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

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 $\omega =\frac{v}{r}$
$\therefore K=\frac{1}{2}\left(\frac{m{r}^2}{2}\right){\left(\frac{v}{r}\right)}^2$
$=\frac{1}{4}m{v}^2$