Calculating Power Output of Rotating Object with Mass M, Radius R, and Speed V

[Hanababa]

An object with a mass M is rotating around an axis, with an arm of radius R at a certain speed V.
Assuming that the movement is constant, and a 100% efficiency, how many watt of electrical power can this device generate ?

For example purpose, we can take M=60 kg, R=5 meters and V=10 meters per second.

I can't find all the formulas and am lost with the units.

Thanks a lot

 

[Lunar_Lander]

Maybe the first thing could be to figure out the rotational energy of the device. First you need the moment of inertia, which you need to do with the mass, the radius and the Huygens-Steiner theorem (also known as Parallel Axis theorem).

Once you figured that out (I leave it to you to find the formulas with this hints, I just found that you should be able to do that), you can calculate the angular velocity, this time you need the given radius and the given velocity. Insert into the formula for $$E_{rot}$$, calculate it and then we can talk again :).

 

[Hanababa]

Huygens-Steiner theorem:
I parallele(moment of inertia) = I(cm) + MR²

Then, K (the rotational energy) is
K = 1/2 * I parallele * V²
So, K = 1/2 * (Icm + MR²) * V²

So remaining questions are:
1. how do you get Icm ?
2. As an energy K should be in Joule, which is a watt.second. I need a result is watt only.

If Icm is indeed equal to 0, we would then have
K = 1/2 * 60 * 5² * 10²
K = 75000 Joule ??

Thank you Lunar Lander

 

[Lunar_Lander]

If Icm is indeed equal to 0, we would then have
K = 1/2 * 60 * 5² * 10²
K = 75000 Joule ??

You are on the right track, but you miss that you can't use v directly. You have to use the angular velocity, which is $$\omega=\frac{v}{r}$$.

In this case here you have this mass rotating around an axis at the distance r=5 m. For this case (a mass rotating around an axis), the HS-theorem reduces to $$I=M \cdot R^{2}$$.

 

[Hanababa]

Right,
so K = 1/2 * M * R² * (w)² with w = V/R
K = 1/2 * M * R² * (V/R)²
K = 1/2 * 60 * 5² * (10/5)²
K = 3000 Joule ??

It must be wrong because on the above formula, we see that the R² are cancelling themselves, which would mean that R is not relevant, cannot be. Sorry, I am a bit lost..

 

[Lunar_Lander]

Your value is correct, as the unit of omega is sec^-1. Thus there are no R² which cancel each other.

So you now arrived at the energy. The equation for getting the power of an electric generator is P_el=P_mech-P_loss. The last one is zero, as you gave 100% efficiency. The electrical power thus should be equal to the power required to spin up the mass. However, there normally should be a time given, as the power is equal to the energy pro unit time. I think someone else needs to pick up here...

But you are welcome Hanababa! :)

 

[Hanababa]

Thanks again,
does it mean that if the device is operating for 1 year, so the value W in watt would be
K = W * 365*24*3600
K = W * 31536000

So, about 31,5 Megawatt of electrical production per year ?