# Rolling Forces

[SOLVED] Rolling Forces

## Homework Statement

A bowler throws a bowling ball of radius R = 11 cm down a lane. The ball slides on the lane with initial speed vcom,0 = 6.0 m/s and initial angular speed 0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.19. The kinetic frictional force fk acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed vcom has decreased enough and angular speed has increased enough, the ball stops sliding and then rolls smoothly.
(a) What then is vcm in terms of ?
___________m·$$\omega$$

(b) During the sliding, what is the ball's linear acceleration?
___________m/s2

(c) During the sliding, what is the ball's angular acceleration?

(d) How long does the ball slide?
___________s

(e) How far does the ball slide?
___________m

(f) What is the speed of the ball when smooth rolling begins?
___________m/s

## The Attempt at a Solution

I am utterly lost in this problem. I assume that since the ball is sliding the force of kinetic friction is applying a torque onto the ball so:

T = FkR
T = Fk(.11m)

and Fk = uN so

T = (.19)(9.8)(.11)m
so
T = .2048 * m

now, I'm lost. Any pointers in the right direction?

Last edited:

Ok so I got that angular acceleration during sliding is 42.3 rad/s by the fact that, T = I$$\alpha$$ and also equal to the equation above.
I realize that when the ball stops slipping that the Acceleration of the Center of mass = $$\alpha$$ * R and that all those equations start to apply where Velocity of the Center of mass = $$\omega$$ * R . But how do I figure out the linear acceleration during the sliding?