Question:medium
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
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
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Rod about one end: $I = \frac{1}{3}ml^2$.
Updated On: Apr 24, 2026
- $ml^2\omega$
- $\frac{1}{4}ml^2\omega$
- $2ml^2\omega$
- $\frac{1}{3}ml^2\omega$
- $\frac{1}{2}ml^2\omega$
Show Solution
The Correct Option is D
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