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Time to stop a flywheel

  1. Jul 7, 2009 #1
    I have a question in regards to time it take to stop a flywheel.
    I have a 10 foot dia x 3 foot thick solid flywheel that weighs approx 20,000 lbs.
    It is rotating at about 2 rpm I have a electric brake that is rated to apply 9500 in-lb of torque.

    How long (sec) or how many rads will it take to stop?

    If possible please show math, the diameter, mass and speed can vary.

  2. jcsd
  3. Jul 7, 2009 #2


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    This sounds like homework. YOU show the math.
  4. Jul 7, 2009 #3


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    Staff: Mentor

    Wow, that is a BIG flywheel !! What kind of machine is it on?

    As negitron says, if this is homework/coursework for school, we cannot work out the answer for you. Per the PF Rules (see the link at the top of the page), you must show the relevant equations and do the bulk of the work. If it is not for homework/coursework, and it still looks like it is from school, those rules still apply.

    Having said that, the key concepts involved are the "Moment of Inertia" of the flywheel, and the rotational kinematic equations of motion. You can read about both of those at wikipedia.org as background:



    EDIT -- BTW, the brake will apply a torque to the flywheel (at some radius out from the central axis of rotation), and this torque will generate a deceleration (negative acceleration) to slow the flywheel
  5. Jul 8, 2009 #4
    I was trying to simplify the question. It is actually a large industrial belt winding machine. The machine winds industrial belting up to 4' wide while an operator inspects it for damage, up to a diameter of 10' (it varies speed based on diameter to maintain a 30-40 fpm liner feed. It currently freewheels to a stop but because of a recent OSHA inspection, HR/Safety is telling me if the e-stop is hit, it has to stop within 2 seconds. I had the 9500 in-lb brake in the crib and before I put it on I was hoping to figure out if it will work.

    Hope this clarifies

    Thanks for the assistance,
    Last edited: Jul 8, 2009
  6. Jul 8, 2009 #5
    I found a simplifed formula

    Time (s) = Inertia (lb-ft^2) * Delta RPM / (308 * Torque (lb-ft))

    Inertia = .5 * Mass (lbs) * radius ^2 (ft)


    Inertia = .5 * 20000lbs * 5ft ^2 = 250000 lb-ft^2

    Convert 9245 in-lb to 770.4 ft-lb

    Time = 250000 lb-ft^2 * 1.5 rpm / (308 * 770.4 ft-lb) = 1.58 sec

    I'm under 2sec so I should be OK
  7. Jul 8, 2009 #6


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    First you should compute the energy of the flywheel. The diameter is not enough as flywheels are usually heavier at the far edge that in the middle so you need to find out it's wheight ( maybe already given) and the distance to the surface center of mass of a cut-out section ( only one half of the wheel )to the axis L . Knowing it's mass and speed ( the speed is 2*pi*L*Rpm/60 ) energy is mv2/2.

    Then the friction will linearily take away energy til it stops ...but... I'll have to continue some other time.
  8. Jul 8, 2009 #7


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    Staff: Mentor

    Lok makes a good point. Is the flywheel a uniform cylinder, or is its mass more concentrated out near the outer edge?
  9. Jul 8, 2009 #8
    It is a basically a solid rubber cylinder of more or less uniform density, other than a 12" dia particleboard core.

    The weight is approximate based on .036 lbs/in^3 + a, little fudge factor.
  10. Jul 8, 2009 #9


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    If so a solid rubber cylinder, then L is equal to half the diameter. Energy can be easily calculated.

    Friction is a simple force that equals = y*F (y being the coefficent and F the force applied by the brake ). The brake surface is only important for wear etc.

    Knowing the Braking force and the energy stored by the flywheel you can obtain it's stopping distance = E/F ( E - energy of flywheel , F - braking force )

    But the thing is that depending on where the brake is placed ( near the axis or near the edge ) you will get more or less revolutions of the flywheel. For the least amount of revolutions the edge is best.
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