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Question

Immediately after the right string is cut, what is the linear acceleration of the free end of the rod ?
219021_25a2d074a07e4d3080706120c9cd2465.png

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Solution

Torque about A
τA=Iα , where I is moment of inertia about A , is I=ml23
Center of mass of the rod acts its geometrical center i.e. AC=l/2
So mg×l2=ml23α
α=3g2l
Acceleration of free end B is aB=α×BA=3g2l×l
aB=3g2

689006_219022_ans_306629e5448640b9a77c0ea458b4b8f8.png

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