Calculating Angular Acceleration of a Rolling Object

Question

A bowling ball of mass M = 6.50 kg, radius R = 10.0 cm, and moment of inertia I = $(2/5)MR^2$ is given an initial center of mass velocity $v_0 = 3.00 m/s$ that is parallel to a horizontal surface. The ball initially slips alon the floor, but eventually rolls w/o slipping. The coefficient of kinetic friction between the ball and the floor is $\mu_k = .300$.

Part (a) asks for the acceleration of the center of mass of the ball, which = -2.94 m/s^2.

Part (b) asks for the angular acceleration of the bowling ball.

The correct answer is 73.6 rad/s^2. I used the approach below, but I'm not sure why it's wrong:

$a_t = r \alpha$

$\alpha = \frac{a_t}{r} = -29.4 rad/s$

Why is the above approach wrong?

Answer 1

Your approach is wrong because the ball is slipping - therefore the relationship between $a_t$ and $r\alpha$ is not constant (your expression $a_t = r\alpha$ does not hold).

Instead, you need to look at the torque on the ball: you know the normal force and the coefficient of friction, so you know the force being applied off center

$$F = \mu m g$$

which gives rise to a torque $\Gamma$

$$\Gamma = \left|\ \vec F \times \vec r\ \right| = \mu m g r$$

The angular acceleration $\alpha$ is given by

$$\alpha = \frac{\Gamma}{I}$$

The the angular momentum of the sphere $I=\frac25 mr^2$, you get

$$\alpha = \frac{m g \mu r}{\frac25 m r^2} = \frac{5\cdot 0.3 \cdot 9.81}{2\cdot 0.1} = 73.6 \text{ rad/s}$$

I wrote a couple of related answers... See this one and this other one.