1. Physics 430: Lecture 10 Energy of Interaction Dale E. Gary NJIT Physics Department

2. October 3, 2008 4.9 Energy of Interaction of Two Particles  Up to now we have been discussing the energy of a single particle. Now let’s look at the energy of two particles.  Imagine two isolated particles, alone in the universe (no external forces) interacting with each other. Particle 2, at position r 2 from the origin, exerts a force F 12 on particle 1, while the particle 1 at position r 1 exerts an equal and opposite force F 21 =  F 12 on particle 2. The distance between the two particles is then r = r 1  r 2 as shown in the figure.  For definiteness, consider a gravitational interaction, so that, as usual, r1r1 r2r2 r = r 1  r 2 1 2 O  From the definition of r and the vector difference of the two position vectors, we can rewrite this as:  The fact that this depends only on the difference of the two position vectors means that the force is translationally invariant.

3. October 3, 2008 Energy of Interaction of Two Particles-2  This fact of translational invariance—the fact that the force of interaction depends only on the distance between two particles, and not on their absolution position, allows us to greatly simplify the discussion by choosing a point, r 2, say, as the origin (even though the position of particle 2 is changing, i.e. the origin need not be a fixed point).  With this choice, our earlier discussion of force on a single particle applies. For example, if the force F 12 on particle 1 is conservative, then where the operator with respect to the coordinates r 1 = (x 1, y 1, z 1 ).  When the curl above is zero (i.e. the force is conservative) then we can define a potential energy  Note that U(r 1 ) is defined throughout space, but the force at particle 1 is the gradient of U(r 1 ) evaluated at the point where particle 1 is, i.e. at r 1.

4. October 3, 2008 Energy of Interaction of Two Particles-3  The foregoing is specific to the case when particle 2 is at the origin, but we can always translate to some other origin such that particle 2 is at position r 2, so that:  The trick is that we do not have to modify the operator, because a derivative like is unchanged by adding a constant to x 1.  We can then find the force on particle 2 by particle 1 as  What this says is that we can consider a single potential energy for the interaction between particles 1 and 2, and evaluate its gradient at the position of each particle to find the force on each particle.  Now consider the two particles moving through space under their mutual interaction, so that particle 1 moves through a distance dr 1 under the force F 12 and particle 2 moves through a distance dr 2 under the force F 21.

5. October 3, 2008 Total Energy of Interaction  By the work-KE theorem, there will be a change in kinetic energy of each of the particles:  We simply add these to find the total work done: where the last equality is because again F 21 =  F 12. This can be rewritten  If we rename (r1 – r2) as r, we finally have:  Let’s pause to appreciate this result. It says that if the force of interaction between two particles is conservative, then we can define a potential energy throughout space. As the particles move, the potential energy changes, but the change is just the negative of the change in kinetic energy that results from the interaction force.  Obviously, then, the total energy is conserved, i.e.  Notice that there are two kinetic energies, one for each particle, but only one potential energy arising from the configuration of the particles.

6. October 3, 2008 Elastic Collisions  Elastic collisions are ones that take place through conservative forces, so that no energy is lost (to heat or other mechanism). A collision between a proton and an electron, for example, occurs through the conservative electrostatic force. Collisions between two billiard balls is largely elastic, because the balls are made to act like stiff springs when they collide. Example 4.8: An Equal-Mass Elastic Collision  Statement of the problem: Consider an elastic collision between two particles of equal mass m 1 = m 2 = m, as shown in the figure. Prove that if particle 2 is initially at rest then the angle between the two outgoing velocities is  = 90 o.  Solution: Conservation of momentum gives: The fact that the collision is elastic means that Squaring the momentum equation, after eliminating m, gives: Comparing these, we see that  1 2 v1v1 v1v1 v2v2 2 1 v 1 and v 2 are 90 degrees apart

7. October 3, 2008 4.10 Generalizing to Multiple Particles  In the case of many particles (the text goes through the argument explicitly for 4 particles, and then for N particles), the same thing holds.  Taking the system of N particles interacting through conservative forces, one can consider the “internal potential energy” of the system, U int, and any “external potential energy”, U ext, imposed from outside.  The total potential energy is then where the sums are over all of the particles and the  >  in the second of the double sum ensures that we do not double count an interaction, e.g. U 12 and U 21.  The total kinetic energy is much simpler, just  One can calculate the force on particle  by meaning calculate the gradient of U at the position of the particle . The total energy E = T + U is conserved.

8. October 3, 2008 Rigid Bodies  The subtlety of can be illustrated with an example of rigid body rotation.  For a macroscopic rigid body, the number of particles (atoms) may be vast. However, because the body is rigid, the interaction forces between atoms is constant, and the particles do not shift position appreciably to change the internal potential energy.  Thus, the internal potential energy is constant and we can completely ignore it.  However, if the body is NOT rigid, and the shape or configuration of atoms changes, then there may be an appreciable change in internal kinetic energy that DOES have to be taken into account.  An example is a star that may expand or contract during its lifetime. The change in configuration of the internal layers of the star affect the potential energy, and hence also the kinetic energy.  Since total energy is conserved, for example, an expanding star must become cooler.

9. October 3, 2008 Example 4.9: A Cylinder Rolling Down an Incline  Statement of the problem: A uniform rigid cylinder of mass M, radius R rolls without slipping down a sloping track as shown in the figure. Use energy conservation to find its speed v when it reaches a vertical height h below its point of release.  Solution: The cylinder is a rigid body, so we may ignore its internal potential energy. The external potential energy is where y is the vertical position of its CM measured from any convenient location. The KE of a rotating (rolling) cylinder is where I is the moment of inertia, I = ½MR 2, and  is the angular velocity of rolling. Because the cylinder is rolling without slipping,  = v/R. Thus, the final kinetic energy is If the cylinder starts from rest, the change in kinetic energy is equal to minus the change in potential energy So the final speed is h