Question

The angular momentum of a uniform rod of mass $m$ and length $l$, rotating in a horizontal circle about one of its ends with an angular velocity $\omega$ is

• A. $ml^2\omega$
• B. $\frac{1}{4}ml^2\omega$
• C. $2ml^2\omega$
• D. $\frac{1}{3}ml^2\omega$
• E. $\frac{1}{2}ml^2\omega$

Hint:

Rod about one end: $I = \frac{1}{3}ml^2$.

Answer 1

The Correct Option is D

Concept:
• Angular momentum $L = I\omega$

Step 1: Moment of inertia of rod about end
\[ I = \frac{1}{3}ml^2 \]

Step 2: Compute angular momentum
\[ L = I\omega = \frac{1}{3}ml^2\omega \] Final Conclusion:
Option (D)