Question

A 8 g bullet is fired at 450 m/s into a solid cylinder of mass 25.4 kg and a radius 0.2 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 9 cm from the center.

What is the initial angular momentum of the system?

Find the angular velocity of the system after the bullet strikes and adheres to the surface of the cylinder.

Answer 1

m = mass of bullet = 0.008 kg

v = speed of bullet before collision= 450 m/s

r = distance of bullet line of motion from the axis = 0.09 m

R = radius of cylinder = 0.2 m

M = mass of cylinder = 25.4 kg

I = moment of inertia of cylinder = (0.5) MR² = (0.5) (25.4) (0.2)² = 0.508 kgm²

w = angular velocity

I_b = moment of inertia of bullet after adhering to surface = mR²

L_i = initial angular momentum of bullet = mvr = 0.008 x 450 x 0.09 = 0.324 kgm/s

using conservation of angular momentum

L_i = L_f

0.324 = (I + I_b) w

0.324 = (0.508 + mR²) w

0.324 = (0.508 + (0.008) (0.2)²) w

w = 0.64 rad/s