The moment of inertia of a body about a given axis is $12 \mathrm{~kg}-\mathrm{m}^2$. Initially, the body is…
The moment of inertia of a body about a given axis is $12 \mathrm{~kg}-\mathrm{m}^2$. Initially, the body is at rest. In order to produce a rotational kinetic energy of $15000 \mathrm{~J}$, an angular acceleration of $10 \mathrm{rads}^{-2}$ must be applied about that axis for a duration of
2 s
4 s
10 s
5 s
Solution
Given, moment of inertia, $I=12 \mathrm{~kg}-\mathrm{m}^2$
Rotational kinetic energy, $E_k=\frac{1}{2} I \omega^2$
$
15000 \mathrm{~J}=\frac{1}{2} \times 12 \times \omega^2
$
So,
$
\omega=\sqrt{\frac{30000}{12}}=\sqrt{2500}=50 \mathrm{rad} / \mathrm{s}
$
Angular acceleration,
$
\alpha=10 \mathrm{rad} / \mathrm{s}^2
$
As
$
\alpha=\frac{\omega}{t}
$
So,
$
\text { time }=\frac{\omega}{\alpha}=\frac{50}{10}=5 \mathrm{~s}
$