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

Answer: (A)

$E=\frac{1}{2}I{\omega }^2\because L=I\omega$
$E=I\omega \Rightarrow E=\frac{L^2}{2I}$
Now, M.O.I. is given by
$I=m{k}^2$
$L_i=L_f$
Angular momentum must remain same always now the mass is half and radius of gyration is double.
The M.O.I is given by
$E^{\prime }=\frac{L^2}{2(2mk{)}^2}$
As we compare it with initial energy
$E^{\prime }=\frac{E}{2}$