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Question

A uniform rod of mass m and length l can rotate in a vertical plane about a smooth horizontal axis hinged at point H. Find the force exerted by the hinge just after the rod is released as shown in the figure.


A
mg7
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B
mg105
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C
mg5
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D
mg5
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Solution

The correct option is B mg105

Taking torque equation about hinge H,
τnet=Iα
mgcos37×l2=ml23α
α=6g5l
[torque due to hinge reaction is zero]

Let the hinge force components be N1 and N2
Tangential acceleration of COM of the system
at=l2α=3g5

Writing force equation perpendicular to the rod,
mgcos37N1=mat
mg×45N1=m×3g5
N1=mg5
Since, angular velocity (ω)=0 at the time of release and net force along the rod is zero since there is no acceleration towards the center,
N2=mgsin37=3mg5

Therefore, hinge force (N) is
N=N21+N22=(mg5)2+(3mg5)2=mg105
Hence, option (b) is the correct answer.

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