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Question

Physics Question on System of Particles & Rotational Motion

A slender uniform rod of mass MM and length ll is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ\theta with the vertical is :

A

3g2lsinθ\frac{3g}{2 l} \sin \, \theta

B

2g3lsinθ\frac{2g}{3l} \sin \, \theta

C

3g2lcosθ\frac{3g}{2l} \cos\, \theta

D

2g2lsinθ\frac{2g}{2l} \sin\, \theta

Answer

3g2lsinθ\frac{3g}{2 l} \sin \, \theta

Explanation

Solution

Torque at angle θ\theta
τ=Mgsinθ.\tau = Mg \, \sin \, \theta .