Question

A 10.0 g bullet is fired horizontally into the rim of a stationary hardwood

disk that has a mass of 3.00 kg and a radius of 50.0 cm as shown to the right.
The disk is free to rotate about its central axis without friction. Estimate the
impact speed of the bullet if the disk rotates with an angular velocity of
3.00 rads/s after the bullet becomes imbedded in the rim. Assume that the
penetration distance of the bullet is negligible compared with the radius of
the disk.

Answer 1

To estimate the impact speed of the bullet, we can use the principle of conservation of angular momentum. The angular momentum of the system before the bullet becomes embedded in the rim is equal to the angular momentum of the disk after the collision.

Step-by-step explanation:

To estimate the impact speed of the bullet, we can use the principle of conservation of angular momentum. The angular momentum of the system before the bullet becomes embedded in the rim is equal to the angular momentum of the disk after the collision. The angular momentum of a rotating object is given by the product of its moment of inertia and angular velocity.

Since the bullet is fired horizontally, its initial angular momentum is zero. After becoming embedded in the rim, the bullet adds its mass to the moment of inertia of the disk. We can use the equation for the moment of inertia of a disk, I = (1/2)MR^2, where M is the mass of the disk and R is its radius. Solving for the initial angular velocity, ω_i :

I*ω_i + m*0 = (I + mR^2)*ω_f

Substituting the known values:

(1/2)(3.00 kg)(0.50 m)^2 * 0 + (0.010 kg + 3.00 kg)(0.50 m)^2 * 3.00 rad/s = (1/2)(3.00 kg)(0.50 m)^2 * ω_f

Simplifying:

0.00 kg·m^2/s + 0.250 kg·m^2/s = (0.75 kg·m^2) * ω_f

ω_f ≈ 0.33 rad/s

Therefore, the impact speed of the bullet is approximately 0.33 rad/s.