Question

A circular ring of mass $m$ and radius $r$ is rolling on a smooth horizontal surface with speed $v$. Its kinetic energy is:

• A. $ \frac{1}{8}mv^{2} $
• B. $ \frac{1}{4}mv^{2} $
• C. $ \frac{1}{3}mv^{2} $
• D. $ mv^{2} $

Answer 1

Answer: (B)

Kinetic energy of rotation of a body having moment of inertia $I$ and angular velocity $\omega$ is given by
$K=\frac{1}{2} I \omega^{2}$
For a circular ring of radius $r$,
and mass $m$ moment of inertia $(I)$
about its diameter is given by
$I=\frac{M r^{2}}{2} $
$\therefore K=\frac{1}{2}\left(\frac{M r^{2}}{2}\right)^{2} \omega^{2}$
Also, $v=r \omega$,
therefore $\omega=\frac{v}{r} $
$\therefore K=\frac{1}{2}\left(\frac{m r^{2}}{2}\right)\left(\frac{v}{r}\right)^{2}$
$=\frac{1}{4} m v^{2}$