A uniform thin rod of length 0.500 m and mass 4.0 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 10 rad/s immediately after the collision, what is the bullet's speed just before impact? Solution with diagram on page 4m problem #11.55: http://www.ifm.liu.se/edu/coursescms/TFYA16/lessons/Le-8-extra.pdf [Broken] Moment inertia of system: I = Irod + mr^2 Solution: conservation of ang momentum: rmvsinθ = (1/12 ML^2 + mr^2)ω So conservation of ang momentum is Linitial = Lfinal but this problem uses: ang momentum of a particle = ang momentum of a rigid body? What i dont understand: Why is the Moment inertia of system I = Irod + mr^2? what i did was before looking up the solution was: ML^(2)/12 ω = rmvsinθ + MR^(2)ω because ang momentum inital was just the rod rotating by itself and then comes the bullet so ang momentum of a bullet plus ang momentum of the rod. rmvsinθ = (Irod + mr^2)ω doesnt make sense to me. Also can someone explain to me what a rigid body is? because i know rigid body is L = Iω. Are rigid bodies like a rod, ring, disk or shell?