Question

A small hole is made in a disc of mass $M$ and radius $R$ at a distance $R/4$ from the centre. The disc is supported on a horizontal peg through this hole. The moment of inertia of the disc about horizontal peg is

• A. $\frac{M{R}^2}{9}$
• B. $\frac{5}{16}M{R}^2$
• C. $\frac{9}{16}M{R}^2$
• D. $\frac{5}{4}M{R}^2$

Answer 1

The correct option is C $\frac{9}{16}M{R}^2$

Moment of Inertia of disc about its center is $\frac{1}{2}M{R}^2$

Therefore by parallel axis theorem,

$I=I_{CM}+M{d}^2=\frac{1}{2}M{R}^2+M{\left(\frac{R}{4}\right)}^2=\frac{1}{2}M{R}^2+\frac{1}{16}M{R}^2=\frac{9}{16}M{R}^2$