Question

A spherical ball of mass \(m\) and radius \(r\) rolls without slipping on a rough concave surface of large radius \(R\). It makes small oscillations about the lowest point. The time period is:

• A. \(2\pi\sqrt{\dfrac{7(R-r)}{5g}}\)
• B. \(2\pi\sqrt{\dfrac{5(R-r)}{7g}}\)
• C. \(2\pi\sqrt{\dfrac{2(R-r)}{5g}}\)
• D. \(2\pi\sqrt{\dfrac{2(R-r)}{7g}}\)

Hint:

Centre moves on radius \((R-r)\); rolling adds a factor \(7/5\) to the effective inertia, like a pendulum.

Answer 1

The Correct Option is A

Treat the ball rolling in the bowl like a physical pendulum and use the torque equation about the contact point.

Effective length. The centre of the ball travels on a circle of radius \(L=(R-r)\). Small angular displacement about the bottom is $\theta$.

Restoring force. The tangential component of gravity is $mg\sin\theta \approx mg\theta$, which drives the ball back to the lowest point.

Effective mass factor. Because the sphere rolls without slipping, its rotational inertia adds to its resistance to acceleration. The linear acceleration of a rolling solid sphere down an incline of effective slope is reduced by the factor \[\frac{1}{1+\frac{I}{mr^{2}}}=\frac{1}{1+\frac{2}{5}}=\frac{5}{7}.\] So the tangential acceleration of the centre is \[a=\frac{5}{7}\,g\sin\theta\approx\frac{5}{7}g\theta .\] Equation of motion. With arc coordinate $s=L\theta$, $a=L\ddot\theta$, giving \[L\ddot\theta=-\frac{5}{7}g\theta\;\Rightarrow\;\ddot\theta+\frac{5g}{7L}\theta=0.\] Period. Angular frequency $\omega=\sqrt{\dfrac{5g}{7L}}$ with $L=R-r$, therefore \[T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{7(R-r)}{5g}}.\] \[\boxed{T=2\pi\sqrt{\dfrac{7(R-r)}{5g}}}\]