In the figure, a 5.12 g bullet is fired into a 0.177 kg block attached to the en

Question

In the figure, a 5.12 g bullet is fired into a 0.177 kg block attached to the end of a 0.203 m nonuniform rod of mass 0.585 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 0.0558 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 3.43 rad/s, what is the bullet's speed just before impact?

Explanation / Answer

a) 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 = 0.203 m from "A" then;

I = (0.0558) + Mr^2 + mr^2

= 0.0558 + 0.177*.203^2 + 0.00512*0.203^2

= 0.0633 kg m^2

b) 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

= 0.0633*3. 43/(0.00512*0.203)

= (.205)(4.5)/(.003)(.6)   

= 208.897 m/s

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