Question

A thin spherical shell of mass m and radius R lying on a rough surface is hit sharply by a cue horizontally as shown. Find h so that shell does not slip over the surface.

Answer 1

R/3

Answer 2

1. Condition for Pure Rolling: For no slipping, the velocity of the point of contact must be zero, which implies $v = R\omega$.

2. Angular Impulse-Momentum (about point of contact): Take the point of contact (P) with the ground as the pivot. The friction impulse produces no torque about P. The cue impulse $J_{cue}$ acts at a height $(2R-h)$ from P. The change in angular momentum about P is $I_P \omega$.

3. Moment of Inertia: For a thin spherical shell, $I_{CM} = \frac{2}{3} mR^2$. By the parallel axis theorem, $I_P = I_{CM} + mR^2 = \frac{5}{3} mR^2$.

4. Equation Formation: $J_{cue} (2R-h) = \frac{5}{3} mR^2 \omega$.

5. Substitution: Substitute $\omega = v/R$ into the equation: $J_{cue} (2R-h) = \frac{5}{3} mvR$.

6. Linear Impulse-Momentum (about Center of Mass): For the specific height $h$ that causes immediate pure rolling, the friction impulse $J_{friction}$ is zero. Thus, $J_{cue} = mv$.

7. Solve for h: Substitute $J_{cue} = mv$ into the equation from step 5: $mv (2R-h) = \frac{5}{3} mvR$.

8. Result: Cancel $mv$ and solve for $h$: $2R-h = \frac{5}{3}R \implies h = 2R - \frac{5}{3}R = \frac{R}{3}$.