Power in a uniform circular motion

Question

I've a circular motion and I want to find the power that the motion is generating. The formula's I got:

$$\text{P}\left(t\right)=\text{W}'\left(t\right)=\text{E}'\left(t\right)$$

Where $\text{P}$ is the power, $\text{W}$ is work and $\text{E}$ is energy.

But I don't know how to continue with known formula's? The wheel (that spin in a circular motion, has a certain velocity and revolutions per minut, but I want to keep them as a variable)

Thanks in advance for any help.

EDIT

What I now have:

$$\text{E}=\text{K}=\frac{\text{I}\times\omega^2}{2}=\frac{\text{I}\times\left(2\pi\times\frac{\text{n}}{\text{t}_\text{n}}\right)^2}{2}=\frac{\left(\text{m}\times\text{r}^2\right)\times\left(2\pi\times\frac{\text{n}}{\text{t}_\text{n}}\right)^2}{2}=\frac{2\pi^2\text{m}\text{n}^2\text{r}^2}{\text{t}_\text{n}^2}$$

Where $\text{n}$ is the number of turns of the wheel and $\text{t}_\text{n}$ is the time (number of seconds) it takes for the wheel to spin $\text{n}$ times.

The wheel I use:

See the picture in the article

Answer 1

The wheel I use:

A 100 % accurate calculation of the inertia moment of this type of wheel is hard to do but getting a reasonable estimate is perfectly doable.

To do so we can use the simple principle that like masses, inertia moments are additive and subtractable. Thus by breaking the wheel up into constituent parts and calculating the inertia moment of each part, we can then sum to get a total $I$.

We can use this list of inertia moments for great help.

Part 1: these flanges can be considered flat discs with a central disc cut out of them:

$$I_1=m\Big(\frac{R^2}{2}-\frac{r^2}{2}\Big),$$

where $R$ is the large radius and $r$ the smaller one. $m$ is the mass of the flange.

Part 2: the spokes.

Each spoke contributes approx.:

$$I_2\approx \frac{mL^2}{3},$$

with $L$ the length of a spoke and $m$ it's mass.

Part 3: the scoops.

Treat each scopp approx. as a point mass, so each contributes:

$$I_3=\frac{mr^2}{2},$$

with $r$ the distance of a scoop from the centre of the wheel and $m$ the mass of a scoop.

Part 4: the central axle.

Assuming it's a hollow cylinder:

$$I_4=mr^2,$$

where $m$ is its mass and $r$ its radius.

Part 5: looks like a cylinder that is between the flanges and holds the scoops in place.

$$I_5=mr^2,$$

where $m$ is its mass and $r$ its radius.

Total $I$:

Add up all the above, using the right number of parts for each section.

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It might be easier to actually measure $I$. Consider the following diagram:

A rope is attached to the top of the wheel and a mass $m$ dangles off it. At $t=0$ the mass is released.

The mass $m$ causes torque $\tau$ and thus angular acceleration $\alpha$, acc. Newton's second law (applied to pure rotation):

$$\tau=I\alpha$$

With $\tau=mgR$,

$$mgR=I\alpha$$

So,

$$\alpha =\frac{mgR}{I}$$

Using this equation of motion it can further be deduced that:

$$I= \frac{mgR}{\pi}t^2,$$

where $t$ is the time for the weight to hit the floor.

Answer 2

A constantly spinning wheel doesn't generate any power. Power is energy change per time; energy added or lost per time. As long as it spins constantly (with constant angular velocity $\omega$), no energy is added or removed.

The expression you have found is kinetic energy, but the power is zero unless it changes:

$$P=E'=K'=0$$

The little mark $'$ indicates differentiation to time. Since time is not included in the expression for $K$, you are differentiating a constant value. Which gives zero.

If power was generated, then it means that kinetic energy changes with time. Then time $t$ would have been included in the expression.

Answer 3

If you are trying to calculate the power in Watts which a water wheel can generate, there is a much simpler answer, given in the reference below, based on the rate at which the water loses gravitational PE.

$$P=\eta Q\rho gH$$

where $Q$ is the volume flow rate of the water ($m^3/s$), $\rho$ is the density of water ($kg/m^3$), and $H$ is the height through which it falls. $\eta$ is the efficiency which is around 60% (mechanical) for an overshot wheel - and can be over 80%. After gearing up, the efficiency for generating electricity is less than 20%.

Reference : Waterwheels - British Hydropower Association