# Problem regarding concept of conservation of angular momentum [duplicate]

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Two cylinders of radii r1, and r2 having moments of inertia I1, and I2, about their respective axes. Initially, the cylinders rotate about their axes with angular speeds w1, and w2 as shown in the figure. The cylinders are moved closer to touch each other keeping the axes parallel. The cylinders first slip over each other at the contact but the slipping finally ceases due to the friction between them. Find the angular speeds of the cylinders after the slipping ceases.

I applied conservation of momentum here but I'm unable to obtain the right answer. Taking both cylinders as system, since only friction acts and these forces contribute to internal torques so with absence of external torques I conserved angular momentum of the system but the answer is incorrect.

My question is why cant we conserve angular momentum in such a scenario? How is there external torque and what forces are providing the external torque?

## marked as duplicate by John Rennie, Emilio Pisanty, Kyle Kanos, Jon Custer, sammy gerbilFeb 20 '18 at 19:42

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• Hi Ola. Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. – John Rennie Feb 20 '18 at 8:29
• Are the axles (axes of rotation) constrained to move only along a left/right line? If the answer is yes then there are are external torques acting on the system. – Farcher Feb 20 '18 at 9:23
• @Farcher Can you please explain how there are external torques acting on the system – Hola Feb 20 '18 at 9:42
• I just want to know the CONCEPTUAL ERROR of using conservation of angular momentum in this problem, nothing else. Please can you help me with why I can't conserve angular momentum / why is external torque acting on the system and which are these forces contributing to external torque, thats all. My question is only regarding THE CONCEPT. – Hola Feb 20 '18 at 9:48

## 1 Answer

If there are no forces acting on the axles on the disc then the force diagram is as shown in the left hand diagram with internal frictional force $F_{12}$ and $F_{21}$ acting on the discs.
If the system was like this then as each of the discs has one fore on it the centre of mass of each of the discs will undergo a translational acceleration.

To stop the centres of mass moving two external force $F_{1E}$ and $F_{2E}$ must act on the discs as shown in the right hand diagram.
Together these two external forces form a external couple and hence an external torque acting on the system of two discs.

• Ok but acc to diagram, where is any horizontal force for translation here? Wont the normal from the ground = weight of cylinder + friction for first one and for second, normal + friction = weight? – Hola Feb 20 '18 at 11:12
• @Ola I have neglected gravitational forces as they would just add an extra unnecessary complication to the problem or you could think of the discs as being horizontal and you are looking at a plan of the system. The translational motion in the left hand diagram would be up and down. – Farcher Feb 20 '18 at 11:37
• 'or you could think of the discs as being horizontal and you are looking at a plan of the system. The translational motion in the left hand diagram would be up and down' I didnt understand can you please explain again? – Hola Feb 20 '18 at 14:00
• @Ola The up and down refers to the directions as seen on my diagram. The reference to horizontal discs was there to remove the necessity to consider gravitational attraction. – Farcher Feb 20 '18 at 14:19
• The discs are already horizontal right. How does that remove the necessity to consider gravity, I'm unable to follow. Horizontal wrt what? – Hola Feb 20 '18 at 14:47