A solid sphere of mass $2 \mathrm{~kg}$ is rolling without slipping on a horizontal surface with a velocity…
A solid sphere of mass $2 \mathrm{~kg}$ is rolling without slipping on a horizontal surface with a velocity $5 \mathrm{~ms}^{-1}$. The rotational kinetic energy of the sphere is
$25 \mathrm{~J}$
$12.5 \mathrm{~J}$
$10 \mathrm{~J}$
$20 \mathrm{~J}$
Solution
Mass of solid sphere, $m=2 \mathrm{~kg}$ velocity, $v=5 \mathrm{~m} / \mathrm{s}$
Rotational kinetic energy is given by
$
\begin{aligned}
& \text { K. } E=\frac{1}{2} I \omega^2 \\
& \left(\text { Here } I=\frac{2}{5} \mathrm{~m} \mathrm{r}^2 ; \mathrm{v}=\mathrm{r} \omega\right) \\
& =\frac{1}{2} \times \frac{2}{5} \mathrm{mr}^2 \times \frac{\mathrm{v}^2}{\mathrm{r}^2} \\
& =\frac{1}{5} \mathrm{mv}^2=\frac{1}{5} \times 2 \times 5^2=10 \mathrm{~J}
\end{aligned}
$