A uniform thin rod of length 042 m and mass 46 kg can rotate

A uniform thin rod of length 0.42 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 3.1 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 = 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? Axis Number Units

Solution

this is based on the principle of law of conservation of angular momentum

i.e total intial angualr momentum = total final angular momentum

initial angular momentum I1 = mvr

here r = L/2 sin theta

so

L1 =   l/2 * mv sin theta ------------------1

for final angular momentum L2 = I2 Wf

I2 =   1/12* ML^2

so

L2i = 1/12 ML^2 Wf

and also

after Bullet lodges, L2 = 1/4 *mL^2

total Lf =   (1/12 ML^2 Wf*) + (1/4 *mL^2* wf)-------------2

equating 1 and 2

(0.42/2 * 3.1*10^-3 * v * sin 60) = (1/12 * 4.6 * 0.42^2*12) + (1/4 * 3.1*10^-3 * 0.42^2*12)

solving for V

v = 1.4*10^3 m/s   or 1442 m/s