A uniform thin rod of length 0.75 m and mass 4.5 kg can rotate in a horizontal p

Question

A uniform thin rod of length 0.75 m and mass 4.5 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is a rest when a 3.0-g bullet traveling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet velocity makes an angle of 60° with the rod (see the figure). If the bullet lodges in the rod and the angular velocity of the rod is 9.0 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

Explanation / Answer

given that

mass of rod M = 4.5 kg

mass of bullet m = 0.003 kg

length of rod = radius of vertical circle = r = 0.75/2 = 0.375 m

w = 9 rad /s

theta = 60 degree

Using co-nservation of angular momentum

Angular momentum of bullet, relative to center of rod, just before impact, is

Li = m*v*r*Sin(60)

Angular momentum of bullet & rod immediately after impact is;

Lf = [m*r^2 + (M/3)*r^2]*w

Li = Lf

m*v*r*sin(60) = (m*r^2 +(M/3)*r^2)*w

put all the values

0.003*v*0.375*0.86 = [ 0.003*(0.375)^2 + 4.5/3*(0.375)^2 ]*9

0.0097*v = [ 0.000421 + 0.210 ]*9

0.0097*v = 1.90

v = 1.90 / 0.0097

v = 199.10 m/s

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