Question

Four holes of radius $R$ each are cut from a thin square plate of side $4R$ and mass $M$. The moment of inertia of the remaining portion about $z-$ axis (out of the plane) is

• A. $\frac{\pi }{12}M{R}^2$
• B. $(\frac{4}{3}-\frac{\pi }{4})M{R}^2$
• C. $(\frac{8}{3}-\frac{10\pi }{16})M{R}^2$
• D. $(\frac{4}{3}-\frac{\pi }{6})M{R}^2$

Answer 1

The correct option is C $(\frac{8}{3}-\frac{10\pi }{16})M{R}^2$

Assume mass of one of the removed disc is $m$.
Since mass is uniformly distributed,
$\frac{M}{4R\times 4R}=\frac{m}{\pi R^2}$
$\Rightarrow m=\frac{\pi M}{16}$ .....(i)
$I_{\text{remaining}}=I_{\text{plate}}-I_{\text{removed}}$
$=I_{\text{plate}}-4\times I_{\text{disc}}$


As we know,
MOI of plate $I_{\text{plate}}=\frac{M}{12}(a^2+b^2)$
[Here , $a=b=4R$]
$\Rightarrow I_{\text{plate}}=\frac{M}{12}\left[\right(4R{)}^2+\left(4R{)}^2\right]=\frac{8}{3}M{R}^2$
and $I_{\text{disc}}$ about $z$ axis (by parallel axis theorem)
$I_{\text{disc}}=\frac{m{R}^2}{2}+m(\sqrt{2}R{)}^2$

So,
$I_{\text{remaining}}=\frac{8}{3}M{R}^2-4\times (\frac{1}{2}\times \frac{\pi M}{16}\times R^2+\frac{\pi M}{16}\times 2R^2)$
$(\because m=\frac{\pi M}{16})$
$\Rightarrow I_{\text{remaining}}=\frac{8}{3}M{R}^2-4\times \frac{5}{2}\times \frac{\pi M}{16}\times R^2=M{R}^2(\frac{8}{3}-\frac{20\pi }{32})$
or $I=(\frac{8}{3}-\frac{10\pi }{16})M{R}^2$