A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is: (Moment of inertia of rod about A is ) from Physics Laws of Motion Class 11 Manipur Board
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A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is: (Moment of inertia of rod about A is fraction numerator straight m l squared over denominator 3 end fraction)

  • fraction numerator 3 straight g over denominator 2 straight l end fraction
  • fraction numerator 2 straight l over denominator 3 straight g end fraction
  • fraction numerator 3 straight g over denominator 2 straight l squared end fraction
  • mg straight l over 2

A.

fraction numerator 3 straight g over denominator 2 straight l end fraction

The moment of inertia of the uniform rod about an axis through the end and perpendicular to length is
           straight I space equals space fraction numerator straight m l squared over denominator 3 end fraction
where m is mass of rod and l its length.
Torque left parenthesis straight t space equals space Iα right parenthesis acting on centre of gravity of rod is given by
                  straight t equals mg l over 2
or               Iα space equals space mg l over 2
or space space fraction numerator straight m l squared over denominator 3 end fraction straight alpha space equals space mg l over 2
therefore space space space space straight alpha space equals fraction numerator 3 straight g over denominator 2 l end fraction


 

The moment of inertia of the uniform rod about an axis through the end and perpendicular to length is
           straight I space equals space fraction numerator straight m l squared over denominator 3 end fraction
where m is mass of rod and l its length.
Torque left parenthesis straight t space equals space Iα right parenthesis acting on centre of gravity of rod is given by
                  straight t equals mg l over 2
or               Iα space equals space mg l over 2
or space space fraction numerator straight m l squared over denominator 3 end fraction straight alpha space equals space mg l over 2
therefore space space space space straight alpha space equals fraction numerator 3 straight g over denominator 2 l end fraction


 

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