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): \( \sqrt{\frac{3g \sin \alpha}{2L}} \).

Answer 2

Applying the principle of energy conservation, we equate the initial potential energy of the rod to its rotational kinetic energy when it reaches the horizontal position: \[ PE_{\text{initial}} = KE_{\text{rot}}. \] Since the center of mass of the rod is at a height of \( L \sin \alpha \) above the final position, the initial potential energy is: \[ MgL \sin \alpha = \frac{1}{2} I \omega^2. \] For a uniform rod pivoted at one end, the moment of inertia is: \[ I = \frac{1}{3} M (2L)^2 = \frac{4}{3} M L^2. \] Substituting this into the energy equation gives: \[ gL \sin \alpha = \frac{2}{3} L^2 \omega^2. \] Rearranging to solve for the angular velocity, we get: \[ \omega = \sqrt{\frac{3g \sin \alpha}{2L}}. \]
Final answer: The angular velocity is \[ \boxed{\sqrt{\frac{3g \sin \alpha}{2L}}}. \]