A disc of radius R is made to oscillate about | vedclass

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(A) The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgL}}$,where $I$ is the moment of inertia about the axis of rotation a

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$2\pi \sqrt{\frac{3R}{2g}}$

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(A) The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgL}}$,where $I$ is the moment of inertia about the axis of rotation and $L$ is the distance from the pivot to the center of mass.For a disc of mass $m$ and radius $R$ oscillating about an axis passing through its periphery,the distance from the pivot to the center of mass is $L = R$.Using the parallel axis theorem,the moment of inertia about the periphery is $I = I_{cm} + mR^2 = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2$.Substituting these values into the time period formula:$T = 2\pi \sqrt{\frac{\frac{3}{2}mR^2}{mgR}} = 2\pi \sqrt{\frac{3R}{2g}}$.

$A$ disc of radius $R$ is made to oscillate about a horizontal axis passing through its periphery. Its time period would be

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