Question

A body having a moment of inertia about its axis of rotation equal to $3kg{m}^2$ is rotating with angular velocity of $3rad{s}^{-1}$. Kinetic energy of this rotating body is same as that of a body of mass $27kg$ moving with velocity $v$. The value of $v$ is

• A. $1m{s}^{-1}$
• B. $0.5m{s}^{-1}$
• C. $2m{s}^{-1}$
• D. $1.5m{s}^{-1}$

Answer 1

Answer: (A)

We know that
The kinetic energy $=K=\frac{1}{2}m{v}^2=\frac{1}{2}I{\omega }^2$
where $m=27kg$ (mass of the body)
$\omega =3rad/s$ (angular velocity)
$I=3kg-m^2$ (moment of inertia)
$v=?$
Or $m{v}^2=I{\omega }^2$
$v^2=\frac{I{\omega }^2}{m}$
$v^2=\frac{3\times 3^2}{27}$
$v^2=\frac{27}{27}=1$
$v=\sqrt{1}=1m{s}^{-1}$