Four holes of radius $5 \mathrm{~cm}$ are cut from a thin square plate of $20 \mathrm{~cm}$ and mass $1 \mathrm{~kg}$. The moment of inertia of the remaining portion about $Z$-axis is

Question

Aicharya · Accepted Answer

Area mass density, $\sigma=\frac{M}{16 R^2}$ $ \left(\because \text { Area }=4 R \times 4 R=16 R^2\right) $ Mass of each hole $m_1=\sigma \pi R^2=\frac{M}{16 R^2} \pi R^2=\frac{\pi M}{16}$ Distance between centre of plate and centre of hole $ x=\frac{\sqrt{(2 R)^2+(2 R)^2}}{2}=\frac{2 \sqrt{2} R}{2} . $

Moment of inertia of one hole at about Z-axis $ I_1=\frac{1}{2} m_1 R^2+m_1 x^2=\frac{5 \pi}{32} M R^2 $ Moment of inertia of whole plate about Z-axis, $ I=\frac{M(4 R)^2}{6}=\frac{8}{3} M R^2 $ Required moment of Inertia $ I_0=I-4 I_1,=\left[\frac{8}{3}-4\left(\frac{5 \pi}{32}\right)\right] M R^2=\left[\frac{8}{3}-\frac{5 \pi}{8}\right] M R^2 $ Given, $R=5 \mathrm{~cm}$ and $M=1 \mathrm{~kg}$ So, $I_0=\left[\frac{8}{3}-\frac{5 \pi}{8}\right] 1 \times 25 \times 10^{-4}=0.0017 \mathrm{~kg}-\mathrm{m}^2$

Four holes of radius $5 \mathrm{~cm}$ are cut from a thin square plate of $20 \mathrm{~cm}$ and mass $1…

Solution