A solid sphere of mass M and radius R is placed at the top of a fixed inclined plane of height h and angle \theta. The sphere is released from rest and rolls down the incline without slipping. The incline is rough enough to prevent slipping.
After reaching the bottom of the incline, the sphere moves onto a horizontal surface with negligible friction. It then collides elastically with a smooth vertical wall perpendicular to its direction of motion. After collision, it rolls back up the incline.
(a) Find the speed of the center of mass of the sphere just before hitting the wall.
(b) After rebounding elastically, what maximum height does the sphere reach on the incline?
(c) How much time (in terms of M, R, h, \theta) elapses between the initial release and the return to the top?
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We have a solid sphere of mass M and radius R placed at the top of an inclined plane. The plane has height h and makes an angle theta with the horizontal. The sphere is released from rest and rolls down without slipping, then collides elastically with a vertical wall before rolling back up.
To find the speed before hitting the wall, we use conservation of energy. The initial potential energy M g h is converted to translational and rotational kinetic energy. For a solid sphere, the moment of inertia is two-fifths M R squared. Using the rolling condition v equals R omega, we get v equals the square root of ten-sevenths g h.
For part b, we analyze the elastic collision with the wall. The collision reverses the translational velocity but keeps the angular velocity unchanged. The total kinetic energy after collision equals M g h. When the sphere rolls back up the incline, this energy converts to potential energy. By conservation of energy, the maximum height reached is exactly h, the same as the initial height.
For part c, we calculate the total time. First, we find the acceleration down the incline using Newton's laws. The gravitational component minus friction equals mass times acceleration. The friction provides the torque for rolling. This gives us acceleration equals five-sevenths g sine theta. The time down the incline is the square root of fourteen h over five g sine squared theta. By symmetry, the time up equals the time down, so the total time is twice this value.
To summarize our analysis of the rolling sphere problem: The speed before hitting the wall is the square root of ten-sevenths g h. After the elastic collision, the sphere returns to the same height h. The total time for the round trip is two times the square root of fourteen h over five g sine squared theta. This problem demonstrates energy conservation in rolling motion and elastic collisions.