In the figure, a 7.43 g bullet is fired into a 0.435 kg block attached to the en

Question

In the figure, a 7.43 g bullet is fired into a 0.435 kg block attached to the end of a 0.556 m nonuniform rod of mass 0.494 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 0.0512 kg·m2. 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 2.94 rad/s, what is the bullet's speed just before impact?

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=.556 m from "A" then;
I = (.0512) + Mr^2 + mr^2
= (.0512) + (.435)(.30) + (.00743)(.30)
= .183 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
= (.183)(2.49)/(.00743)(.556)
= 110.30 m/s

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