Question

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

• A. $1 \,ms ^{-1}$
• B. $0.5\, ms ^{-1}$
• C. $2\, ms ^{-1}$
• D. $1.5\, ms ^{-1}$

Answer 1

Answer: (A)

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