Conservation of Angular Momentum?

Question

A uniform thin rod of length 0.50 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 = 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?
m/s

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

j

Answer 2

each spinoff of mass with appreciate to action has an result on the fee and/or direction of the interior of reach gravitational field. Even issues like compression and shear rigidity do this slightly. The mass-capability contribution to the gravitational field is the only contribution that's large adequate for us to observe many of the time (the DC element, you're able to desire to declare). So angular momentum is a minor source of gravity, which motives gravity to be below totally a substantial stress.