A spherical ball of mass m and radius r rolls without slipping on a...

Let $\theta$ be the angular displacement of the center of the ball from the vertical. The center of the ball moves in a circle of radius $R - r$.

The velocity of the center of the ball is $v = (R - r)\dot{\theta}$.

For pure rolling without slipping, the angular velocity of the ball about its center is $\omega = \dfrac{v}{r} = \dfrac{(R - r)\dot{\theta}}{r}$.

The total kinetic energy of the ball is:

$K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

Substituting $I = \dfrac{2}{5}mr^2$ for a solid sphere:

$K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right)\left(\dfrac{v}{r}\right)^2 = \dfrac{7}{10}mv^2 = \dfrac{7}{10}m(R - r)^2\dot{\theta}^2$

The potential energy of the ball relative to its lowest position is:

$U = mg(R - r)(1 - \cos\theta)$

For small angles, $1 - \cos\theta \approx \dfrac{\theta^2}{2}$, so:

$U \approx \dfrac{1}{2}mg(R - r)\theta^2$

By conservation of mechanical energy, $E = K + U = \text{constant}$:

$\dfrac{7}{10}m(R - r)^2\dot{\theta}^2 + \dfrac{1}{2}mg(R - r)\theta^2 = \text{constant}$

Differentiating with respect to time $t$:

$\dfrac{7}{5}m(R - r)^2\dot{\theta}\ddot{\theta} + mg(R - r)\theta\dot{\theta} = 0$

$\ddot{\theta} + \dfrac{5g}{7(R - r)}\theta = 0$

Comparing with the standard equation of SHM $\ddot{\theta} + \omega_0^2\theta = 0$, we get the angular frequency:

$\omega_0 = \sqrt{\dfrac{5g}{7(R - r)}}$

The time period of oscillation is:

$T = \dfrac{2\pi}{\omega_0} = 2\pi\sqrt{\dfrac{7(R - r)}{5g}}$