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

• A. \(0.12 \, \text{N m}\)
• B. \(0.6 \, \text{N m}\)
• C. \(1 \, \text{N m}\)
• D. \(0.8 \, \text{N m}\)

Hint:

Always convert moment of inertia to SI units before calculation. The rate of change of angular momentum is torque.

Answer 1

The Correct Option is C

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 \[ \boxed{1 \, \text{N m}} \]