Question

A uniform thin rod of length 0.500 m and mass 4.00 kg is attached to a pivot at one end. The other end of the rod is attached to a uniform sphere with a mass of 2.00 kg and a radius of 0.100 m. The rod and sphere initially hang vertically. A bullet with a mass of 5.00 g is fired horizontally at a speed of 200 m/s into the center of the sphere. The bullet embeds itself in the sphere. What is the angular speed of the rod, sphere, and bullet after the collision?

Answer 1

angular momentum:
l=rmvsinQ

The rotational inertia on the rod:
I=ML^2/12

The parallel-axis theorem:
I=Irod+mr^2

Therefore:
rmvsinQ=(ML^2/12+mr^2)w
v=(4)(0.50)^2+(0.005)(0.275)^2(11.0)/(0.250)(0.005)sin(60.0)
v=25618.6 m/s