20. In the figure below, a 3.0 g bullet is fired into a 0.30 kg block attached t

Question

20. In the figure below, a 3.0 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

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Explanation / Answer

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=.6 m from "A" then;
I = (.06) + Mr^2 + mr^2
= (.06) + (.3)(.36) + (.003)(.36)
= .169 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
= (.169)(4.5)/(.003)(.6)
= 422.7 m/s

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