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

Answer 1

Answer: (D)

If M is mass of the square plate before cutting the holes, then mass of portion of each hole,
$m=\frac{M}{16R^{2}}\times\pi R^{2}=\frac{M \pi}{16}$
$\therefore \,$ Moment of inertia of remaining portion about Z axis
$I=I_{\text{square}}-4 I_{\text{hole}}$
$=\frac{M}{12}\left(16 R^{2}+16 R^{2}\right)-4\left[\frac{mR^{2}}{2}+m\left(\sqrt{2}R\right)^{2}\right]$
$=\frac{M}{12}\times32 R^{2}-10m R^{2}$
$=\frac{8}{3} MR^{2}-\frac{10}{16}MR^{2} \pi$
$I=\left(\frac{8}{3}-\frac{10\pi}{16}\right) MR^{2}.$