A uniform rod of length $ L $ is hinged at one end and initially held in horizontal position. After it is released, find its angular speed when the rod becomes vertical.

Question

A uniform rod of length $ L $ is hinged at one end and initially held in horizontal position. After it is released, find its angular speed when the rod becomes vertical., $ (g L)^{1 / 2} $, $ \left(\frac{2 g}{L}\right)^{1 / 2} $, $ \left(\frac{3 g}{L}\right)^{1 / 2} $, $ \left(\frac{4 g}{L}\right)^{1 / 2} $

MARKS App · Accepted Answer

The gain in kinetic energy, ∆ K E = 1 2 m l 2 3 ω 2 Loss in potential energy, - ∆ U = m g h loss , where h loss = L 2 is the loss of height of the center of mass. By conservation of mechanical energy, ∆ K E = - ∆ U m g L 2 = 1 2 m L 2 3 ω 2 ⇒ ω = 3 g L 1 2

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