A uniform thin rod of length 0.45 m and mass 4.3 kg can rotate in a horizontal p

Question

A uniform thin rod of length 0.45 m and mass 4.3 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.3 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle theta = 60 degree with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 12.0 rad/s immediately after the collision, what is the bullet's speed just before impact? the tolerance is +/-2%

Explanation / Answer

initial angular momentum Li = m*v*(l/2)*sin60

= 3.3e-3*v*0.225*sin60v = 6.43e-4 v

final momentum = [ (1/12)*M*l^2 + m*(l/2)^ ] *w

Lf = ( ((1/12)*4.3*0.45*0.45) + (3.3e-3*0.225*0.225) )*12

Lf = 0.87275475 kg m^2/s

Lf = Li

6.43e-4 v = 0.87275475

v = 1357.3 m/s

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