In the figure, a 7.11 g bullet is fired into a 0.306 kg block attached to the en

Question

In the figure, a 7.11 g bullet is fired into a 0.306 kg block attached to the end of a 0.664 m nonuniform rod of mass 0.210 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 0.0670 kg-m2. Treat the block as a particle. What then is the rotational inertia of the block-rod-bullet system about point A? If the angular speed of the system about A just after impact is 2.40 rad/s, what is the bullet's speed just before impact? Number Units Number Units .

Explanation / Answer

A) rotational inertia of the system is I = I1+I2

I1 = 0.0670 kg-m^2

I2 is the moment of inertia of the block and bullet = (m+M)*r^2 = (0.00711+0.306)*0.664*0.664 = 0.138 kg-m^2

then I = 0.067+0.138 =0.205 kg-m^2

B) w = 2.4 rad/s

using law of copnservation of energy

energy of the system before impact = energy of the system after impact

0.5*m*v^2 = 0.5*I*w^2

0.5*0.00711*v^2 = 0.5*0.205*2.4*2.4 = 0.5904

v = 12.88 m/s

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