Question

A disc of radius $R / 2$ is cut from a uniform disc of radius $R$ as shown in the figure. The mass of the disc (with cavity) is $M$. The moment of inertia of this disc about an axis $X X$ ' which passes through the center $0$ of the disc and is in the plane of the disc will be

• A. $\frac{7 MR ^{2}}{12}$
• B. $\frac{13 MR ^{2}}{24}$
• C. $\frac{13 MR ^{2}}{48}$
• D. $\frac{11 MR ^{2}}{36}$

Answer 1

Answer: (C)

$I _{ Z }=\frac{ M _{1} R ^{3}}{ Z }\left[\frac{ M _{2}( R / 2)^{3}}{ Z }+\left(\frac{ R }{2}\right)^{2}\right]$
Whose $M _{1}=\sigma \pi R ^{2}$
$M _{2}=\sigma \pi\left(\frac{ R }{2}\right)^{2}$
$\sigma=\frac{ M }{\pi R ^{2}-\pi R ^{2} / y }$