A sphere of radius 'r' is kept on a concave m | vedclass

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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 i

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$2 \pi[(R-r) / g]^{\frac{1}{2}}$

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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 is provided by the component of gravity acting along the tangent to the path.The restoring torque about the center of curvature $O$ is given by:$	au = -mg(R-r) \sin 	heta$For small oscillations,$\sin 	heta \approx 	heta$,so $	au \approx -mg(R-r) 	heta$.Using the rotational analog of Newton's second law,$	au = I \alpha$,where $I$ is the moment of inertia about the axis of rotation passing through $O$.Since the sphere is rolling without slipping,the effective moment of inertia about $O$ is $I = I_{cm} + m(R-r)^2 = \frac{2}{5}mr^2 + m(R-r)^2$.However,for a simple oscillation problem where we consider the motion of the center of mass,we use the equation of motion for the center of mass: $F_{restoring} = -mg \sin 	heta = m a_{cm}$.Since $a_{cm} = (R-r) \alpha$,we have $m(R-r) \alpha = -mg 	heta$.$\alpha = -\left(\frac{g}{R-r}ight) 	heta$.Comparing this with the $S$.$H$.$M$. equation $\alpha = -\omega^2 	heta$,we get $\omega^2 = \frac{g}{R-r}$.The time period is $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{R-r}{g}}$.

$A$ sphere of radius '$r$' is kept on a concave mirror of radius of curvature '$R$'. The arrangement is kept on a horizontal table. If the sphere is displaced from its equilibrium position and left,then it executes $S$.$H$.$M$. The period of oscillation will be ($g =$ acceleration due to gravity).

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