In the figure, a 6.09 g bullet is fired into a 0.405 kg block attached to the en

Question

In the figure, a 6.09 g bullet is fired into a 0.405 kg block attached to the end of a 0.442 m nonuniform rod of mass 0.987 kg. The block-ród-bullet system then rotates in the plane of the figure, about a fixed axis at A. The rotational inertia of the rocd alone about A is 0.0727 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 2.18 rad/s, what is the bullet's speed just before impact? Rod Block Bullet (a) Number Units (b) Number Units

Explanation / Answer

total rotational inertia = Irod + I bullet + I block

Itotal = Irod + mr^2 + Mr^2

Itotal = 0.0727 + (0.00609 * 0.442^2) + (0.405 * 0.442^2)

Itotal = 0.153 kgm^2

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part B

use Intial ANg Momentum = final Ang momentum

MVR = I W

v = IW/MR

v = (0.153 * 2.18)/(0.00609 * 0.442)

v = 123.91 m/s

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