Question

A circular disc of radius R and thickness R/6 has moment of inertia I about an axis passing through its center and perpendicular to its plane. It is melt and recast into a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is:

• A. $\frac{I}{5}$
• B. $\frac{2I}{5}$
• C. $\frac{4I}{5}$
• D. $\frac{I}{10}$

Answer 1

The correct option is A $\frac{I}{5}$

given radius of the disc =$R$

thickness of the disc = $\frac{R}{6}$

volume of the disc $V_1=\pi R^{2}t=\pi R^2\frac{R}{6}=\pi \frac{R^3}{6}$

let the radius of the sphere be $R^{\prime }$

volume of the sphere $V_2=\frac{4}{3}\pi {R^{\prime }}^3$

since volume remains same

$\pi \frac{R^3}{6}=\frac{4}{3}\pi {R^{\prime }}^3⟹R^3=8{\left(R^{\prime }\right)}^3⟹R=2R^{\prime }$

let $I_1$ be the M.O.I of disc =$\frac{M{R}^2}{2}$

let $I_2$ be M.O.I of the sphere = $\frac{2}{5}M{\left(R^{\prime }\right)}^2=\frac{2}{5}M{\left(\frac{R}{2}\right)}^2$

$I_2=I_1\left(\frac{1}{5}\right)$