Questions
A horizontal heavy uniform bar of weight W is supported at its ends by two men. At the instant, one of the men lets go off his end of the rod, the other feels the force on his hand changed to
detailed solution
Correct option is D
Let the mass of the rod is M⇒ Weight (W)=MgInitially for the equilibrium F+F=Mg⇒F=Mg/2When one man withdraws, the torque on the rod τ=Iα=Mgl2⇒ Ml23α=Mgl2 As I=Ml2/3⇒ Angular acceleration, α=32gland linear acceleration,a=l2α=3g4Now if the new normal force at A is F′ then Mg−F′=Ma⇒ F′=Mg−Ma=Mg−3Mg4=Mg4=W4Talk to our academic expert!
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A uniform rood AB of length and mass is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is , the initial angular acceleration of the rod will be
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