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
mg√7
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B
mg√105
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C
mg5
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D
mg√5
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Solution
The correct option is Bmg√105
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, mgcos37∘−N1=mat ⇒mg×45−N1=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=mg√105
Hence, option (b) is the correct answer.