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Help! flywheel

  1. Apr 25, 2005 #1
    Question:
    A flywheel 0.600 m in diameter pivots on a horizontal axis. A rope is wrapped around the outside of the flywheel, and a steady pull of 40.0 N is exerted on the rope. The flywheel starts from rest, and 5.00 m of rope are unwound in 2.00 s.

    Okay i have tried finding the angular velocity by first finding the linear velocity=5/2 then putting it into v=omega*r and making omega the subject. Then after finding omega I substituted it into the equation of (omega - 0)/2 but its wrong :(

    Where have I gone wrong? Can someone help me out? thanks in advance
     
  2. jcsd
  3. Apr 25, 2005 #2

    OlderDan

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    You have not actually stated the question you are trying to answer. Something about acceleration I bet. In any case, you have probably confused average velocity with final velocity. The information given in the problem gives you the average velocity. From that you have to decuce final velocity and probably use that to find acceleration.
     
  4. Apr 25, 2005 #3
    What is the angular acceleration of the flywheel?
     
  5. Apr 25, 2005 #4

    OlderDan

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    So.. what did you actually calculate? What was your answer?
     
  6. Apr 25, 2005 #5
    i got 4.17 as angular accerleration but it was wrong
     
  7. Apr 25, 2005 #6
    Acceleration = Velocity / Time.. you found the velocity, you have the time..
     
  8. Apr 25, 2005 #7
    would that give me the tangential acceleration?
    the i can work out a_tan = r*alpha?
    alpha = angular acceleration
     
  9. Apr 25, 2005 #8
    Alternatively if oyu can find the angle swept during the 2s, you can find the acceleration using the relationship

    [tex] \theta = \frac{\alpha t^2}{2} [/tex]
     
  10. Apr 25, 2005 #9

    OlderDan

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    Either you have confused the average and final velocity, or used the given diameter instead of the radius in your calculation.
     
  11. Apr 25, 2005 #10
    so to work out theta i would use s=r*theta with s=5 and r=0.3?
     
  12. Apr 25, 2005 #11

    OlderDan

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    That would be correct for using

    [tex] \theta = \frac{\alpha t^2}{2} [/tex]

    You can also do it using your original velocity calculation, but you need to understand that is the average velocity, which is 1/2 the final velocity.
     
  13. Apr 25, 2005 #12
    alright - i tried using the equation above with theta but I still get the wrong answer for the acceleration: i did -
    5/0.3 = (alpha(4))/2 and got 8.33
     
  14. Apr 25, 2005 #13

    OlderDan

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    That looks right.. what are they giving as the correct answer?

    By velocity it would be average velocity = 5m/2 sec = 2.5m/sec. So omega = v/r = 5m/2sec/.3m = 8.33/sec. The final angular velocity would be twice that and the time to reach that angular velocity is 2 seconds

    alpha = 2*8.33/sec/2sec = 8.33/sec^2
     
    Last edited: Apr 25, 2005
  15. Apr 25, 2005 #14
    thanks!! i just realised that i did something wrong haha
     
  16. Apr 28, 2005 #15
    how would u work out the final kinetic energy here?
    i tried using k_e=.5*I*(omega)^2

    but I is an unknown, as to find I, u use I=kmr2, so I=(1/2)*m*(.03)^2 (as this is a solid cylinder)
    and i hav no idea how to work out m.

    wait can u use f=ma? so 40 n = m* a, where a =r*alpha, or=2.499

    so 40=m*2.499 and m=16.01kg?

    ___NVm got the answer
     
    Last edited: Apr 28, 2005
  17. Apr 28, 2005 #16
    You would need the mass to find kinetic energy.
     
  18. Apr 28, 2005 #17

    OlderDan

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    If you have found the angular deceleration correctly, you now have everything needed to figure out I, and you need to do it.
     
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