Say you had two fixed, non-identical objects rotating around one axis at different speeds relative to one another, and they can be coupled together by friction (like how a clutch in a car works). How would I calculate the frictional force required for them to stick together instantaneously?
For example:
Object A and B are two solid cylinders rotating about a single axis (imagine how they would rotate if they were rolling) Object A is rotating at 10000 RPM, and object B is rotating at 1000 RPM; Object B has a larger radius, is made of a denser material, and has more mass and a larger moment of inertia. The two object come into contact with each other (picture the clutch of a car).
If the frictional force is sufficient, they should "stick" and the angular momentum should be conserved (I1ω1+I2ω2=Inetω); but if it isn't, they should slip.
What would be the frictional force required to "stick" the two objects together instantaneously?
Answer
Forces cause accelerations. The rotational analogue of this is that torques cause rotational accelerations. "Instantaneous" doesn't exist. You need to figure out the maximum amount of time you want it to take.
Say you determine that before and after coupling, one object will have some change in angular velocity, $\Delta \omega$. You will have already calculated or measured the moment of inertia, $I$. Further, you decide that for the purposes of your device, "instantly" means in less than 0.1 seconds.
$$ \Delta \omega = \alpha t$$ $$ \Delta \omega = \frac {\tau}{I} t$$ $$ \frac{\Delta \omega I}{t} = \tau$$
So you'll plug in for the change in rotational velocity, moment of inertia, and time required and you'll get the necessary torque. The faster you need it to couple, the higher torques/forces you'll have to generate.
Given the geometry of how you're applying it, you can turn the torque into a force. If it's driven by friction, you can use that and the coefficient of friction to determine the required normal force to produce that much friction.
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