Ballistic Cylinder: Angular Velocity After Bullet Impact

[lew44]

Homework Statement

A 13.0 g bullet is fired at 578.1 m/s into a solid cylinder of mass 24.1 kg and a radius 0.11 m. The cylinder is initially at rest and is mounted on fixed vertical axis that runs through it's center of mass.
The line of motion of the bullet is perpendicular to the axle and at a distance 2.20 cm from the center. Find the angular velocity of the system after the bullet strikes and adheres to the surface of the cylinder.

Homework Equations

conservation of angular momentum

The Attempt at a Solution

I tried,

.013 x (.022^2) (578.1/.11) = ((.013)(.022^2) + (.5)(24.1)(.11^2)) x W

I got .227rad/s as an answer but that was incorrect. Any ideas?

 

[Filip Larsen]

If you think of the bullet as an impulsive force on the cylinder you should be able to figure out how much of that force is radial to the cylinder (which is absorbed by the bearing) and how much is tangential (which will make the cylinder rotate). If I do the calculation I end up with a angular speed of around 1.13 rad/s.

The right-hand side of your equation looks ok to me, but I'm not sure I understand how you arrived at the left-hand side of your equation. I get something similar but not quite the same. Perhaps you missed taking only the tangential part of the impulse.