Question

In the figure, a 8.28 g bullet is fired into a 0.996 kg block attached to the end of a 0.611 m nonuniform rod of mass 0.490 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 A is 0.0857 kg-m^2. Treat the block as a particle, (a) What then is the rotational inertia of the block-rod-bullet system about point A? (b) If the angular speed of the system about A just after impact is -4.66 rad/s, what is the bullet's speed just before impact? (a) Number Units (b) Number Units

Answer 1

The total moment of inertia is
I = I(rod) + I(block) + I(bullet)

Since the block and bullet are treated as "point" masses a distance r=.611 m from "A" then;
I = (.0857) + Mr^2 + mr^2
= (.0857) + (.49)(.611)^2 + (.00828)(.611)^2
= .272 Kg-m^2

Conservation of angular momentum.
The bullets angular momentum = py = mvy = (just before impact)= mvr
The system angular momentum after impact is Iw
mvr = Iw
v = Iw/mr
= (.272)(4.66)/(.00828)(.611)
= 250.54m/s