A uniform thin rod of length 0.60 m and mass 3.5 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0 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 θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 14 rad/s immediately after the collision, what is the bullet's speed just before impact? (Answer is in m/s)

Question

Physics

Jaden Best

6 August, 01:04

A uniform thin rod of length 0.60 m and mass 3.5 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0 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 θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 14 rad/s immediately after the collision, what is the bullet's speed just before impact? (Answer is in m/s)

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Answers (1)

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Jack Wilkerson · Answer 1

The moment of inertia for the rod is

mr*L^2/12

After collision

I=mr*l^2/12+mb*r^2

I=0.073

The momentum of the rod after collision is

I*w

=1.022

The momentum of the bullet that is radial

to the rod is

vr=v*sin (60)

vr/r=w

so the radial momentum of the bullet is

3*vr / (0.25*1000)

vr*0.012

applying conservation of momentum

vr*0.012=1.022

vr=85 m/s

and

v=85/sin (60)

v=98 m,/s

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