Question:
The angular momentum of a wheel having a rotational inertia of
\[
0.2\,\text{kg m}^2
\]
about its symmetric axis decreases from
\[
4 \text{ to } 2\,\text{kg m}^2\text{s}^{-1}
\]
in \(4\,\text{s}\). The average power of the wheel is:
The angular momentum of a wheel having a rotational inertia of
\[
0.2\,\text{kg m}^2
\]
about its symmetric axis decreases from
\[
4 \text{ to } 2\,\text{kg m}^2\text{s}^{-1}
\]
in \(4\,\text{s}\). The average power of the wheel is:
Show Hint
For rotational motion,
\[
L=I\omega
\]
and
\[
K=\frac{1}{2}I\omega^2.
\]
First find angular velocities from angular momentum, then compute rotational kinetic energies.
Updated On: Jun 24, 2026
- \(7.5\,\text{W}\)
- \(15\,\text{W}\)
- \(5\,\text{W}\)
- \(12\,\text{W}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use the relation between angular momentum and angular velocity.
Angular momentum is \[ L=I\omega \] Given, \[ I=0.2\,\text{kg m}^2 \] Initial angular momentum: \[ L_1=4\,\text{kg m}^2\text{s}^{-1} \] Final angular momentum: \[ L_2=2\,\text{kg m}^2\text{s}^{-1} \] Thus, \[ \omega_1=\frac{L_1}{I}=\frac{4}{0.2}=20\,\text{rad/s} \] \[ \omega_2=\frac{L_2}{I}=\frac{2}{0.2}=10\,\text{rad/s} \]
Step 2: Find initial and final rotational kinetic energies.
Rotational kinetic energy is \[ K=\frac{1}{2}I\omega^2 \] Initial kinetic energy: \[ K_1=\frac{1}{2}(0.2)(20)^2 \] \[ K_1=0.1\times 400 \] \[ K_1=40\,\text{J} \] Final kinetic energy: \[ K_2=\frac{1}{2}(0.2)(10)^2 \] \[ K_2=0.1\times 100 \] \[ K_2=10\,\text{J} \]
Step 3: Find change in energy.
Energy lost: \[ \Delta K=40-10=30\,\text{J} \] Time taken: \[ t=4\,\text{s} \] Average power: \[ P=\frac{\Delta K}{t} \] \[ P=\frac{30}{4} \] \[ P=7.5\,\text{W} \]
Step 4: Final conclusion.
Hence, the average power of the wheel is \[ \boxed{7.5\,\text{W}} \]
Angular momentum is \[ L=I\omega \] Given, \[ I=0.2\,\text{kg m}^2 \] Initial angular momentum: \[ L_1=4\,\text{kg m}^2\text{s}^{-1} \] Final angular momentum: \[ L_2=2\,\text{kg m}^2\text{s}^{-1} \] Thus, \[ \omega_1=\frac{L_1}{I}=\frac{4}{0.2}=20\,\text{rad/s} \] \[ \omega_2=\frac{L_2}{I}=\frac{2}{0.2}=10\,\text{rad/s} \]
Step 2: Find initial and final rotational kinetic energies.
Rotational kinetic energy is \[ K=\frac{1}{2}I\omega^2 \] Initial kinetic energy: \[ K_1=\frac{1}{2}(0.2)(20)^2 \] \[ K_1=0.1\times 400 \] \[ K_1=40\,\text{J} \] Final kinetic energy: \[ K_2=\frac{1}{2}(0.2)(10)^2 \] \[ K_2=0.1\times 100 \] \[ K_2=10\,\text{J} \]
Step 3: Find change in energy.
Energy lost: \[ \Delta K=40-10=30\,\text{J} \] Time taken: \[ t=4\,\text{s} \] Average power: \[ P=\frac{\Delta K}{t} \] \[ P=\frac{30}{4} \] \[ P=7.5\,\text{W} \]
Step 4: Final conclusion.
Hence, the average power of the wheel is \[ \boxed{7.5\,\text{W}} \]
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