Question:
The angular velocity of a body changes from \(6 \, \text{rad/s}\) to \(21 \, \text{rad/s}\) in a time of \(1.5 \, \text{s}\). If the moment of inertia of the body is \(100 \, \text{g m}^2\), then the rate of change of angular momentum of the body is
The angular velocity of a body changes from \(6 \, \text{rad/s}\) to \(21 \, \text{rad/s}\) in a time of \(1.5 \, \text{s}\). If the moment of inertia of the body is \(100 \, \text{g m}^2\), then the rate of change of angular momentum of the body is
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Always convert moment of inertia to SI units before calculation. The rate of change of angular momentum is torque.
Updated On: Jun 6, 2025
- \(0.12 \, \text{N m}\)
- \(0.6 \, \text{N m}\)
- \(1 \, \text{N m}\)
- \(0.8 \, \text{N m}\)
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The Correct Option is C
Solution and Explanation
Step 1: Given:
\[
\omega_1 = 6 \, \text{rad/s},
\omega_2 = 21 \, \text{rad/s},
\Delta t = 1.5 \, \text{s},
I = 100 \, \text{g m}^2 = 0.1 \, \text{kg m}^2 \] Step 2: Angular momentum \( L = I \omega \) Step 3: Rate of change of angular momentum: \[ \frac{dL}{dt} = I . \frac{\Delta \omega}{\Delta t} = 0.1 . \frac{21 - 6}{1.5} = 0.1 . \frac{15}{1.5} = 0.1 . 10 = 1 \, \text{N m} \] % Final Answer \[ \boxed{1 \, \text{N m}} \]
\omega_2 = 21 \, \text{rad/s},
\Delta t = 1.5 \, \text{s},
I = 100 \, \text{g m}^2 = 0.1 \, \text{kg m}^2 \] Step 2: Angular momentum \( L = I \omega \) Step 3: Rate of change of angular momentum: \[ \frac{dL}{dt} = I . \frac{\Delta \omega}{\Delta t} = 0.1 . \frac{21 - 6}{1.5} = 0.1 . \frac{15}{1.5} = 0.1 . 10 = 1 \, \text{N m} \] % Final Answer \[ \boxed{1 \, \text{N m}} \]
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