A uniform thin rod of length 0.75 m and mass 5.0 kg can rotate in a horizontal p

Question

A uniform thin rod of length 0.75 m and mass 5.0 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 11.0 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

Explanation / Answer

Using the Mass moment of Inertia:

I rod = mL^2/12 = 5.0*(0.75^2)/12 = 0.234375
I bullet = md^2 = (0.003)(0.75/2)^2 = .000421
I rod + bullet = I rod + I bullet = 0.262167+ 0.000242
I rod + bullet = 0.234796

Angular momentum: L = Iw
w is the angular velocity

L of bullet the instant before being lodged in the rod:
Lb = .000421*v
v is the velocity of the bullet in the radial direction.

L after the bullet is lodged
Lrb = 0.234796*11= 2.582

Conservation of evergy: Lb = Lrb

.000421*v = 2.582
v = 6134.83m/s

Now we need to find the velocity in the direction the bullet was travelling

Vb = v/sin(45)= 8675.96 m/s

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