#060 In the figure, a 5.94 g bullet is fired into a 0.632 kg block attached to t

Question

#060

In the figure, a 5.94 g bullet is fired into a 0.632 kg block attached to the end of a 0.440 m nonuniform rod of mass 0.384 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 0.0434 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 3.38 rad/s, what is the bullet's speed just before impact?

Please take your time to ensure accuracy.

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 = (0.0434) + Mr^2 + mr^2
= (.06) + (.632)(.1939) + (.00594)(.1936)
= 0.1835 Kg-m^2

Conservation of angular momentum

mvr = Iw
v = Iw/mr
= 0.1835 )(3.38)/(.00594)(.44)
= 237.31 m/s

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