Angular Velocity Calculation for a Ballistic Cylinder with a Fired Bullet

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The discussion revolves around calculating the angular velocity of a solid cylinder after a bullet strikes and adheres to it. The bullet's mass is 14.0 g, and it is fired at a velocity of 396.1 m/s into a cylinder with a mass of 18.1 kg and a radius of 0.45 m. The conservation of angular momentum is applied to find the angular velocity, but the initial calculations are incorrect due to a missing square in the distance used. Participants clarify that the initial angular momentum should be calculated as 0.5 times the mass times the radius squared and correct the bullet's mass to 0.014 kg. Accurate calculations are essential for determining the correct angular velocity of the system post-collision.
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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
 
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Conservation of momentum (mv = (M+m)v') will get a linear speed, which can be converted to rotational speed if I am remembering right...:rolleyes:
 
(.14)(.09)(880.2) =((.14)(.09^2) + (1/2)(18.1)(.45^2))

You lost a square on the .09 on the left side...
 
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 .
 
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Can you explain to me how you got the .5 for the initial L?
 
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