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(B) When the sphere is displaced by a small angle $	heta$,the center of the sphere moves along a circular arc of radius $(R - r)$.The restoring force

Vedclass · Accepted Answer

$2\pi \sqrt{\frac{R - r}{g}}$

Vedclass · Answer

(B) When the sphere is displaced by a small angle $	heta$,the center of the sphere moves along a circular arc of radius $(R - r)$.The restoring force acting on the sphere is $F = -mg \sin 	heta$.Since $	heta$ is small,$\sin 	heta \approx 	heta = \frac{x}{R - r}$,where $x$ is the linear displacement.The restoring force is $F = -mg \left( \frac{x}{R - r} ight)$.The acceleration is $a = \frac{F}{m} = -\left( \frac{g}{R - r} ight) x$.Comparing this with the standard $S.H.M.$ equation $a = -\omega^2 x$,we get $\omega^2 = \frac{g}{R - r}$,so $\omega = \sqrt{\frac{g}{R - r}}$.The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{R - r}{g}}$.

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