posted by Darlene .
A uniform thin rod of length 0.3m and mass 3.5kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3g bullet traveling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet's velocity makes an angle of 60 degrees with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 9 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?
Use the law of conservation of angular momentum, measured about the vertical axis. Initially, the rod has no angular momentum but the bullet has angular momentum m V (L/2) cos 60 about that axis. Afterwards, the rod with embedded bullet has moment of inertia I = (1/12)ML^2 + m (L/2)^2 and its moment of inertia is I w. This is enough information to solve for m, the initial velocity V of the bullet.
M = rod mass = 3.5 kg
m = bullet mass = .003 kg
L = rod length = 0.3 m
w = final angular velocity of rod = 9 rad/s
I had been using that equation but stupid me was trying to solve for w instead of v! But is it definitely cosine? I had been thinking it would use sine. Thank you sooo much!