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Help with Flywheel problem.

  1. Oct 28, 2004 #1
    Hi people.
    I wonder if anyone could check a problem I had been set as homework and [possibly point me in the right direction for the second part.

    A solid disc flywheel, mass 900kg, 0.75m radius accelerated at 0.8rads/s(sq) from rest to 480 RPM in 63 secs.

    a) Calculate the number of revolutions during the acceleration period.
    I tried it like this:
    used the formula theta = w.t + 1/2.alpha.t(sq)

    then once i got the answer in radians divide it all by 2pi to get the revolutions. Is that correct? (252 revs)

    on a previous question I calculated the torque to be 202 Nm
    This next question is causing me some real problems, if anyone could point me in the right direction i would be gratefull.

    d) if 20% of the enery is used during a press operation calculate the reduction in speed (RPM) of the flywheel.

    Last edited: Oct 29, 2004
  2. jcsd
  3. Oct 28, 2004 #2

    Doc Al

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    Staff: Mentor

    That's fine. You could also have used the average speed: [itex]\theta = \omega_{ave} \Delta t[/itex].
    The KE of the flywheel is [itex]{KE} = 1/2 I \omega^2[/itex]. So set up a ratio of the energies before and after the press operation and see what happens to [itex]\omega[/itex].
  4. Oct 29, 2004 #3
    Thanks .
    I have ahd a look at the problem again and I am still abit puzzeled.
    I get the torque before the [press operation to be 202 Nm and after it has dropped to 162 Nm (the 20% drop).
    What I am not too sure about is does the torque in Nm equal the KE?

    If it does then i can use KE = 1/2.I.w(sq)

    If this is not the case i still dont know where to go. I assume the value `I` cannot change.

    ohhh hang on right if torque=I.alpha then the torque after operation = I.alpha so if I transpose to find alpha (because i know the new torque) then I can get the new deceleration??? I am still not sure about what I have just said above so any advice would help.

  5. Oct 29, 2004 #4

    Doc Al

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    Staff: Mentor

    No, the torque does not equal the KE. (Although the units are the same!)

    You need not calculate torque to solve this part of the problem. Set up the ratio of the KE before and after:
    [tex]{KE}_1/{KE}_2 = \omega_1^2/\omega_2^2[/tex]
    Let the ratio be [itex]{KE}_1/{KE}_2 = 1/0.8[/itex], and let [itex]\omega_1[/itex] be the original speed of 480 rpm (no need to worry about units since we are using ratios). Solve for [itex]\omega_2[/itex].
  6. Oct 30, 2004 #5
    Thankyou Doc Al

    That has helped loads. I get the new speed to be 429 RPM which .
    If I could ask just one more question just to clarify things. I would just tlike you to explain where the "1" comes from in "KE1/KE2 = 1/0.8". Sorry if its obvious. Is it just how ratios are sorted out?
    thanks loads for your help. and im sure that will be it now.
  7. Oct 30, 2004 #6

    Doc Al

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    No problem. Since KE2 = 0.8 KE1, then KE1/KE2 = KE1/(0.8KE1) = 1/0.8. Let me know if that doesn't make sense.
  8. Oct 31, 2004 #7
    Hi again.
    Sorry for being dumb, but I am still not 100% sure.
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