Can somebody help me with this physics problem?

Question

A solid bowling ball of radius R and mass M is released from the hand of a bowler, at floor-level, with linear speed V but angular speed omega=0. There is a coefficient of kinetic friction of mu between the ball and the floor. How much time and distance is required for the ball to begin rolling without slipping? what is V and omega at that time? You should notice something very interesting. Please don't just give the answer and show how it was solved! Thank you.

? · Accepted Answer

There is one force in the direction of motion, the kinetic friction. The equations of motion for translation and rotation are

m dv/dt = -μ  m g

I dω/dt = μ m g R,

with I= 2/5 m R^2 the moment of inertia of the ball.

The solutions are

v(t) = v(0) - μ g t

ω(t) = ω(0) + (5μg/(2R)) t

In this problem ω(0) = 0.

Pure rolling sets in when v(t) = ω(t) R:

v(0) - μ g t =  (5μg/2) t

t = 2v(0)/ (7μg)

At this time we have

v = 5v(0)/7
ω = 5v(0)/(7R)

During this time of uniform deceleration the distance traveled is

d = v(0) t - 1/2 (μ g) t^2
= 2v(0)^2/ (7μg) - 1/2 (μ g ) (2v(0)/ (7μg))^2
= 12 v(0)^2 /(49μ g)

This distance does not depend on the radius of the ball. Some people consider that interesting...

More importantly note that once pure rolling sets in, the gliding friction changes into rolling friction. This changes the equations of motion from that point in time onward, allowing pure rolling. If you would not take this into account, the equations of motion above would lead you to believe that the rotational speed would increase whilst the speed was still decreasing beyond the point of pure rolling.

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Can somebody help me with this physics problem?

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