Power of a Rotating Mass.

No.

It is insignificant, unless you are working to Joules.

You don't want power, you want energy. Power is energy divided by time.

First, you need to figure out the moment of inertia, I.

Energy = 1/2 I w^2 . (w is the rotation speed in radians per second.)

https://en.wikipedia.org/wiki/Rotational_energy

This not much different than your April question about power and torque...and the answers are still the same. One piece of information that you failed to include, and which has a major impact on the answer, is how the mass is distributed around the center of rotation.

Is it a uniform disc, does it taper, is all the mass concentrated near the shaft or as far from the shaft as possible. Look up flywheels and moment of inertia and you'll learn the distinction (and the answer to your question).

Its a uniform disc

That's so easy the equation can be looked up. Try Kempe's Engineers' Yearbook, any edition.

Even if you do want to account for bearing friction, the second formula is incorrect. The 0.001 will be added or subtracted somewhere, not used to multiply the whole formula.

Personally I'd be more concerned with balancing the disc or whatever it is.

And the above is correct, as you describe it you're dealing with potential energy (of the kinetic variety) not power. Unless there is a force involved in here somewhere.

But again, I see this as either a vibrations issue first with the details given, or a HW question that's being disguised. If it's the latter it's a very simple dynamics problem with the answer given above.

The answer given in comment 2 is correct i.e. Energy E = 1/2*I*w^2

When you calculate Moment of Inertial 'I' the distribution of masses is automatically included. For instance consider a small mass 'm1' located at a distance of 'r1' from the center of rotation. Let the speed of rotation be 'w'. The kinetic energy of the mass will equal 1/2*m1*(r1*w)^2 = 1/2*m1*r1^2*w^2. Similarly for other masses m2,m3,m4... at distances r2,r3,r4.....etc.The total energy of the rotating mass will then equal

1/2 *(m1*r1^2+m2*r2^2+m3*r3^2+....)*w^2 = 1/2*I*w^2. Note that 'w' is constant for the rotating mass.

"ball bearings" LOL, just exactly what kind of ball bearings are you intending to use, THAT will determine whether or not consider not only their "coefficient of friction" but also:

are they lubed or dry?

are they lubed with oil or grease?

are they sealed, shielded or open? bearing manufacturers include factors for seal drag and lube viscosity for a reason - because it matters.

are they single row or double row DGBBs?

are they mounted or inserts?

So regardless of the correct formula; you need to examine your bearings more than coefficient of friction.

You seem to have confused responders with your terminology.

I would say it is clearer if you ask one of the following.

1. How much power do I need to supply to maintain rotational velocity given a certain level of friction?

2. How much power can this spinning disk supply which will then depend on how quickly it slows down and therefore the rate at which it gives up energy to something else.

3. What is the potential energy of the disk at a specific rotational velocity?

Any of these questions make sense:

Power relates to the rate at which energy is supplied or given up dependent on viewpoint, most real life situations involve friction and so power is usually required to maintain constant motion.

Often when people ask what is the power of a mechanism they mean how much power can it supply although its tricky as a 200W sound system will supply power to its suroundings but in turn requires power from the electrical socket.

Usually when people want to know how much power a mechanism requires they will talk of power requirement or state how much power is required to keep a disk spinning with bearings having a friction coefficient of......

Better to give a description of what you are asking as suggested above to avoid ambiguity.