1. Analyzing Forces and Torque

The sphere rests on the circular edge of the hole. The normal force from the edge points towards the center of the sphere. Friction acts tangentially at the contact circle, creating a torque that opposes rotation.

Let $\phi$ be the angle the radius to the contact point makes with the vertical axis.

$$\sin \phi = \frac{r_{hole}}{R} \quad \text{and} \quad \cos \phi = \frac{\sqrt{R^2 – r_{hole}^2}}{R}$$

2. Torque Calculation

Vertical Equilibrium: The vertical component of the total normal force supports the weight of the sphere. The normal force is directed towards the center, making an angle $\phi$ with the vertical.

$$N_{total} \cos \phi = mg \implies N_{total} = \frac{mg}{\cos \phi}$$

Friction Torque: Friction $f = \mu N_{total}$ acts tangentially at distance $r_{hole} = R \sin \phi$ from the axis of rotation.

$$\tau = f \cdot r_{hole} = (\mu N_{total}) (R \sin \phi)$$ $$\tau = \mu \left( \frac{mg}{\cos \phi} \right) R \sin \phi = \mu mg R \tan \phi$$

Using the geometry of the hole ($\tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{r_{hole}/R}{\sqrt{R^2 – r_{hole}^2}/R}$):

$$\tau \propto \tan \phi = \frac{r_{hole}}{\sqrt{R^2 – r_{hole}^2}}$$

3. Relationship for Stopping Time

The angular deceleration $\alpha$ is constant: $\alpha = \tau / I$. The time to stop is $\Delta t = \omega_0 / \alpha$. Thus:

$$\Delta t \propto \frac{1}{\tau} \propto \frac{\sqrt{R^2 – r_{hole}^2}}{r_{hole}}$$

Let’s verify the ratio for the two cases:

$$\frac{\Delta t_2}{\Delta t_1} = \frac{\frac{\sqrt{R^2 – r_2^2}}{r_2}}{\frac{\sqrt{R^2 – r_1^2}}{r_1}} = \left( \frac{r_1}{r_2} \right) \sqrt{\frac{R^2 – r_2^2}{R^2 – r_1^2}}$$

4. Numerical Calculation

Given: $R = 10$, $r_1 = 8$, $r_2 = 6$, $\Delta t_1 = 9.0$ s.

Case 1 ($r_1=8$): $\sqrt{10^2 – 8^2} = \sqrt{100 – 64} = 6$. Factor $\propto 6/8 = 3/4$.

Case 2 ($r_2=6$): $\sqrt{10^2 – 6^2} = \sqrt{100 – 36} = 8$. Factor $\propto 8/6 = 4/3$.

Ratio:

$$\frac{\Delta t_2}{\Delta t_1} = \frac{4/3}{3/4} = \frac{16}{9}$$ $$\Delta t_2 = 9.0 \times \frac{16}{9} = 16 \text{ s}$$