A bullet with a mass of 13.0 g is fired at a wooden rod that hangs vertically do
ID: 2262511 • Letter: A
Question
A bullet with a mass of 13.0 g is fired at a wooden rod that hangs vertically down from a pivot point that passes through the upper end of the rod. The bullet embeds itself in the exact center of the rod and the rod/bullet system swings up, reaching a maximum angular displacement of 42.0 degrees from the vertical. The rod has a mass of 375 g, a length of 1.10 m, and we can assume the rod rotates without friction about the pivot point. Assume the bullet is traveling horizontally when it hits the rod, and use g = 10.0 m/s2.
(a) What is the rotational inertia of the rod/bullet system after the collision, rotating about the pivot point at the upper end of the rod?
(b) What is the angular speed of the rod/bullet system immediately after the collision?
(c) What is the bullet's speed just before it hits the rod?
Explanation / Answer
(a) rot inertia = rod + bullet = (1/3) M L^2 + m r^2 =
= (1/3) * 0.375 * 1.10^2 + 0.013 * 0.55^2 = 0.1551825 kg-m^2
(b) use cons of energy
(1/2) I w^2 = m g h
(1/2) * 0.1551825 * w^2 = (0.375+0.013) * 10 * 0.55*(1-cos42)
w^2 = 7.0643
w = angular speed =2.6578776976 rad/s
(c) ang momentum of bullet = ang mom of system
mvr = I w
0.013 * v * 0.55 = 0.1551825 * 2.6578776976
v = 57.686m/s
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