A uniform thin rod of length 0.4 m and mass 0.5 kg can rotate in a...

A uniform thin rod of length 0.4 m and mass 0.5 kg can rotate in a horizontal plane about a vertical axis on the left end of the rod. The rod is at rest when a 10.0-g bullet traveling in the horizontal plane of the rod is fired into the right end of the rod at an angle 90o with the rod. The bullet lodges in the rod and the angular velocity of the rod is 10 rad/s immediately after the collision. The rotational inertia of a rod about axis through center perpendicular to length L is (1/12)*ML2.

Find the rotational inertia of the rod-bullet system when the bullet lodges in the rod. Keep 5 decimal places.

What is the angular momentum after the collision? Keep 4 decimal places.

Find the bullet’s initial velocity.

Solutions

Expert Solution

The rotational inertia of the rod is . => =>

The rotational inertia of the bullet is => =>

The rotational inertia of the rod-bullet system is =>

The angular momentum of the system after collision is =>

There is no net torque about the axis of rotation. The final angular momentum of the rod-bullet system after collision is equal to initial angular momentum of the bullet with respect to the axis of rotation.

if v is the speed of the bullet then initial angular momentum is L = 0.01 x v x 0.4 = 0.004 v

By conservation of angular momentum, 0.004 v = 0.2827 => v = 70.675 m/s

11 ML? 12

We were unable to transcribe this image

11 = 0.00667kg - m?

We were unable to transcribe this image

genius_generous answered 1 year ago

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