In the figure, a 6.37 g bullet is fired into a 0.412 kg block attached to the en

Question

In the figure, a 6.37 g bullet is fired into a 0.412 kg block attached to the end of a 0.393 m nonuniform rod of mass 0.395 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 0.0601 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 9.80 rad/s, what is the bullet's speed just before impact?

Explanation / Answer

I=Icm+mh^2

=0.0601kgm^2+(0.412kg*0.393^2m)

=0.123kgm^2

=123kgm^2.

(B) p=Iw/r

=(123kgm*9.8)/0.393

=3.06kgm/s

due to conversation of momentum this was also the momentum of the bullet just before impact.

v=p/m=3.06/0.00637

=4.8km/s.

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