Question

A uniform rod of length \( 2L \) is placed with one end in contact with the earth and is then inclined at an angle \( \alpha \) to the horizontal and allowed to fall without slipping at the contact point. When it becomes horizontal, its angular velocity will be:

• A. \( \sqrt{\frac{3g \sin \alpha}{2L}} \)
• B. \( \sqrt{\frac{2L}{3g \sin \alpha}} \)
• C. \( \sqrt{\frac{6g \sin \alpha}{L}} \)
• D. \( \sqrt{\frac{L}{g \sin \alpha}} \)

Hint:

Energy conservation helps in rotational motion problems involving falling rods.

Answer 1

The Correct Option is A

Using energy conservation: \[ PE_{\text{initial}} = KE_{\text{rot}}. \] The center of mass is initially at height \( L \sin \alpha \), so: \[ MgL \sin \alpha = \frac{1}{2} I \omega^2. \] For a rod pivoted at one end: \[ I = \frac{1}{3} M (2L)^2 = \frac{4}{3} M L^2. \] \[ gL \sin \alpha = \frac{2}{3} L^2 \omega^2. \] Solving for \( \omega \): \[ \omega = \sqrt{\frac{3g \sin \alpha}{2L}}. \] % Final Answer Thus, the correct answer is option (1): \( \sqrt{\frac{3g \sin \alpha}{2L}} \).