Angular Velocity Calculation for a Ballistic Cylinder with a Fired Bullet

[Punchlinegirl]

A 14.0 g bulllet is fired at 396.1 m/s into a solid cylinder of mass 18.1 kg and a radius 0.45 m. The cylinder is initially at rest and is mounted on a 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 9.00 cm from the center. Find the angular velocity of the system after the bullet strikes and adheres to the surface of the cylinder.
First I converted the velocity to angular velocity by dividing by the radius.
I used conservation of angular momentum
$$(MR^2) \omega = (MR^2 + (1/2)MR^2) \omega$$
(.14)(.09)(880.2) =((.14)(.09^2) + (1/2)(18.1)(.45^2)) \omega [/tex]
Solving for omega gave me .545 which wasn't right.. can someone tell me what I'm doing wrong? Thanks

 

[stuplato]

Conservation of momentum (mv = (M+m)v') will get a linear speed, which can be converted to rotational speed if I am remembering right...

 

[stuplato]

(.14)(.09)(880.2) =((.14)(.09^2) + (1/2)(18.1)(.45^2))

You lost a square on the .09 on the left side...

 

[lightgrav]

initial L should be .5 [m.kg.m/s] , (easiest by r x p) ... your omega_i is off.
Also, 14 gram = 0.014 kg .

 

[Punchlinegirl]

Can you explain to me how you got the .5 for the initial L?