Question

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

• A. $\frac{9}{16}{MR}^2$
• B. $\frac{5}{4}{MR}^2$
• C. $\frac{13M{R}^2}{32}$
• D. $\frac{5}{16}{MR}^2$

Answer 1

The correct option is A $\frac{9}{16}{MR}^2$

$I_A+M(\frac{R}{4}{)}^2$ $=I_B$

$I_A=\frac{M{R}^2}{2}$

$I_B=\frac{M{R}^2}{2}+\frac{M{R}^2}{16}$

$=\frac{9M{R}^2}{16}$