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Units And Measurement

Question
CBSEENPH11020769

A uniform rod AB of length l and mass m 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 ml2/3, the initial angular acceleration of the rod will be:

  • 2g/3l

  • mgl/2

  • 3gl/2

  • 3g/2l

Solution

D.

3g/2l

The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is 
l = ml3/3
where m is a mass of rod and l is the
straight tau space equals space mg straight iota over 2
Iα space equals space mg straight l over 2
ml squared over 3 straight alpha space equals space mg straight l over 2
straight alpha space equals space fraction numerator 3 straight g over denominator 2 straight l end fractionlength.
Torque acting on centre of gravity of rod is given by