Question

A uniform rod of mass M = 5.02kg and length L = 1.08m can pivot freely (i.e., we ignore friction) about a hinge attached to a wall, as seen in the figure below.



The rod is held horizontally and then released. At the moment of release, determine the angular acceleration of the rod. Use units of rad/s^2.

Mg L L2

Answer 1

At the moment of release the torque on the rod about the hinge is;

T = Mg(L/2)

Moment of inertia of the rod about the hinge is,

I = ML²/3

We have,

T = Iα, where α is angular acceleration of the rod.

or,  Mg(L/2) = [ ML²/3]α

or, g/2 = [L/3]α

or, α = 3g/2L = 3(9.8 m/s²)/(2X1.08 m)

or, α = 13.61 rad/s² is  the angular acceleration when the rod is released.

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