Question

A ring with (mass 4 kg, and a radius 1 m) is horizontally mounted with a pivot at the center, and rotates counterclockwise with an angular speed 1 rad/s. a bullet (mass 0.2 kg, speed 40 m/s) is shot and collides with the ring at its radius 1 m, and then remains lodged. what is the initial moment of inertia of the bullet?

_____ kgm²

Answer 1

The initial moment of inertia of the bullet is 8 kgm².

Step-by-step explanation:

To find the initial moment of inertia of the bullet, we can use the principle of conservation of angular momentum.

The angular momentum before the collision is the same as after the collision, since there is no external torque acting on the system.

The angular momentum of the bullet before the collision is given by L = mvr, where m is the mass of the bullet, v is its speed, and r is the radius at which it collides with the ring.

The ring and bullet together can be considered as a system with a new moment of inertia I.

The angular momentum of the system after the collision is given by L = Iω, where ω is the angular speed of the system after the collision.

Since the angular momentum before and after the collision is the same, we have mvr = Iω.

Solving for I, we get I = (mvr) / ω. Plugging in the given values, we have:

I = (0.2 kg)(40 m/s)(1 m) / (1 rad/s)

= 8 kgm².