A uniform thin rod of length 0.400 m and mass 4.40 kg can rotate in a horizontal plane about a vertical axis through its center. the rod is at rest when a 3.0 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 17 rad/s immediately after the collision, what is the bullet's speed just before impact?

The rod has a mass of m = 4.4 kg and a length of L = 0.4 m.

Its polar moment of inertia is

J = (mL²) / 12

= (1/12) * [ (4.4 kg) * (0.4 m) ²]

= 0.05867 kg-m²

The mass of the bullet is 0.3 g.

If its velocity is v m/s, then its linear momentum is

P = (0.3 x 10⁻³ kg) * (v m/s)

Its linear momentum perpendicular to the rod is

P*sin (60°) = 2.5981 x 10⁻⁴ v (kg-m) / s

The angular momentum about the center of the rod when the bullet strikes is

T = (2.5981 x 10⁻⁴ v (kg-m) / s) * (0.2 m) = 5.1962 x 10⁻⁵ v (kg-m²) / s

Because the bullet lodges into the end of the rod, the combined polar moment of inertia is

J + (0.3 x 10⁻³ kg) * (0.2 m) ² = 0.05867 + 1.2 x 10⁻⁵ = 0.0587 kg-m²

The initial angular velocity is ω = 17 rad/s.

Because angular momentum is conserved, therefore

5.1962 x 10⁻⁵ v (kg-m²) / s = (0.0587 kg-m²) * (17 rad/s)

v = 19204 m/s

Answer: 19204 m/s