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I had a doubt regarding conservation of angular momentum in the following problem.

'A rod AB of length l and mass m is hinged at the topmost point A and it is free to rotate about this point. If a ball of mass m/2 moving at speed u strikes the other end and sticks to it , find the angular speed with which the rod will move after the collision.'

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The book said that we can conserve angular momentum of the (bullet + rod) system about the hinge point. I understood the fact that the torque due to the hinge reaction is zero as well as the torque due to the weight of the rod. But by doubt is whether the torque due to weight of the bullet about the hinge point is also zero since the necessary condition to conserve angular momentum is that net external torque about that point should be zero. Does air resistance or any resistive force come into play so as to make the net torque equal to zero?

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At the very moment that the bullet sticks to the rod, the rod remains vertical and therefore the weight of the bullet is downward along the line passing through the hinge. The torque then is zero. You can, therefore, apply the conservation of momentum to the moment just before and just after the collision.

Following the collision, the rod begins to rise due to the impulse transferred to it and as it is no longer vertical, the torque due to its weight will no longer be zero. Therefore you cannot apply the conservation of angular momentum for the motion following the collision. (But you can apply conservation of energy now!)

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