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Flywheel spinning

  1. Nov 18, 2012 #1
    At what rate is the flywheel spinning when the power comes back on?
    A high-speed flywheel in a motor is spinning at 450rpm when a power failure suddenly occurs. The flywheel has mass 44.0kg and diameter 77.0cm. The power is off for 32.0s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 170 complete revolutions.

    1. At what rate is the flywheel spinning when the power comes back on?

    2. How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on?

    3. How many revolutions would the wheel have made during this time?

    So it seems like if I get the first part the rest will be easy. This is what I have for the first part but it doesn't seem right.


    170 revs = 1068.141502221 rad
    450 rpm=47.1 rad/s
    t = 32.0 sec

    1068.141502221 = 47.1 * 32 +0.5 * α * 32^2
    So α = 0.528992424 rad/s?
     
    Last edited: Nov 18, 2012
  2. jcsd
  3. Nov 18, 2012 #2

    Dick

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    You solved that last equation wrong. You got a positive value of α. That would say it's speeding up. Try it again.
     
  4. Nov 18, 2012 #3

    ehild

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    α is the angular acceleration. You need to calculate the rpm at t=32 s.


    ehild
     
  5. Nov 18, 2012 #4

    Dick

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    True. I probably shouldn't have stopped there. Finding α is only a step along the way. And you have the units of α wrong. Should be rad/s^2.
     
  6. Nov 18, 2012 #5
    How do I get it to be negative unless it should be -.5? Oh right it is rad/s^2.
     
  7. Nov 18, 2012 #6

    Dick

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    Show how you got α = 0.528992424 rad/s^2.
     
  8. Nov 18, 2012 #7
    I divided right from left

    Formula I used: θ = ωi * t + ½ * α * t^2
     
  9. Nov 18, 2012 #8

    Dick

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    No complaints about the equation you derived from but in numbers that becomes: 1068.1 =α 512.0 + 1507.2. How does "divided right from left" give you a positive α?
     
  10. Nov 18, 2012 #9
    α = (1068.1 - 47.1 * 32) / ( 0.5 * 32^2)

    Mhm not sure what I was doing before but now I did this and got negative; -0.857617188 rad/s^2

    To get it to rad/s I use ωf = ωi + α * t

    Therefore it's (47.1 + ( -0.857617188 * 32) = 19.656 rad/s
     
  11. Nov 18, 2012 #10

    Dick

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    That's better. Can you get the final angular velocity now?
     
  12. Nov 18, 2012 #11
    Sorry I edited the one before


    To get it to rad/s I use ωf = ωi + α * t

    Therefore it's (47.1 + ( -0.857617188 * 32) = 19.656 rad/s
     
  13. Nov 18, 2012 #12

    Dick

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    Ok, so the rest is easy, yes?
     
  14. Nov 18, 2012 #13
    I'm doing it right now, I'll post my answers just to be sure. Btw thanks a lot!
     
  15. Nov 18, 2012 #14
    2. ωf = ωi + α * t

    0 = 47.1 -0.857617188 * t

    So 47.1/ ( 0.857617188 ) = 54.9 seconds


    3. θ = ωi * t + ½ * α * t^2

    41.7 * 54.9 + 0.5 * -0.857617188 * 54.9^2

    2289.22 - 1292.4 = 996.82

    996.82 radians / 2 * pi = 158.64

    It says number 3 is wrong?
     
    Last edited: Nov 18, 2012
  16. Nov 19, 2012 #15

    ehild

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    You made a mistake (in red).

    ehild
     
  17. Nov 19, 2012 #16
    Blah human blunder, lol thanks a ton.

    2585.79 - 1292.4 = 1293.39

    1293.39/ (2 * pi) = 205.849 revs

    Thanks again!
     
  18. Nov 19, 2012 #17

    ehild

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    You could have saved lot of work by using the formula θ=0.5(ωiif)t, which is the same as

    number of revolutions= 0.5(fi+ff)t (f is revolutions/s).

    fi=450/60=7.5 s-1
    170=0.5(7.5+ff))32--->ff=3.125 s-1, ωf= 19.6 rad/s


    ehild
     
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