# Conservation of angular momentum

1. Jan 4, 2014

### Karol

1. The problem statement, all variables and given/known data
4 small balls of mass m each are connected with light rods like in the picture. the system is rotating with angular velocity ω0.
When the system is in the rightmost position the upper ball disconnects from the system.
What is the angular velocity of the 3 remaining balls

2. Relevant equations
Initial angular velocity round point O:
$$L_0=4m\omega_0 l^2$$
Because of preservation of linear momentum, the velocity of the C.O.M of the 3 balls is
$$V=\frac{\omega_0 l}{3}$$

3. The attempt at a solution
I am told the angular momentum is preserved, and i know the upper ball moves at
$$V=\omega_0 l$$
And the 3 remaining balls must have angular momentum of
$$L=3m\omega_0 l^2$$
And this implies they rotate at angular velocity ω0, but they rotate around their center of mass, so the directions of velocities and distances from point O change.
I know the answer is they rotate at ω0, but why?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Jan 4, 2014

### ehild

The initial angular momentum around point O is
$$L_0=4m\omega_0 l^2$$

Without external torque, the angular momentum is conserved, it is the same with respect to the same axis after the ball is released. The released ball still has it angular momentum ml2ω0, and the other three has 3ml2ω with respect to the initial axis of rotation at O.

At the same time, the released ball moves with v=ω0l velocity to the left and the other three connected ball shifts to the right while rotating about the centre of mass. But the CM also has angular momentum with respect to O, and its angular momentum and the angular momentum of the rotation of the three connected balls around the CM adds up to 3ml2ω .

ehild