Question

The moment of inertia of a thin circular disc of mass $M$ and radius $R$ about any diameter is

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

Answer 1

The correct option is B $\frac{M{R}^2}{4}$

R.E.F image

Solution We know that moment of inertia of

disk about $A_1\left(I_1\right)=\frac{M{R}^2}{2}$

Applying Perpendicular Aois

theorem,

$I_1=I_2+I_3$

$\frac{M{R}^2}{2}=I_2+I_3$

By the symmetry of disk, we can

conclude that

$I_2=I_3$

So, $\frac{M{R}^2}{2}=2I_2$

$\Rightarrow I_2=I_3=\frac{M{R}^2}{4}$

Option $-B$ is correct