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.