Question

A uniform disc of radius $R$ lies in $xy$ plane with its centre at origin. Its moment inertia about the axis $x=2R$ and $y=0$ is equal the moment of inertia about the axis $y=d$ and $z=0$. Where $d$ is equal to:

• A. $\frac{4}{3}R$
• B. $\frac{\sqrt{17}}{2}R$
• C. $\sqrt{13}R$
• D. $\frac{\sqrt{15}}{2}R$

Answer 1

The correct option is A $\frac{4}{3}R$

Here

$I_1$ = moment of inertia of disc about an axes

parallel to diameter = $I$ + $M{r}^2$

= $\frac{M{R}^2}{4}+M(2R{)}^2$

= $\frac{M{R}^2}{4}+M\left(4R^2\right)$

= $\frac{17M{R}^2}{4}$

$I_2$ = moment of inertia of disc about an axis

perpendicular to plane passing through

Y = 2d

= $I_0+M{r}^2$

= $\frac{M{R}^2}{2}+M{d}^2$

But from the given data $I_1=I_2$

$\frac{17M{R}^2}{4}=\frac{M{R}^2}{2}+M{d}^2$

$\frac{17M{R}^2}{4}-\frac{M{R}^2}{2}+M{d}^2$

$\frac{17R^2}{4}-\frac{2R^2}{4}=d^2$

$\frac{15R^2}{4}=d^2$

d = $\frac{\sqrt{1}5R}{2}$