Question

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

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

Answer 1

The correct option is D $(\frac{8}{3}-\frac{10\pi }{16}){MR}^2$

Given : Radius of each circular hole $=R$

Side length of the square sheet $L=4R$

Moment of inertia of the complete sheet about z-axis $I_z=\frac{M{L}^2}{6}=\frac{M(4R{)}^2}{6}=\frac{8}{3}M{R}^2$

Mass of the complete sheet is given as $M$

Area of one circular hole $A=\pi R^2$

$\therefore$ Mass of one circular hole $m_1=\frac{M}{(4R{)}^2}\times \pi R^2=\frac{\pi M}{16}$

Moment of inertia of the hole about its center i.e O, $I_o=\frac{1}{2}{m}_1{R}^2=\frac{1}{2}\times \frac{\pi M}{16}{R}^2=\frac{\pi }{32}M{R}^2$

$\therefore$ Moment of inertia of one hole about z axis, $I_{1_Z}=\frac{\pi }{32}M{R}^2+m_1{d}^2$

where $d=OZ=\sqrt{2}R$

$⟹$ $I_{1_Z}=\frac{\pi }{32}M{R}^2+\frac{\pi M}{16}(\sqrt{2}R{)}^2=\frac{5\pi }{32}M{R}^2$

Thus moment of inertia about z- axis due to 4 circular holes $I_{T_Z}=4\times I_{1_Z}=\frac{10\pi }{16}M{R}^2$

Net moment of inertia of the remaining sheet about z-axis, $I_Z^{\prime }=I_Z-I_{T_Z}$

$⟹$ $I_Z^{\prime }=(\frac{8}{3}-\frac{10\pi }{16})M{R}^2$