# Flywheel storing kinetic energy

1. Nov 15, 2011

### QuarkCharmer

1. The problem statement, all variables and given/known data
There is a 2.30 m diameter flywheel, with a mass of 75.0 kg. For safety/structural (or whatever) reasons, the acceleration at a point on it's rim cannot exceed 3430 m/s^2.

What is the max KE that can be stored in the flywheel?

2. Relevant equations
$KE = \frac{1}{2}Iω^{2}$
$I = kmr^{2}$
The constant k, for a solid uniform disk is $\frac{1}{2}$

3. The attempt at a solution

$$KE = \frac{1}{4}(75.0)(1.15^{2})ω^{2}$$

I'm not sure how to go about getting this in terms of radial acceleration?

I was thinking that I could somehow use the relationship that linear acceleration (a) = radial acceleration (α) times the radius.
$$a = αr$$

But that is not panning out at all. Can someone give me some direction please?

Last edited: Nov 15, 2011
2. Nov 15, 2011

### LawrenceC

Hint: What causes a flywheel to fail? Flywheels are generally heavy so they do not accelerate angularly at a high rate. So what else is there?

3. Nov 15, 2011

### Delphi51

You need Fc = ma = mv²/r

4. Nov 15, 2011

### grzz

I think one can assume that the angular velocity is constant. Then a point on its rim will have no tangential acceleration but only centripetal acceleration.

Then one can use v = r x omega

5. Nov 15, 2011

### SteamKing

Staff Emeritus
Hint: What's the centripetal acceleration at the outer edge of the flywheel? There's a neat formula for it.

6. Nov 15, 2011

### QuarkCharmer

I don't see how to use that in this situation. I don't know the centripital force, or the linear velocity, nor do I understand how to get the maximum kinetic energy out of the equation, or substitute that into the formula for the radial kinetic energy?

7. Nov 15, 2011

### LawrenceC

But you certainly know the maximum acceleration which is in what direction?

8. Nov 15, 2011

### QuarkCharmer

Oh, wait. The acceleration is in m/s^2, by "radial" they are talking about the centripital acceleration, not the "rotational acceleration (rad/s^2). So:

$$a_{c}=\frac{v^{2}}{r}$$
$$v = ωr$$
$$a_{c} = \frac{(ωr)^{2}}{r}$$
$$a_{c} = r(ω)^{2}$$
$$3430 = (1.15)ω^{2}$$
$$\frac{3430}{1.15} = ω^{2}$$
$$ω_{max} = 54.61$$

Now:

$KE_{max} = \frac{1}{2}Iω^{2}$
$KE_{max} = \frac{1}{2}I(54.61)^{2}$
$I = 0.5(75)(1.15)^{2} = 49.59375$

$KE_{max} = \frac{1}{2}(49.59375)(54.61)^{2}$

$KE_{max} = 73950.53$ J

Does that make sense?

9. Nov 15, 2011

### LawrenceC

What you did makes sense. I did not check your value for I but using your value for I in my equations, I get essentially the same answer.

Flywheels fail because the internal forces (which are due to centripital forces) cause stresses that exceed what the material can withstand.

10. Nov 15, 2011

### QuarkCharmer

Great, thanks! I am sure that my I value is correct, I pulled it from the tables.

Appreciate the help, I can't believe I didn't catch that they were not talking about rotational acceleration.