A uniform thin rod of length 0.11 m and mass 4.6 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 5.0 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 12.0 rad/s immediately after the collision, what is the bullet's speed just before impact?

Explanation:

This problem is based on conservation of rotational momentum.

Moment of inertia of rod about its center

= 1/12 m l² , m is mass of the rod and l is its length .

= 1 / 12 x 4.6 x .11²

I = .004638 kg m²

The angular momentum of the bullet about the center of rod = mvr

where m is mass , v is perpendicular component of velocity of bullet and r is distance of point of impact of bullet fro center .

5 x 10⁻³ x v sin60 x .11 x .5  where v is velocity of bullet

According to law of conservation of angular momentum

5 x 10⁻³ x v sin60 x .11 x .5 = ( I + mr²)ω , where ω is angular velocity of bullet rod system and  ( I + mr²) is moment of inertia of bullet rod system .

.238 x 10⁻³ v = ( .004638 + 5 x 10⁻³ x .11² x .5² ) x 12

.238 x 10⁻³ v = ( .004638 + .000015125 ) x 12

.238 x 10⁻³ v = 55.8375 x 10⁻³

.238 v = 55.8375

v = 234.6 m /s

i think it is non rem sleep