The moment of inertia of a body about a given axis is $1.2 \mathrm{~kg} \mathrm{~m}^{2}$. Initially, the…

The moment of inertia of a body about a given axis is $1.2 \mathrm{~kg} \mathrm{~m}^{2}$. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 joule, an angular acceleration of 25 radian $/ \mathrm{sec}^{2}$ must be applied about that axis for a duration of
  1. 4 seconds
  2. 2 seconds
  3. 8 seconds
  4. 10 seconds

Solution

$
\begin{array}{l}
\text { (b) } \mathrm{I}=1.2 \mathrm{~kg} \mathrm{~m}^{2}, \mathrm{E}_{\mathrm{r}}=1500 \mathrm{~J}\\
\alpha=25 \mathrm{rad} / \mathrm{sec}^{2}, \omega_{1}=0, \mathrm{t}=?\\
\text { As } \mathrm{E}_{\mathrm{r}}=\frac{1}{2} \mathrm{I} \omega^{2}\\
\omega=\sqrt{\frac{2 \mathrm{E}_{\mathrm{r}}}{\mathrm{I}}}=\sqrt{\frac{2 \times 1500}{1.2}}=50 \mathrm{rad} / \mathrm{sec}\\
\text { From } \omega_{2}=\omega_{1}+\alpha t\\
50=0+25 \mathrm{t}, \quad \therefore \quad \mathrm{t}=2 \text { seconds }
\end{array}
$

Asked in: JEE Mains - Rotational Motion - Test 3

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