The figure below shows the overhead view of a uniform rod of length 0.450 m and

Question

The figure below shows the overhead view of a uniform rod of length 0.450 m and mass 4.53 kg that can rotate in a horizontal plane about the vertical axis p, through its center. The rod is initially at rest when a bullet of mass 2.80 g traveling in the rotation plane is fired into one end of the rod. The angle between the rod and the path of the bullet is Theta = 55.0 degree. The speed of the bullet before the impact is v = 2.1 km/s. What is the angular momentum of the bullet-rod system after the impact? Assume that the bullet becomes lodged in the rod.

Explanation / Answer

Here ,

as there is no external torque acting on the rod-bullet system ,

inital angular momentum = final angular momentum

NOw,

final angular momentum = m*v*l*sin(theta)/2

final angular momentum = 0.00280 * 0.450 * 2100 * sin(55)/2

final angular momentum = 1.084 Kg.m^2/s

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