# Bowling Question - Rotational Motion

1. Nov 15, 2008

### gingerelle

1. The problem statement, all variables and given/known data

A uniform, spherical bowling ball of mass m and radius R is projected horizontally along the
floor at an initial velocity v0 = 6.00 m/s. The ball is not rotating initially, so w0 = 0. It
picks up rotation due to (kinetic) friction as it initially slips along the floor. The coefficient of
kinetic friction between the ball and the floor is μk. After a time ts, the ball stops slipping and makes a transition to rolling without slipping at angular speed w(s) and translational velocity v(s).
Thereafter, it rolls without slipping at constant velocity.

If a standard bowling alley is 19.0 m long and the ball slips for half its length, what is the
value of μk?

2. Relevant equations

1]]] L = Iw
2]]] F = ma
3]]] Torque = r cross F
4]]] x(f) = x(initial) + v0t + 1/2at^2
5]]] theta(f) = theta (i) + w(0)t + 1/2angularacceleration(t^2)

3. The attempt at a solution

So, basically, I found the linear acceleration equation and the angular acceleration of the ball, as well as Ts (and I know these answers are correct). But I just wanted to check if I'm allowed to do the following:

I added equation 4 and equation 5 together, and made it equal to 9.5 metres. But before that, I converted equation 5 into metres by multiplying both sides by (pi X R)/180 degrees ...since 360 degrees turn = 1 circumference travelled.

after I did that, I just subbed time (Ts) equation in (from what I calculated before) and when I isolated uk, I got around 0.0995 (coefficient of friction). But basically, I was just wondering if that's...physically correct..adding the two equations and making it equal 9.5m (half of 19.0 metres, which is the distance that the ball skidded for). thanks and please reply.

2. Nov 15, 2008

### asleight

$$\vec{v}_f^2=36m/s-19m\cdot\vec{a}$$. This final velocity is when it starts picking up angular velocity (id est, $$\vec{v}=\vec{r\omega}$$).