# Conservation of angular momentum exercise [closed]

## Exercise:

A disk of radius $R$ and moment of inertia $I_1$ rotates with angular velocity $\omega_0$. The axis of a second disk, of radius $r$ and moment of inertia $I_2$ is at rest. The axes of the two disks are parallel. The disks are moved together so that they touch. After some initial slipping the two disks rotate together. Find the final rate of rotation of the smaller disk. ## Attempt:

$L_{1_0} = L_1 + L_2 \rightarrow I_1\omega_0 = I_1\omega_1 + I_2\omega_2$

$\omega = \frac{v}{r} \rightarrow v = \omega r$

$\omega_1 R = \omega_2 r \rightarrow \omega_1 = \frac{r}{R}\omega_2$

$I_1\omega_0 = I_1\frac{r}{R}\omega_2 + I_2\omega_2 \rightarrow \omega_2 = \frac{I_1\omega_0}{\frac{r}{R}I_1 + I_2}$

$$\omega_2 = \frac{I_1\omega_0}{\frac{r}{R}I_1 + I_2}$$

## Request:

Is my solution correct? If not, where and why?

## closed as off-topic by AccidentalFourierTransform, John Rennie, garyp, sammy gerbil, Bill NFeb 23 '17 at 2:51

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• Angular momentum cannot be conserved as is assumed in your solution. The system starts with the clockwise angular momentum of disc 1. After contact with disc 1, disc 2 has to have an anti-clockwise angular momentum. To conserve angular momentum disc 1 would need to rotate faster in the clockwise direction which is impossible as then the kinetic energy of the system of two discs would increase with no input of energy. – Farcher Feb 23 '17 at 8:14 Looking at the system above, you can write for the left disk $$I_2 \dot{\omega_2}=Fr$$ and for the second one $$I_1 \dot{\omega_1}=FR$$ where $F$ is the (unknown) friction force. Getting $F$ from the second equation $$F=\frac{I_1}{R} \dot{\omega}_1$$and putting in the first we get $$\frac{I_2}{r} \dot{\omega}_2-\frac{I_1}{R}\dot{\omega}_1=0$$ This means that tha quantity $$\frac{I_2}{r} \omega_2-\frac{I_1}{R}\omega_1$$ is conserved. Setting it equal to the initial value we obtain $$\frac{I_2}{r} \omega_2-\frac{I_1}{R}\omega_1=-\frac{I_1}{R}\omega_0$$

In the final situation there is no slipping, so $\omega_2 r=-\omega_1 R$ and substituting $\omega_1$ in the previous equation we get $$\frac{R I_2}{r} \omega_2+\frac{r I_1}{R}\omega_2=-I_1\omega_0$$ which gives the final solution $$\omega_2=-\frac{I_1 r R }{R^2 I_2+r^2 I_1}\omega_0$$ The angular momentum of the system is not conserved because there are external forces applied on the axes of the disks, and they apply a torque on the system.

• In think this derivation using the friction force between the wheels and assuming that the axes of the disks are fixed and don't start to rotate around each other is correct. When the whole experiment is done in free space with a device for the pivots of the axes of known (or negligible) moment of inertia the conservation of angular momentum can probably still be used leading to an additional rotation of the axes of the disks around each other. – freecharly Feb 23 '17 at 5:00
• How is the angular momentum not conserved .....The disks was rotating and it sets the other in rotation. A angular momentum should be conserved about the center of mass. – Shashaank Feb 23 '17 at 12:08
• This is not an isolated system, there are external torques acting on it. – GCLL Feb 23 '17 at 16:48
• @Shashaank There are forces acting on the axles equal in magnitude to the frictional forces. – Farcher Feb 23 '17 at 22:45
• @GCLL Sorry , I couldn't understand . Could you tell which force in particular is giving the torque – Shashaank Feb 24 '17 at 5:31

The solution is not correct because the angular velocities must have opposite signs after contact. Thus $$\omega_1 = -\frac{r}{R}\omega_2$$ which yields the correct angular velocity of the smaller disk $$\omega_2 = \frac{I_1\omega_0}{I_2-\frac{r}{R}I_1}$$

• Good point. I totally missed that. However, could there be more errors? The answer in the book is $\omega_2 = \frac{rRI_1}{r^2I_1+R^2I_2}$ (the book's answer could be wrong; it's had mistakes before). – Fine Man Feb 22 '17 at 20:30
• @Sir Jony - There seems to be a problem with the use of conservation of angular momentum. When you assume equal disks, there can be no solution because then both disks rotate in opposite sense which always should result in total angular momentum zero. The formula in you comment is dimensionally incorrect, probably a factor $\omega_0$ is missing. – freecharly Feb 22 '17 at 20:46
• Right, I forgot to multiply the RHS by $\omega_0$. Thanks for catching that. – Fine Man Feb 22 '17 at 20:55
• @Sir Jony - In order to get the correct answer you have probably to consider that the contact of the disks in the described manner will also produce a rotation of the disk axes around each other. Otherwise the paradox mentioned above likely cannot be resolved. – freecharly Feb 22 '17 at 20:58
• I assume that the axes simply represent pivot points. Other than that, I think they're meant to be ignored. – Fine Man Feb 22 '17 at 20:59