Question

The rotational kinetic energy of a body is $E$. In the absence of external torque, if mass of the body is halved and radius of gyration doubled, then its rotational kinetic energy will be :-

• A. $0.5E$
• B. $0.25E$
• C. $E$
• D. $2E$

Answer 1

The correct option is A $0.5E$

Rotational kinetic energy of the body $E=\frac{1}{2}I{w}^2$

Using $L=Iw$

We get $E=\frac{L^2}{2I}$

Since moment of inertia $I=m{k}^2$ where $k$ is radius of gyration

kinetic energy $E=\frac{L^2}{2m{k}^2}$

In the absence of external torque, $L=constant$

$⟹E\propto \frac{1}{m{k}^2}$

So, $\frac{E_2}{E_1}=\frac{m_1{k}_1^2}{m_2{k}_2^2}$

$⟹\frac{E_2}{E}=\frac{m{k}^2}{\frac{m}{2}\times (2k{)}^2}=\frac{1}{2}$

$⟹E=0.5E$