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

Step 1: The centre of the ball moves on a circle of radius \((R-r)\), since the ball of radius \(r\) stays in contact with the concave surface of radius \(R\). Let \(\theta\) be the angle the line joining the centre of curvature to the ball centre makes with the vertical.

Step 2: Use energy method. The ball rolls without slipping, so its kinetic energy is translation plus rotation: \[KE=\tfrac{1}{2}mv^{2}+\tfrac{1}{2}I\omega^{2}.\] For a solid sphere \(I=\tfrac{2}{5}mr^{2}\) and rolling gives \(v=r\omega\), so \[KE=\tfrac{1}{2}mv^{2}+\tfrac{1}{2}\left(\tfrac{2}{5}mr^{2}\right)\frac{v^{2}}{r^{2}}=\tfrac{1}{2}mv^{2}\left(1+\tfrac{2}{5}\right)=\tfrac{7}{10}mv^{2}.\] Step 3: The centre speed is \(v=(R-r)\dot{\theta}\). The height of the centre above the lowest position is \(h=(R-r)(1-\cos\theta)\), so potential energy \(U=mg(R-r)(1-\cos\theta)\).

Step 4: Total energy is constant: \[E=\tfrac{7}{10}m(R-r)^{2}\dot{\theta}^{2}+mg(R-r)(1-\cos\theta).\] Differentiate with respect to time and cancel \(\dot{\theta}\): \[\tfrac{7}{5}m(R-r)^{2}\ddot{\theta}+mg(R-r)\sin\theta=0.\] Step 5: For small oscillations \(\sin\theta\approx\theta\): \[\ddot{\theta}+\frac{5g}{7(R-r)}\theta=0.\] This is SHM with \(\omega^{2}=\dfrac{5g}{7(R-r)}\), so \[T=2\pi\sqrt{\frac{7(R-r)}{5g}}.\] \[\boxed{T=2\pi\sqrt{\dfrac{7(R-r)}{5g}}}\]