Understanding Net Internal Torque and Kinetic Friction in a System

[annamal]

In a system, the net internal torque should be 0.
If we have two fly wheels, one spinning with angular velocity w, and the other at rest and the flywheel at rest is dropped onto the other flywheel, the two fly wheels reach the same angular velocity due to friction between the two wheels. I am confused how the net internal force would be 0. What is the force equal and opposite to the kinetic friction?

My follow up question is how is this kinetic friction considered an internal force if it is dwindling the speed of the two flywheels?

 

[nasu]

It has to be another kinetic friction, acting on the second object in interaction.

 

[annamal]

It has to be another kinetic friction, acting on the second object in interaction.

My follow up question is how is this kinetic friction considered an internal force if it is dwindling the speed of the two flywheels? There seems to be a net external force.

 

[Baluncore]

My follow up question is how is this kinetic friction considered an internal force if it is dwindling the speed of the two flywheels?

Friction between the flywheels slows one flywheel while it accelerates the other.
Once they have the same angular velocity, it is effectively one flywheel again.

 

[jbriggs444]

My follow up question is how is this kinetic friction considered an internal force if it is dwindling the speed of the two flywheels? There seems to be a net external force.

It is an internal force since it applies between two bodies that are both considered part of the "system" under consideration.

The one flywheel speeds up. The other flywheel slows down. The torque of the one on the other is in one direction. The torque of the other on the one is in the opposite direction. The two torques add together like vectors and cancel each other out.

 

[nasu]

My follow up question is how is this kinetic friction considered an internal force if it is dwindling the speed of the two flywheels? There seems to be a net external force.

You seem to believe that internal forces cannot change the kinetic energy of the system.

This is wrong. The change in KE of a system is equal to the net work of both internal and external forces.
The internal forces can convert the KE in other forms of energy reducing the relative motion of the parts to a stop. Internal forces can also increase the internal energy of the system, as in the explozion of a completely isolated object. So, you don't need an external force to account for the wheel slowing down.

Also, the forces in a Newton's third law pair are always describing the same interaction, between the same two objects. The pair of a friction force is another friction force. And the pair of an internal force is another internal force. If the two interacting objects are part of the system, both forces are internal. If one of the two object is external, then both forces are external.

 

[annamal]

It is an internal force since it applies between two bodies that are both considered part of the "system" under consideration.

The one flywheel speeds up. The other flywheel slows down. The torque of the one on the other is in one direction. The torque of the other on the one is in the opposite direction. The two torques add together like vectors and cancel each other out.

Since the net internal force in a system should be 0, it confuses me that the flywheels can slow down due to friction which results in a net force.

 

[jbriggs444]

Since the net internal force in a system should be 0, it confuses me that the flywheels can slow down due to friction which results in a net force.

One of the flywheels sped up. How is that "slowing down"?

The total angular momentum of the system was preserved. You had one flywheel spinning with total angular momentum $L = I\omega$. You ended with two flywheels spinning at half the rate with total angular momentum $L=(2I)\frac{\omega}{2}$

As @nasu notes, the friction did drain kinetic energy. Which is fine -- the disks heated up where friction caused their relative velocities to equalize. The initial kinetic energy would have been $\text{KE}=\frac{1}{2}I\omega^2$ and the final kinetic energy would have been half that, $\text{KE}=\frac{1}{2}(2I){(\frac{\omega}{2})}^2$

The analogous thing for linear momentum is a moving blob of putty on an icy tabletop impacting a stationary blob of putty on the same tabletop. The two move away at half speed. Total linear momentum is conserved while kinetic energy is reduced by half. The only force involved is an internal force between the two blobs of putty.

 

[Lnewqban]

In a system, the net internal torque should be 0.
If we have two fly wheels, one spinning with angular velocity w, and the other at rest and the flywheel at rest is dropped onto the other flywheel, the two fly wheels reach the same angular velocity due to friction between the two wheels. I am confused how the net internal force would be 0. What is the force equal and opposite to the kinetic friction?

My follow up question is how is this kinetic friction considered an internal force if it is dwindling the speed of the two flywheels?

A system is in equilibrium when the summations of both, forces and torques acting on it (from within or from outside its boundaries), equal zero.

In practical terms, friction is used to connect two rotating heavy parts mainly to reduce the internal forces in the mechanism.
That is achieved by reducing the acceleration rate of the driver-driven parts by increasing the time in which the full coupling happens.

If parts were robust enough, you could replace friction with a dog clutch, obtaining the same result of slowing the driver part and speeding the driven part.

Please, see:
https://en.wikipedia.org/wiki/Dog_clutch#:~:text=A dog clutch (also known,speed and will never slip.

The angular momentum accumulated in the driver flywheel (work done by a previous external torque, now removed, on our driving flywheel) can only do so much work, which is the product of rotational angle and torque.
When, suddenly or slowly, that limited accumulated kinetic energy, is required to rotate additional mass or moment of inertia, it can only do it by reducing the angular velocity.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html#alm

https://courses.lumenlearning.com/boundless-physics/chapter/conservation-of-angular-momentum/

In our idealized problem, the original angular momentum (disregarding the waste of energy of excessive friction, leaving the system as heat) is conserved.
Its amount should be the same prior to and after the coupling of both flywheels is completed.

 

[nasu]

Since the net internal force in a system should be 0, it confuses me that the flywheels can slow down due to friction which results in a net force.

The net internal force is zero but the net work of the internal forces does not have to be zero. The change in KE is given by the work and not just by force alone.
The friction does not result in a net force. There are two friction forces, each acting on one of the wheels. Their resultant is zero. If you have static friction, then there is no work done and the kinetic energy does not change. One object slows down and the other accelerates. If you have kinetic friction, some KE is dissipated and the total KE decereases.

If you just drop an object on top of a moving one you start with some kinetic friction until the two objects reach the same velocity and move together. It is like an innelastic collision. Some KE is dissipated. In innelastic collisions you have the same thing: the net force is zero but the net work is not. Internal forces decerase the total KE of the system.