In the figure below, a 4.0 g bullet is fired into a 0.80 kg block attached to th
ID: 1505614 • Letter: I
Question
In the figure below, a 4.0 g bullet is fired into a 0.80 kg block attached to the end of a 0.60 m nonuniform rod of mass 0.50 kg. The block-rod-bullet system then rotates in the plane of the figure about a fixed axis at A. The rotational inertia of the rod alone about that axis at A is 0.060 kg middot m^2. Treat the block as a particle. What then is the rotational inertia of the block-rod-bullet system about points? kg middot m^2 If the angular speed of the system about A just after impact is 4.5 rad/s, what is the bullet's speed just before impact?Explanation / Answer
a)
L = length of rod = 0.6 m
Irod = 0.06 kgm2
Ibullet = mb L2 = (0.004) (0.6)2 = 0.00144 kgm2
Iblock = mblock L2 = 0.80 (0.6)2 = 0.288
Total moment of inertia of system = Isys = Irod + Ibullet + Iblock = 0.06 + 0.00144 + 0.288 = 0.35 kgm2
b)
using conservation of angular momentum
mb v L = Isys W
(0.004) v (0.60) = 0.35 (4.5)
v = 656.25 m/s