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A 12.0g bullet is fired at 570.1m/s into a solid cylinder of mass 16.1kg and a radius 0.470m. 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 9.40 cm from the center. Find the angular velocity of the system after the bullet strikes and adheres to the surface of the cylinder.

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Added Sun, 21 Jun '15
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this is a problem of conservation of angular momentum

the bullet has angular momentum with respect to the cylinder; the amount of this angular momentum is m v d where m is the bullet mass, v is its velocity and d is its perpendicular distance from the cylinder's rotation axis

the angular momentum of the cylinder after impact will be I*w

where I is the moment of inertia of the cylinder/bullet system and w the angular velocity

the moment of inertia of the system will be I(cylinder)+I(due to bullet)

I(cylinder)=1/2MR^2 where M, R are the mass and radius of the cylinder,
and the contribution to the system's I from the bullet is mR^2 where m is the mass of the bullet

since angular momentum is conserved, we have

angular momentum before impact = angular momentum after impact

m v d = (1/2MR^2+mR^2) w

w= m v d/R^2(1/2M+m)

using values given:

w=(0.012*570.1*0.094)/(0.47^2*(16.1/2+0.012)
w=0.361rad/s
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