Angular momentum question HELP ASAP

Question

A bullet of mass 200 g is shot with 𝑣 = 100 𝑚 𝑠 toward a block of mass 2 kg that is placed on a frictionless table. The block is connected to rod of negligible mass and length 2 m, which is pinned at point P. The bullet, which is shot perpendicular to the rod, enters the block, and becomes embedded in it. The system then rotates about point P with angular speed 𝜔. What is 𝜔? Show your detailed work.

The answer is supposed to be 4.5 rad/s but I'm not sure how to do it. Please help! The teachers solution says that the initial angular momentum is mvr but how does the bullet have an initial angular momentum? Isn't it only linear momentum in the beginning?

Answer 1

It is correct that the bullet has non zero linear momentum before the collision; because of that, it also has non zero angular momentum with respect to pin point P ( if you chose another reference point the bullet is pointing to, for instance the big mass M at the end of the rod, then the angular momentum would be 0).

With respect to P, the total angular momentum is conserved, so L_initial = L_final.

L_initial = angular momentum of bullet = mv * L

L_final = I * ω, I = inertia of the (mass + bullet) system with respect to P (I = (m+M)L^2)

, ω = angular velocity

By imposing conservation: mvL = (m+M)L^2 * ω, you can solve for ω = 50/11 rad/s ~ 4.5 rad/s

PS Please note that the angular momentum would not be necessarily conserved if you chose one other reference point other than P (for instance, the initial location of the mass M).