Question

The moment of inertia of a solid disc made of thin metal of radius $R$ and mass $M$ about one of its diameters is given by $\frac{{MR}^2}{4}$. What will be moment of inertia about this axis if the disc if folded in half about this diameter?

• A. $\frac{{MR}^2}{8}$
• B. $\frac{{MR}^2}{2}$
• C. $\frac{{MR}^2}{4}$
• D. ${MR}^2$

Answer 1

The correct option is A $\frac{{MR}^2}{8}$

Let this disc be of mass $M$

Each of the half portions has mass $\frac{M}{2}$

Now$,$

$MOI$ of whole disc of mass $\frac{M{R}^2}{4}$

This means that $MOI$ due to either of the half parts is $=\frac{M{R}^2}{8}$

Now$,$ ever if the disc is folded about diameter $AB,$ the $MOI$ will not charge because the ever then also the distance of the corresponding part of disc remains same from the axis $AB$ before and after folding$.$

Therefore$,$ correct option is $\frac{M{R}^2}{8}$

Hence$,$

Option $\left(A\right)$ is correct