In the figure, a 6.35 g bullet is fired into a 0.979 kg block attached to the en

Question

In the figure, a 6.35 g bullet is fired into a 0.979 kg block attached to the end of a 0.616 m nonuniform rod of mass 0.864 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 0.0345 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 6.07 rad/s, what is the bullet's speed just before impact?

Explanation / Answer

the rotational inertia of the block-rod-bullet system about point A is

I = I rod + ( m+ M) L^2 = 0.0345 kg·m2.+ ( 6.35 * 10^-3 kg+ 0.979)( 0.616)^2 = 0.408 kg m^2

apply conservation of angular momentum

mL = I w

v = I w/ mL

= 0.408 kg m^2(6.07)/6.35 * 10^-3 kg ( 0.616)

=6.33 * 10^2 m/s

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