In the figure, a 8.71 g bullet is fired into a 0.256 kg block attached to the en

Question

In the figure, a 8.71 g bullet is fired into a 0.256 kg block attached to the end of a 0.125 m nonuniform rod of mass 0.911 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 0.0465 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 9.91 rad/s, what is the bullet's speed just before impact?

Explanation / Answer

a) the rotational inertia of the block-rod-bullet system about point A = I_rod + I_block + I_bullet

= 0.0465 + 0.256*0.125^2 + 0.00871*0.125^2

= 0.0506 kg.m^2

b) let v is the speed of bullet.

Apply conservation of angular momentum

angular momentum before the collision = angular momentum after the collision

m*v*L = I*w

v = I*w/(m*L)

= 0.0506*9.91/(0.00871*0.125)

= 461 m/s

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