A bullet of mass m = 0.024 kg is moving with a velocity 2000 m/s toward a statio

Question

A bullet of mass m = 0.024 kg is moving with a velocity 2000 m/s toward a stationary cylinder of mass M = 50 kg and radius R = 3 m. The bullet passes through the cylinder a distance 2.4 m from the center of the cylinder and emerges on the other side of the cylinder with a speed of 200 m/s. This sets the cylinder rotating about an axis through its center. Assuming no friction, what is the angular velocity of the cylinder What type of collision is this? Work out the details of how you arrived at your answer

Explanation / Answer

let vi = 2000 m/s and vf = 200 m/s

m = 0.024 kg

r = 2.4 m

a) Apply conservation of angular momentum

initial angular momentum = final angular momentum

m*vi*r = m*vf*r + I*w

m*vi*r = m*vf*r + 0.5*M*R^2*w

w = (m*vi*r - m*vf*r)/(0.5*M*R^2)

= (0.024*2000*2.4 - 0.024*200*2.4)/(0.5*50*3^2)

= 0.4608 rad/s

b) Ki = 0.5*m*vi^2

= 0.5*0.024*2000^2

= 48000 J

Kf = 0.5*m*vf^2 + 0.5*I*w^2

= 0.5*m*vf^2 + 0.5*0.5*M*R^2*w^2

= 0.5*0.024*200^2 + 0.25*50*3^2*0.4608^2

= 504 J

Kf < Ki

so, it is an inelastic collision.

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