Solution:
Torque of gravitational force mg about O is mgcosθ ×L/2
Torque = I × α
mgcosθ ×L/2 = (mL²/3) × α
α = (3gcosθ)/2
A uniform pole of length l and mass m is pivoted on the ground with a frictionless hinge O. The pole is free to rotate without friction about an horizontal axis axis passing through O and normal to plane of the page. The pole makes an angle θ with the horizontal. The pole is released from rest in the position shown, then linear acceleration of the free end (P) of the pole just after its release would be
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