A uniform disc of radius R and mass M is free | vedclass

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(C) The disc oscillates as a physical pendulum.The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{Mgd}}$, where $I$ is the mo

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

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(C) The disc oscillates as a physical pendulum.The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{Mgd}}$, where $I$ is the moment of inertia about the pivot point, $M$ is the mass, and $d$ is the distance from the center of mass to the pivot point.The moment of inertia $I$ of the disc about an axis passing through the edge $A$ is calculated using the parallel axis theorem: $I = I_{CM} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.The distance from the center of mass to the pivot point $A$ is $d = R$.Substituting these values into the formula, we get:$T = 2\pi \sqrt{\frac{\frac{3}{2}MR^2}{MgR}} = 2\pi \sqrt{\frac{3R}{2g}}$.

$A$ uniform disc of radius $R$ and mass $M$ is free to oscillate about the axis $A$ as shown in the figure. For small oscillations, the time period is . . . . . . . ($g$ is acceleration due to gravity)

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