In the figure, a 7.26 g bullet is fired into a 0.948 kg block attached to the en

Question

In the figure, a 7.26 g bullet is fired into a 0.948 kg block attached to the end of a 0.186 m nonuniform rod of mass 0.533 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 0.0676 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 7.96 rad/s, what is the bullet's speed just before impact?

Explanation / Answer

(a) Rotational inertia of the block-rod-bullet system
I = Irod + mL2 + ML2
Where Irod is the rotational inertia of the rod = 0.0676 kg-m2
m is mass of bullet = 7.26 g = 7.26*10-3
M is mass of block = 0.948 kg
and L is the length of the rod = 0.186 m
I = 0.0676 +[ (7.26*10-3)*(0.186)2 ]+ [(0.948)*0.1862]
I = 0.1006 kg-m2
(b)
Initial angular momentum of the system = angular momentum of bullet + angular momentum of rod
(Since the rod is at rest therefore the angular momentum of the rod will be ZERO)
= mVL + 0 ----------------(1)
Where V is the velocity of the bullet
Final angular momentum of the system = Iw
where w is the angular velocity = 7.96 rad/s
Now from conservation of momentum
Initial angular momentum = Final angular momentum
mVL =  Iw
(7.26*10-3)*V*(0.186) =  0.1006*7.96
V = 593 m/s

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