A disc of radius $R$ and mass $M$ is pivoted at the rim and is set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be

Question

Dr. Abubacker · Accepted Answer

(d) Time period of a physical pendulum
$T = 2\pi\sqrt{\frac{I_{0}}{mgd}} = 2\pi\sqrt{\frac{\left( \frac{1}{2}mR^{2} + mR^{2} \right)}{mgR}}$
$= 2\pi\sqrt{\frac{3R}{2g}}$ …(i)
$T_{simple\ pendulum} = 2\pi\sqrt{\frac{l}{g}}\ \ \ \ \ \ \ldots(ii)$
Equating (i) and (ii), $l = \frac{3}{2}R$

Q. A disc of radius $R$ and mass $M$ is pivoted at the rim and is set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be

Explanation

Reviewed by Dr. Abubacker

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