Question

A square plate of edge ${\frac{{}^{\prime }a}{2}}^{\prime }$ is cut from a uniform square plate of edge 'a' as shown in figure. Mass of square plate of edge 'a' is $M$. Find out the moment of inertia of the remaining square plate about an axis passing through 'O' (center of square plate of side 'a') and perpendicular to the plane of the plate.

• A. $\frac{5}{192}M{a}^2$
• B. $\frac{27}{64}M{a}^2$
• C. $\frac{9}{64}M{a}^2$
• D. $\frac{M{a}^2}{6}$

Answer 1

The correct option is C $\frac{9}{64}M{a}^2$

Given mass of the square plate of side${}^{\prime }{a}^{\prime }=M$
Let the mass of removed square plate $=m$
For uniform square plate, the surface mass density will be constant.
So, $\frac{M}{a^2}=\frac{m}{{\left(\frac{a}{2}\right)}^2}\Rightarrow m=\frac{M}{4}...\left(i\right)$

Moment of inertia of parent square plate about an axis passing through 'O' (center of square plate of side ${}^{\prime }{a}^{\prime }$) and perpendicular to the plane of the plate. $=\frac{M{a}^2}{6}$
$I_{\text{parent plate}}=\frac{M{a}^2}{6}$
Moment of inertia of removed square plate $\left(I_{\text{removed plate}}\right)$
Using parallel axis theroem $I=I_{com}+m{d}^2$$\because \text{Here}d=\frac{a}{4}$
$I_{\text{removed plate}}=\frac{(-m){\left(\frac{a}{2}\right)}^2}{6}+(-m){\left(\frac{a}{4}\right)}^2$

$=-\frac{5}{48}m{a}^2$
From equation $\left(i\right)$
$I_{\text{removed plate}}=-\frac{5}{48}\times \frac{M}{4}{a}^2=-\frac{5}{192}M{a}^2....\left(ii\right)$

Moment of inertia of the remaining square plate $\left(I_{net}\right)$
$\left(I_{net}\right)=I_{\text{parent plate}}+I_{\text{removed plate}}=\frac{M{a}^2}{6}-\frac{5}{192}M{a}^2$
$=\frac{9}{64}M{a}^2$

So, Moment of inertia of remaining square plate about an axis passing through point ${}^{\prime }{O}^{\prime }$ and perpendicular to the plane of plate is $\frac{9}{64}M{a}^2$.