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Homework Help: Finding time using a given torque, etc.

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.2m diameter and a mass of 250kg. A motor spins up the flywheel with a constant torque of 40Nm. How long does it take the flywheel to reach top angular speed of 1250rpm?


    2. Relevant equations

    Wf = Wi + alpha(delta t)

    3. The attempt at a solution

    I tried using the above equation except I substituted delta W/delta t for alpha. I got 2815.315s, but that seems awfully high, and it wasn't right.
     
  2. jcsd
  3. Oct 14, 2007 #2

    hage567

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    I don't see how you would have got that to work. Did you use the information about torque? Do you know how to find the moment of inertia of a disk? You can find the angular acceleration using those two quantities. Once you've done that, use the equation under "relevant equations" to find the time.
     
  4. Oct 14, 2007 #3
    I = m(r^2) = 250(.6^2) = 90. I did that and I got the exact same answer.
     
  5. Oct 15, 2007 #4
    I'm still unsure about this problem.
     
  6. Oct 15, 2007 #5

    hage567

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    Remember that torque = I * alpha. You found I, you were given the torque in the question. So you can get alpha.

    Also, a flywheel is shaped like a disk, so for I you should use 0.5*M*R^2, not just M*R^2.
     
  7. Oct 15, 2007 #6
    Ok, using the new I = 45, alpha = .889. So then substituting into Wf = Wi + alpha (delta time), delta time = 1406.074. Does that sound right?
     
  8. Oct 15, 2007 #7

    hage567

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    Your angular velocity was expressed as 1250 rpm, (revolutions per minute). You need to change that to radians/second, so you units are consistent and correct. Do you know how to do that?
     
  9. Oct 15, 2007 #8
    Why no I do not....I know how to change it to revolutions/second, but not radians/second
     
  10. Oct 15, 2007 #9

    hage567

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    There are 2*pi radians per revolution.
     
  11. Oct 15, 2007 #10
    Do I change revolutions/minute to revolutions/second then then do the 2pi radians?
     
  12. Oct 15, 2007 #11

    hage567

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    Yes, you must do both.
     
  13. Oct 15, 2007 #12
    Got it. Thanks!!!
     
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