Question

A uniform rod of mass M and length L is hinged at its end. The rod is released from its vertical position by slightly pushing it. What is the reaction at the hinge when the rod becomes horizontal, again vertical.

Answer 1

Answer:

Figure shows the rod at ann angle θθ to the vertical. If we take torques about the pivot we need not be concerned with the force due to the pivot.

The torque due to the weight about OO.

τmg=mgL2sinθτmg=mgL2sinθ ltbr. Using tourqe equation about OO. Here we apply torque equation about fixed axis passing through OO. We can apply torque equation about fixed axis or about centre of mass only.



a. τ=Iα⇒mgL2sinθ=ML23ατ=Iα⇒mgL2sinθ=ML23α

Thus,α=3gsinθ2Lα=3gsinθ2L

when the rod is horizontal θ=π2θ=π2 and α=3g2Lα=3g2L

b. The tangenital linear acceleration of the free end is at=α=3g2