Flywheel storing kinetic energy

  • #1
1,039
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Homework Statement


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?

Homework Equations


[itex]KE = \frac{1}{2}Iω^{2}[/itex]
[itex]I = kmr^{2}[/itex]
The constant k, for a solid uniform disk is [itex]\frac{1}{2}[/itex]


The Attempt at a Solution



[tex]KE = \frac{1}{4}(75.0)(1.15^{2})ω^{2}[/tex]


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.
[tex]a = αr[/tex]

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

Answers and Replies

  • #2
1,198
5
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
Delphi51
Homework Helper
3,407
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You need Fc = ma = mv²/r
 
  • #4
993
13
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
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
Hint: What's the centripetal acceleration at the outer edge of the flywheel? There's a neat formula for it.
 
  • #6
1,039
2
You need Fc = ma = mv²/r
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
1,198
5
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?
But you certainly know the maximum acceleration which is in what direction?
 
  • #8
1,039
2
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:

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

Now:

[itex]KE_{max} = \frac{1}{2}Iω^{2}[/itex]
[itex]KE_{max} = \frac{1}{2}I(54.61)^{2}[/itex]
[itex]I = 0.5(75)(1.15)^{2} = 49.59375[/itex]

[itex]KE_{max} = \frac{1}{2}(49.59375)(54.61)^{2}[/itex]

[itex]KE_{max} = 73950.53[/itex] J

Does that make sense?
 
  • #9
1,198
5
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
1,039
2
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.
 

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