Rotational Motion Question 131

Question: 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

Options:

A) W

B) W2

C) 3W4

D) W4

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Let the mass of the rod is MWeight(W)=Mg

Initially for the equilibrium

F+F=MgF=Mg/2

When one man withdraws, the torque on the rod τ=Iα=Mgl2 Ml23α=mgl2

As I=Ml2/3

Angular acceleration, α=32gl

And linear acceleration

a=12α=3g4

Now if the new normal force at A is F’ then

MgF=Ma

F=MgMa=Mg3Mg4=Mg4=W4.



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