A pendulum is formed by pivoting a disc. What | vedclass

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(B) For a physical pendulum,the time period is given by $T = 2\pi \sqrt{\frac{I}{mg\ell}}$,where $I$ is the moment of inertia about the pivot point an

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$\frac{R}{\sqrt{2}}$

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(B) For a physical pendulum,the time period is given by $T = 2\pi \sqrt{\frac{I}{mg\ell}}$,where $I$ is the moment of inertia about the pivot point and $\ell$ is the distance from the center of mass $(COM)$ to the pivot.Using the parallel axis theorem,$I = I_{COM} + m\ell^2 = \frac{1}{2}mR^2 + m\ell^2$.Substituting this into the formula: $T = 2\pi \sqrt{\frac{\frac{1}{2}mR^2 + m\ell^2}{mg\ell}} = 2\pi \sqrt{\frac{R^2 + 2\ell^2}{2g\ell}}$.For $T$ to be minimum,the term $f(\ell) = \frac{R^2 + 2\ell^2}{\ell} = \frac{R^2}{\ell} + 2\ell$ must be minimum.Differentiating with respect to $\ell$ and setting to zero: $\frac{df}{d\ell} = -\frac{R^2}{\ell^2} + 2 = 0$.Solving for $\ell$: $2\ell^2 = R^2 \implies \ell = \frac{R}{\sqrt{2}}$.

$A$ pendulum is formed by pivoting a disc. What distance from the center of mass should it be pivoted for the minimum time period while performing $SHM$?

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