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a rod rotates about point A .The rod is released from rest in the horizontal position so that it rotates about point A.[moment of inertia about point A =ml^2/3].find initial angular acceleration. [in the question how can we equate torque at A and torque due to mg. torque at A is 0.]

a rod rotates about point A .The rod is released from rest in the horizontal position so that it rotates about point A.[moment of inertia about point A =ml^2/3].find initial angular acceleration. [in the question how can we equate torque at A and torque due to mg. torque at  A is 0.]

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Grade:11

1 Answers

chetan jangir
101 Points
2 years ago
Hey Liza I hope you know how to calculate torque ,Torque due to force about a point is Forcein perpendicular direction *(into)its perpenidular distance about that point (here A) SO we know equation of rotation motion torque net= I(moment of inertia)*angular acceleration .Mg*L/2=Ml^{^{2}}/3\alpha (here \alpha= angular acceleration) Calculating \alpha=3g/2l
Here at A a tension acts but it’s torque is zero as it’s perdicular distance is zero and also it passes through A

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