A hollow bowling ball of mass M = 3 kg and radius R = 20 cm is thrown across a f

Question

A hollow bowling ball of mass M = 3 kg and radius R = 20 cm is thrown across a frictionless surface with a translational speed of 6.0 m/s. Initially the ball is not rolling across the surface, i.e. its motion is only translational. Some time later, the ball encounters a rough surface which instantly causes it to rotate without slipping, a condition which allows us to relate the translational and rotational motions of the ball using ? = VCM / R, where VCM is the velocity of the center of mass. What is the translational velocity of the center of mass in m/s after the bowling ball encounters the rough surface? Imagine the motion of the bowling ball on the smooth surface compared to the rough surface. Are any quantities conserved throughout the motion of the ball? Also, the condition that the ball rolls without slipping means that we have a case of "rolling friction" in which we assume that the friction force does not change the energy of the ball even though the surface is rough.

Explanation / Answer

Energy will be conserved, as the friction force will not cause any work loss...

So, 1/2 mvi^2 = 1/2 mvf^2 + 1/2 I*wf^2
I = 0.66 mr^2
Also, wf = vf/r
So, 1/2 mvi^2 = 1/2 mvf^2 + (1/2) * 0.66 mr^2 * (vf/r)^2

So, 0.5 mvi^2 = 0.5 mvf^2 + 0.33 mvf^2 = 0.8333 mvf^2

so, vf = 4.6477 m/s

wf = vf/r = 23.23 rad / s

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