Question:
The angular momentum of a wheel having a rotational inertia of \( 0.2 \, \text{kg} \, \text{m}^2 \) about its symmetric axis decreases from \( 4 \) to \( 2 \, \text{kg} \, \text{m}^2 \, \text{s}^{-1} \) in \( 4 \, \text{s} \). The average power of the wheel is
The angular momentum of a wheel having a rotational inertia of \( 0.2 \, \text{kg} \, \text{m}^2 \) about its symmetric axis decreases from \( 4 \) to \( 2 \, \text{kg} \, \text{m}^2 \, \text{s}^{-1} \) in \( 4 \, \text{s} \). The average power of the wheel is
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Use the change in rotational kinetic energy to compute average power when angular velocity is indirectly given via angular momentum.
Updated On: Mar 16, 2026
- 7.5 W
- 15 W
- 5 W
- 12 W
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The Correct Option is A
Solution and Explanation
Given:
\[
\begin{align}
\text{Initial angular momentum} = L_1 = 4 \, \text{kg m}^2/\text{s}, \quad
\text{Final angular momentum} = L_2 = 2 \, \text{kg m}^2/\text{s}
\]
\[
\text{Time interval} = \Delta t = 4 \, \text{s}, \quad \text{Moment of inertia} = I = 0.2 \, \text{kg m}^2
\]
Angular momentum \( L = I \omega \Rightarrow \omega = \frac{L}{I} \)
So:
\[
\omega_1 = \frac{4}{0.2} = 20 \, \text{rad/s}, \quad \omega_2 = \frac{2}{0.2} = 10 \, \text{rad/s}
\]
Now, average power is:
\[
\begin{align}
\text{Power} = \frac{\Delta(\text{Rotational K.E.})}{\Delta t} = \frac{\frac{1}{2} I (\omega_1^2 - \omega_2^2)}{\Delta t}
\]
\[
\begin{align}
= \frac{1}{2} \cdot 0.2 \cdot (400 - 100) / 4 = \frac{0.2 \cdot 300}{2 \cdot 4} = \frac{60}{4} = 7.5 \, \text{W}
\]
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