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## Homework Statement

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.

## Homework Equations

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 L

_{int}was 0. The angular momentum for the man is M

_{man}ω(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<m

_{b}V

_{b}cosθ, m

_{b}V

_{b}sinθ>

The book gives the answer as m

_{b}v

_{b}sinθ/(m

_{man}(l/2))

Where did I go wrong?