Question

A uniform disc of radius $R$ lies in the $xy$ plane, with its centre at origin. Its moment of inertia about z-axis is equal to its moment of inertia about line $y=x+c$. The value of $c$ will be

• A. $-\frac{R}{2}$
• B. $\pm \frac{R}{\sqrt{2}}$
• C. $+\frac{R}{4}$
• D. $-R$

Answer 1

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

The line perpendicular to $y=x+c$ passes through center and is in plane of ring.

So Moment of Inertia along line perpendicular to y=x+c is $\frac{M{R}^2}{4}$
Also line $y=x+c$ is at $c/\sqrt{2}$ from center.

So the moment of inertia along line $y=x+c$ is $I_{line}=\frac{M{R}^2}{4}+M(\frac{c}{\sqrt{2}}{)}^2$
Given: $I_{line}=I_z=M\frac{R^2}{2}$
$\Rightarrow \frac{M{R}^2}{4}+M(\frac{c}{\sqrt{2}}{)}^2=M\frac{R^2}{2}$

$\Rightarrow \frac{c^2}{2}=\frac{R^2}{4}$

$\Rightarrow c=\pm \frac{R}{\sqrt{2}}$