Question

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?

Answer 1

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