A rod AB of length $L$ and mass $M$ is hinged at point A . A small bullet of mass B hits the rod with velocity $v$ perpendicular to AB at B. The bullet gets embedded in the rod . Find the angular velocity of the system just after the impact?

We can do this in many ways but each time I get a different answer I try to apply first angular momentum conservation about A of the system then linear momentum conservation but only the answer with angular momentum conservation equal the answer not the other.

So why can't we use linear momentum conservation here that is after impact assume it to be total mass of the rod and bullet times velocity of centre of mass and then multiply by distance from A to find angular velocity which is apparently wrong so why is it so ?


The problem is due to the hinge. Hinges constrain the location of an object (in this case, the end of the rod). It does this through forces.

The way the problem is set up, you don't know the exact forces that are applied by the hinge. In the case of angular momentum about an axis through the hinge, these forces are irrelevant. They supply no torque about that axis.

In the case of linear momentum, the forces are relevant. You would need both the hinge forces (or impulse) and the force (or impulse) from the bullet to find the final linear momentum of the rod. When you do your calculation, what forces on the rod do you use from the hinge?

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    $\begingroup$ I didnt use any force I simply equated the linear momentum of bullet to the linear momentum of rigid body ie mv = (m+M)V where V is velocity of centre of mass of system. And then I found location of centre of mass from A and divided this by velocity to get angular velocity so what did I do wrong ? $\endgroup$ – Matt May 6 '17 at 3:32
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    $\begingroup$ Momentum is conserved only when there are no external forces. Your method would work if the bullet and rod were in free space. But the hinge introduces external forces, so linear momentum is no longer conserved. $\endgroup$ – BowlOfRed May 6 '17 at 4:02

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