A body of mass $0.6 \mathrm{~kg}$ is moving along a circular path of radius $1 \mathrm{~m}$. If the body…
A body of mass $0.6 \mathrm{~kg}$ is moving along a circular path of radius $1 \mathrm{~m}$. If the body moves with $\frac{900}{\pi}$ revolutions per minute, its kinetic energy
$120 \mathrm{~J}$
$270 \mathrm{~J}$
$360 \mathrm{~J}$
$240 \mathrm{~J}$
Solution
We have
$
\begin{aligned}
& \mathrm{V}=\omega \mathrm{r} \\
& =2 \pi \mathrm{fr} \\
& =2 \pi \times \frac{15}{\pi} \times 1 \\
& =30 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
So, Kinetic energy $=\frac{1}{2} \times 0.6 \times 900$
$
=270 \mathrm{~J}
$