Question

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

• A. $\frac{5}{4}R$
• B. $\frac{2}{3}R$
• C. $\frac{3}{4}R$
• D. $\frac{3}{2}R$

Answer 1

The correct option is D $\frac{3}{2}R$



Here, moment of inertia of disc about a point on rim and perpendicular to plane of disc is $I_0=\frac{1}{2}M{R}^2+M{R}^2$

Time period of a physical pendulum.
$T=2\pi \sqrt{\frac{I_0}{Mgd}}$
$T=2\pi \sqrt{\frac{\frac{1}{2}M{R}^2+M{R}^2}{MgR}}$
$T=2\pi \sqrt{\frac{3R}{2g}}\to \left(1\right)$
Also, Time period of simple pendulum,
$T=2\pi \sqrt{\frac{l}{g}}\to \left(2\right)$
equating $\left(1\right)$ and $\left(2\right)$
$\sqrt{\frac{3R}{2g}}=\sqrt{\frac{l}{g}}$
$\Rightarrow l=\frac{3R}{2}⟶\left(d\right)$