# Moment of Inertia (Feynman)

Two balls of equal radius and mass, free to roll on a horizontal plane, are separated by a distance $L$ large compared to their radius. One ball is solid, the other hollow, and they are attracted by a mysterious force. How far will the solid ball roll before it collides with the hollow ball?

Solution:
The acceleration of a rolling object with mass $M$, radius $R$, and moment of inertia $kMR^2$ $(0 < k < 1)$ under force $F$ (parallel to the rolling plane) is $F/(k+1)M$. For the solid sphere $k=2/5$ and for the hollow one $k=2/3$. $F$ and $M$ are the same for both balls, so the ratio of their accelerations must be constant and equal to $(1+2/5)/(1+2/3) = 21/25$. This must also be the ratio of the distances the balls traverse in any given period. (Note: this is independent of the specific properties of the force.) So, the solid ball goes $(25/46)L$ and the hollow ones goes $(21/46)L$.

It is one of the exercises from the Feynman Lectures and I cannot comprehend the solution. My question: why is there a factor $(k+1)$?

$$I' = I + mr^2 = (k+1) m r^2$$
Then if I put the torque $\Gamma = Fr$, I get for the angular acceleration
$$\dot\omega = \frac{\Gamma}{I'} = \frac{Fr}{(k+1)mr^2} = \frac{F}{(k+1)mr}$$
The linear acceleration is of course $\dot v = r\dot\omega$, which is the expression you found.