# Conservation of angular momentum

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1. Oct 4, 2015

### Titan97

1. The problem statement, all variables and given/known data
Two discs of different radii and masses are kept on a smooth horizontal table and both are free to rotate about thier fixed central vertical axis. One of them is given some angular velocity while other is stationary. The rims are brought in contact. There is friction between the rims. Can you conserve angular momentum of system?
If the stationary disc is kept above the other disc such that thier centres coincide, will the angular momentum be conserved of the system be conserved? There is friction on the surface of discs.

2. Relevant equations
None

3. The attempt at a solution
The angular velocity of one disc decreases and the other increases in both cases.
Also,
Case 1: Friction stops when the rims have have equal velocities and till that moment, friction has equal magnitude on both discs.
Case 2: Friction stops when both discs spin with same angular velocity

For case 1, torque acting on both discs are different. Net torque is not zero. So angular momentum of system can't be conserved.
For case 2, angular momentum is conserved since torque due to friction is same on both discs.

Is this correct?

2. Oct 4, 2015

### haruspex

Your answers are correct, but arguably do not get to the heart of the matter.
What is it about the systems that means angular momentum is conserved in the one case but not in the other? Bear in mind that the key is external forces.

3. Oct 4, 2015

### Titan97

In both cases friction is an internal force right?
Angular momentum is conserved when net torque is zero.

In second case, for any disc, I will take a small element of area $rd\theta dr$ at an angle $\theta$ from horizontal.

The mass of this element is $dm=Ard\theta dr$
Here A is Mass per unit area
Torque on this section is $d\tau=r.\mu g.Ard\theta dr$
$$\tau=A\mu g\int_0^{2\pi}\int_0^Rr^2drd\theta$$
Let the upper disc be smaller than lower disc.

Total torque on upper disc is $\tau=\frac{2\pi\mu gA_1R_1^3}{3}$
For the lower disc, its $\tau=\frac{2\pi\mu gA_2R_1^3}{3}$
(Since friction acts only till $R_1$ for the bigger disc)

For torques to be equal, the mass per unit area has to be same.

4. Oct 4, 2015

### haruspex

No, that's not what I meant.
In each case, what external forces and torques might act on the system consisting of the pair of discs?

5. Oct 4, 2015

### Titan97

In the first case, there are hinge forces on the disc. In second case, friction does not act in a single direction. The hinge force is zero.
In both cases, only friction causes torque.

6. Oct 4, 2015

### haruspex

Forces from the axles, yes.
True, but that's internal forces.
Yes.
No. If that were the case a.m. would be conserved in both.
The point is that in the first case there are two axles, with a nonzero linear force exerted by each on the system. The two forces are equal and opposite, but exert a net torque on the system regardless of what axis you choose for taking moments.

7. Oct 5, 2015

### Titan97

That non-zero linear force is the hinge reaction right?

8. Oct 5, 2015

Yes.