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Question

A uniform rod of length L and mass M is pivoted freely at one end. The angular acceleration of the rod when it is at angle θ to the vertical is:

A
3gcosθ2L
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B
2L3gsinθ
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C
3gsinθ2L
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D
32gLsinθ
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Solution

The correct option is C 3gsinθ2L
Torque, T=Iα
where I is moment of inertia, and α is angular acceleration.
Also, T=rFsinθ
mgsinθ(l2)=ml2α3
α=3gsinθ2l

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