Question

A solid sphere of radius ${}^{\prime }{r}^{\prime }$ and mass ${}^{\prime }{m}^{\prime }$ is rotating about an axis passing through its COM with an angular speed ${}^{\prime }{\omega }^{\prime }$ and has rotational kinetic energy $k_1$. If the radius of the sphere is halved and angular speed is doubled, then its rotational kinetic energy becomes $k_2$. Which of the following options is correct ?

• A. $k_1>k_2$
• B. $k_1<k_2$
• C. $k_1=k_2$
• D. $k_1=k_2=0\text{J}$

Answer 1

The correct option is C $k_1=k_2$

Given,
$r_1=r;m_1=m,{\omega }_1=\omega$
$r_2=\frac{r}{2};m_2=m;{\omega }_2=2\omega$
Rotational kinetic energy of solid sphere for $1^{st}$ case:
$k_1=\frac{1}{2}{I}_1{\omega }_1^2=\frac{1}{2}\left(\frac{2}{5}{m}_1{r}_1^2\right){\omega }_1^2$
$\Rightarrow k_1=\frac{1}{5}m{r}^2{\omega }^2$
Rotational kinetic energy of solid sphere for $2^{nd}$ case:
$k_2=\frac{1}{2}{I}_2{w}_2^2=\frac{1}{2}\left(\frac{2}{5}{m}_2{r}_2^2\right)w_2^2$
$\Rightarrow k_2=\frac{1}{5}m{r}^2{w}^2$
$\therefore k_1=k_2\ne 0$