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Flywheel storing kinetic energy

  1. Nov 15, 2011 #1
    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
    [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]

    3. 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: Nov 15, 2011
  2. jcsd
  3. Nov 15, 2011 #2
    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?
  4. Nov 15, 2011 #3


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    You need Fc = ma = mv²/r
  5. Nov 15, 2011 #4
    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
  6. Nov 15, 2011 #5


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    Hint: What's the centripetal acceleration at the outer edge of the flywheel? There's a neat formula for it.
  7. Nov 15, 2011 #6
    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?
  8. Nov 15, 2011 #7
    But you certainly know the maximum acceleration which is in what direction?
  9. Nov 15, 2011 #8
    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]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]


    [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?
  10. Nov 15, 2011 #9
    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.
  11. Nov 15, 2011 #10
    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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