In the figure, a 7.71 g bullet is fired into a 0.999 kg block attached to the en

Question

In the figure, a 7.71 g bullet is fired into a 0.999 kg block attached to the end of a 0.683 m nonuniform rod of mass 0.667 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 1.41 rad/s, what is the bullet's speed just before impact?

Explanation / Answer

I = Irod + mr^2

r = 0.683 m

I = 0.0512 + (1.00671)*0.683^2 = 0.5208 kg-m^2

part b )

mvr = Iw

v = 139.45 m/s

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