Question

A circular disc has a tiny hole in it, at a distance z from its center. Its mass is $M$ and radius $R(R>z)$. A horizontal shaft is passed through the hole and held fixed so that the disc can freely swing in the vertical plane. For small disturbance, the disc performs SHM whose time period is minimum for $z=$

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

Answer 1

The correct option is C $\frac{R}{\sqrt{2}}$

Since we know that
$T=2\pi \sqrt{\frac{I_{support}}{mg{l}_{cm}}}$
here $I_{support}=\frac{1}{2}m{R}^2+m{z}^2$ and $l_{cm}=z$
$\Rightarrow T=2\pi \sqrt{\frac{(\frac{1}{2}m{R}^2+m{z}^2)}{mgz}}=2\pi \sqrt{\frac{(\frac{1}{2}{R}^2+z^2)}{gz}}$
$\Rightarrow T^2=\frac{4{\pi }^2}{g}(\frac{R^2}{2z}+z)$ .......$\left(I\right)$
differentiating equation $\left(I\right)$ with respect to $z$, we have
$\Rightarrow 2T\frac{dT}{dz}=\frac{4{\pi }^2}{g}(-\frac{R^2}{2z^2}+1)$
for maxima or minima, $\frac{dT}{dz}=0$
$\Rightarrow (-\frac{R^2}{2z^2}+1)=0\Rightarrow 2z^2=R^2\Rightarrow z=\frac{R}{\sqrt{2}}$
at $z=\frac{R}{\sqrt{2}}$, $\frac{dT}{dz}$ changes its sign from negative(-) to positive(+), which implies that $z=\frac{R}{\sqrt{2}}$ is the point of minima i.e. for $z=\frac{R}{\sqrt{2}}$ time period of this disc will be minimum.
Alternate solution: $T$ is minimum when
$z=k$ radius of gyration, Therefor $z=\frac{R}{\sqrt{2}}$