A man, mass M, stands on a massless rod which is free to rotate about its center in the horizontal plane. The man has a gun (massless) with one bullet, mass m. He shoots the bullet with velocity Vb, horizontally. Find the angular velocity of the man as a function of the angle θ which the bullet's velocity vector makes with the rod.
Torque = r x F
L = r x mv
T = dL/dt
The Attempt at a Solution
There's no external torque acting on the system of the man and the bullet. Torque = 0 = dL/dt means that L is conserved. The system was at rest meaning Lint was 0. The angular momentum for the man is Mmanω(l/2)^2 post firing. Here's my problem, the bullet's radial vector is changing. r = < l/2 + Vbcos(θ)t, vbsin(theta)t>x<mbVbcosθ, mbVbsinθ>
The book gives the answer as mbvbsinθ/(mman(l/2))
Where did I go wrong?