Physics 321 -- Spring 2005

Homework #4, due at beginning of class Wednesday Feb. 9.

1.         [3pts] Marion & Thornton, problem 9-42. (This is problem 9-40 in 4th edition)
Hint: Read Section 9.8 carefully to get the definition of the coefficient of restitution.  The component of velocity parallel to the surface is unchanged by the collision.  The perpendicular component is not only reversed, but is reduced according to the coefficient of restitution.

2.         [4pts] Marion & Thornton, problem 9-36. (This is problem 9-34 in 4th edition.)
Hint: There are solutions both for alpha > 0 and for alpha < 0.

3.         [5pts] Marion & Thornton, problem 9-43. (This is problem 9-41 in 4th edition.)
Hint: After you have written down the two components of conservation of momentum and the relation between initial and final kinetic energy, you have some serious algebra to do. A good approach is to begin by eliminating the unknown angle by using cos2z + sin2z = 1. This leaves only two equations; you can solve them and then plug back into one of the original equations to find the desired angle. (You may prefer to use Mathematica or one of its competitors instead of hand tools for this.)

4.         [4pts] Marion & Thornton, problem 9-44. (This is problem 9-42 in 4th edition.)
(Assume the diameter of the initial coil is negligibly small, so the rope is vertical and angular momentum can be ignored.)

5.         [4pts] Marion & Thornton, problem 9-19. (Same in 4th edition.)
Hint: You can use Newton’s 2nd Law for the center of mass motion. The motion itself is simple: assume that the links of the chain fall freely when they are above the table, and that they stop without bouncing when they reach it.

6.         [6pts] In class, we solved for the motion of a chain that was initially hanging over the edge of a frictionless horizontal surface. Generalize that problem by assuming that the surface slopes downward toward the edge at an angle theta below the horizontal, and has a coefficient of friction mu. As in the original problem, assume that there is a frictionless barrier in place to keep the part of the chain that is off the table vertical.
(a) Apply Newton's 2nd Law to the vertical part of the chain, to relate the tension at the corner to the acceleration.
(b) Apply Newton's 2nd Law to the part of the chain that is still on the surface, to relate the tension at the corner to the acceleration.
(c) Equate your results from parts (a) and (b) to find the equation of motion of the chain.
(d) Calculate the work done by friction during the time that the length of chain hanging over increases from x0 to x.
(e) Calculate the gravitational potential energy as a function of x.
(f) Use your results from parts (d) and (e) to write a conservation-of-energy equation.
(g) Take the time derivative of your result from part (f) to obtain the equation of motion. If it doesn't agree with your result from part (c), go back and find the mistake!
(h) Solve the equation of motion, assuming the chain starts from rest with a length x0 hanging down.

(Last updated 2/2/2005.)