Taylor 4.10-5.1 (today) and midterm review (Monday).
In elastic collisions, energy and momentum are conserved.
Collisions are treated one at a time between two objects. If you have a situation where three objects collide, it should be treated as a sequence of collisions between two objects, i.e. first collide 1 and 2, then collide 2 and 3, then collide 3 with 1 (if necessary).
I've already covered this topic. Let's look again at the problem of an object rolling down an incline, but this time considering the energy relations.
We analysed this problem using forces and equations of motion earlier. Now, let's treat the problem using energy.
Consider a cylinder (mass M, radius R, and I = ½MR²) starts from rest and rolls without slipping down a hill of arbitrary construction -- just that it goes downhill. What is the velocity of the cylinder after it descends a height h?
Since the cylinder rolls without slipping, it doesn't lose any energy to friction, so the total mechanical energy of the cylinder is conserved. Its mechanical energy is the sum of gravitational potential energy and kinetic energy. Since the cylinder starts from rest, the initial energy is purely gravitational potential energy, Ei = U = Mghi, where hi is the initial height. The final energy is Ef = T + U = ½Mv² + ½Iω² + Mghf. For a rolling cylinder (or wheel), ω = v/R. Equating initial and final energies, we have Mghi = ½Mv² + ¼Mv² + Mghf = (3/4)Mv² + Mghf, and solving for v gives
You learned about Hooke's law for the force exerted by a stretched spring in basic physics, F = -kx. The importance of Hooke's law goes far beyond the potential uses of springs for closing doors. It is the result of the simple observation that for any system near a point of stable equilibrium, the force is approximately a linear restoring force (almost always). Let's see how this happens.
For simplicity, begin with a one-dimensional system. Without loss of generality, we can redefine x so that the equilibrium position is at x=0.There is a potential energy function for the system, and we can expand it in a Taylor series about x=0:
where I've used a prime to represent the derivative with respect to x, in analogy to using a dot to represent a derivative with respect to time. The linear term is zero since ∂U(x=0)/∂x = 0 (the definition of an equilibrium point) and higher order terms are neglected.
The force is the negative of the gradient of the potential. For our one dimensional case the gradient is simply the x derivative, and recalling that in the expansion for the potential, the quantities evaluated at x=0 are constant, we find
Since this is a point of stable equilibrium, the second derivative of the potential energy evaluated at the equilibrium position is positive. Therefore, the force has the same form as the spring force, where the second derivative of U plays the role of k: U''(0) = k. This argument is true aside from some aberrant cases where the second derivative U''(0) = 0.
We conclude that the motion near a point of stable equilibrium is simple harmonic. This is amplified if we draw the potential energy near a point of stable equilibrium. We can apply what we learned about one-dimensional motion to see the turning points, and see that the shape of the PE near the equilibrium is parabolic.