

 

 

Restitution coefficient for straight line motion

 

v A0 = v x A0 i, v B0 = v x B0 i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaicdaaaGccqGH9aqpcaWG2bWaa0baaSqaaiaa dIhaaeaacaWGbbGaaGimaaaakiaahMgacaGGSaGaaGPaVlaaykW7ca aMc8UaaGPaVlaahAhadaahaaWcbeqaaiaadkeacaaIWaaaaOGaeyyp a0JaamODamaaDaaaleaacaWG4baabaGaamOqaiaaicdaaaGccaWHPb aaaa@4D8E@  denote the velocities of A and B just before the collision

v A1 = v x A1 i, v B1 = v x B1 i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaigdaaaGccqGH9aqpcaWG2bWaa0baaSqaaiaa dIhaaeaacaWGbbGaaGymaaaakiaahMgacaGGSaGaaGPaVlaaykW7ca aMc8UaaGPaVlaahAhadaahaaWcbeqaaiaadkeacaaIXaaaaOGaeyyp a0JaamODamaaDaaaleaacaWG4baabaGaamOqaiaaigdaaaGccaWHPb aaaa@4D92@  denote the velocities of A and B just after the collision.

 

The velocities before and after impact are related by

v B1 v A1 =e( v B0 v A0 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaCa aaleqabaGaamOqaiaaigdaaaGccqGHsislcaWG2bWaaWbaaSqabeaa caWGbbGaaGymaaaakiabg2da9iabgkHiTiaadwgadaqadaqaaiaadA hadaahaaWcbeqaaiaadkeacaaIWaaaaOGaeyOeI0IaamODamaaCaaa leqabaGaamyqaiaaicdaaaaakiaawIcacaGLPaaaaaa@4702@

where e is the restitution coefficient.

 

 

 

 

Restitution coefficient for 3D frictionless collisions

 

Define

v A0 v B0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaicdaaaGccaaMc8UaaGPaVlaahAhadaahaaWc beqaaiaadkeacaaIWaaaaaaa@3E6F@  to denote velocities of the particles before collision

v A1 v B1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCODamaaCa aaleqabaGaamyqaiaaigdaaaGccaaMc8UaaGPaVlaahAhadaahaaWc beqaaiaadkeacaaIXaaaaaaa@3E71@  to denote velocities of the particles after collision

 

In addition, we let n be a unit vector normal to the common tangent plane at the point of contact (if the two colliding particles are spheres or disks the vector is parallel to the line joining their centers).

 

The velocities before and after collision are related by

 

( v B1 v A1 )=( v B0 v A0 )(1+e)[ ( v B0 v A0 )n ]n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca WH2bWaaWbaaSqabeaacaWGcbGaaGymaaaakiabgkHiTiaahAhadaah aaWcbeqaaiaadgeacaaIXaaaaaGccaGLOaGaayzkaaGaeyypa0Zaae WaaeaacaWH2bWaaWbaaSqabeaacaWGcbGaaGimaaaakiabgkHiTiaa hAhadaahaaWcbeqaaiaadgeacaaIWaaaaaGccaGLOaGaayzkaaGaey OeI0IaaiikaiaaigdacqGHRaWkcaWGLbGaaiykamaadmaabaWaaeWa aeaacaWH2bWaaWbaaSqabeaacaWGcbGaaGimaaaakiabgkHiTiaahA hadaahaaWcbeqaaiaadgeacaaIWaaaaaGccaGLOaGaayzkaaGaeyyX ICTaaCOBaaGaay5waiaaw2faaiaah6gaaaa@599E@

 

 

 

 

 

 

Angular impulse

 

The Angular Impulse exerted by the force about O during a time interval t 0 t t 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaakiabgsMiJkaadshacqGHKjYOcaWG0bWaaSba aSqaaiaaigdaaeqaaaaa@3E22@  is defined as

A= t 0 t 1 M(t)dt = t 0 t 1 r(t)×F(t)dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyqaiabg2 da9maapehabaGaaCytaiaacIcacaWG0bGaaiykaiaadsgacaWG0baa leaacaWG0bWaaSbaaWqaaiaaicdaaeqaaaWcbaGaamiDamaaBaaame aacaaIXaaabeaaa0Gaey4kIipakiabg2da9maapehabaGaaCOCaiaa cIcacaWG0bGaaiykaiabgEna0kaahAeacaGGOaGaamiDaiaacMcaca WGKbGaamiDaaWcbaGaamiDamaaBaaameaacaaIWaaabeaaaSqaaiaa dshadaWgaaadbaGaaGymaaqabaaaniabgUIiYdaaaa@54B1@

 

 

 

Angular momentum of a particle

h=r×p=r×mv MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiAaiabg2 da9iaahkhacqGHxdaTcaWHWbGaeyypa0JaaCOCaiabgEna0kaad2ga caWH2baaaa@4201@

 

 

Angular impulse MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaieqajugybabaaa aaaaaapeGaa83eGaaa@3723@  Angular Momentum relation for a single particle

 

* Consider a particle that is subjected to a force F(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOraiaacI cacaWG0bGaaiykaaaa@3917@  for a time interval t 0 t t 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaBa aaleaacaaIWaaabeaakiabgsMiJkaadshacqGHKjYOcaWG0bWaaSba aSqaaiaaigdaaeqaaaaa@3E22@ .

* Let r(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCOCaiaacI cacaWG0bGaaiykaaaa@3943@  denote the position vector of the particle

* Let M(t)=r(t)×F(t) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytaiaacI cacaWG0bGaaiykaiabg2da9iaahkhacaGGOaGaamiDaiaacMcacqGH xdaTcaWHgbGaaiikaiaadshacaGGPaaaaa@42A9@  denote the moment of F about the origin

* Let A denote the angular impulse exerted on the particle

* Let h 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiAamaaBa aaleaacaaIWaaabeaaaaa@37CD@  denote the angular momentum at time t 0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG0bWaaSbaaS qaaiaaicdaaeqaaaaa@3755@  

* Let h 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCiAamaaBa aaleaacaaIXaaabeaaaaa@37CE@  denote the angular momentum at time t 1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9 Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG0bWaaSbaaS qaaiaaigdaaeqaaaaa@3756@  

 

In differential form   M= dh dt MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCytaiabg2 da9maalaaabaGaamizaiaahIgaaeaacaWGKbGaamiDaaaaaaa@3B9E@

In integral form A= h 1 h 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaCyqaiabg2 da9iaahIgadaWgaaWcbaGaaGymaaqabaGccqGHsislcaWHObWaaSba aSqaaiaaicdaaeqaaaaa@3C6C@