In the figure below, a 4.0 g bullet is fired into a 0.30 kg block attached to th
ID: 1434777 • Letter: I
Question
In the figure below, a 4.0 g bullet is fired into a 0.30 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·m2. Treat the block as a particle.
(a) What then is the rotational inertia of the block-rod-bullet system about point A?
kg·m2
(b) 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?
m/s
Explanation / Answer
m = 4 g , mb = 0.3 kg ,M =0.5 kg I = 0.06 kg.m^2, r =0.6 m
(a) Moment of inertia of block rod bullet system
If = (M+m)r^2 + I
If = (0.5+0.004)(0.6*0.6) +0.06
If =0.24 kg.m^2
(b) w =4.5 rad/s
From conservation of angular momentum
Li =Lf
mvo*r =If*w
(0.004*vo*0.6) = 0.24*4.5
vo = 450 m/s