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Moment of inertia-yo-yo

  1. Mar 22, 2013 #1
    1. The problem statement, all variables and given/known data

    a yo-yo is spun from rest by pulling on the string with a constant tension of 2.0N. The radius of the inner rod on which the string is strung around is 0.50cm. the tension applied for 5.0seconds after which the yo-yo is observed to spin with an angular velocity of 15 rad/sec

    a)from the information given above, what is the moment of inertia of the yo-yo

    b) what is the total angle the yo-yo has traveled through in these 5.0 seconds

    Now you press your finger against the outer rim of the yo-yo ( which has a radius of 4.0 cm) to bring it to a stop. You apply a constant force of 2.0N directed perpendicular to the rim of the yo-yo. The tension from part a) is no longer being applied to the yo-yo. The coefficient of kinetic friction between you finger and the edge of the yo-yo is 0.80.

    c) how ling does it take for the yo-yo to come to a stop

    2. Relevant equations

    I=1/2Mr^2
    f=ma

    3. The attempt at a solution

    a)
    f=ma
    2=m x 9.8
    2/9.8=m
    m=0.204kg
    I=1/2Mr^2
    I=1/2 x 0.204 x 0.5^2
    I=0.026
    I=0.03

    b) I think you do 15 rad/sec x 5. seconds x 360°
    c) have no idea
     
  2. jcsd
  3. Mar 22, 2013 #2

    Doc Al

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    Assume that the only force you need to worry about is the tension applied to the yo-yo.

    Use: Torque = I * alpha

    What's alpha? What's the torque?
     
  4. Mar 22, 2013 #3

    SteamKing

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    How many radians are in 360 degrees?
     
  5. Mar 22, 2013 #4

    haruspex

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    That would be true if the yo-yo were simply suspended, statically, from the string. But there's acceleration involved here.
    As Doc Al says, you don't need to worry about gravity in this question. That's because you can take moments about the centre of the spindle; the gravitational force acts through that point, so has no moment about it.
    That's true for a uniform disc radius r. The given radius is only for the central spindle. You don't know how big the yo-yo is overall, or how the mass is distributed in it. The point of the question is to calculate I from the external behaviour described.
     
  6. Mar 23, 2013 #5
    sorry, for question c
    c) how ling does it take for the yo-yo to come to a stop

    its ment to be 'how long does it take for the yo-yo to come to a stop'
     
  7. Mar 23, 2013 #6
    i still don't get how to answer question c
     
  8. Mar 23, 2013 #7

    Doc Al

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    First you must solve questions a and b correctly! (You don't really need b to answer c.)

    For question c, you'll have the moment of inertia (from part a) and the forces slowing the yo-yo (applied by the finger) so you can calculate the new angular acceleration (alpha).
     
  9. Mar 23, 2013 #8
    Sois the formula alpha= moment of inertia times the force applies by the finger
     
  10. Mar 23, 2013 #9

    Doc Al

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    No. The formula is just Newton's 2nd law applied to rotation, which I gave before:
    Torque = I * alpha
     
  11. Mar 23, 2013 #10
    But the question says how long does it take for the yoyo to stop. The formula u provided is torque=I x alpha
     
  12. Mar 23, 2013 #11

    Doc Al

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    That's just the first step: Find alpha. Once you have alpha, use some kinematics to solve for the time.
     
  13. Mar 23, 2013 #12
    what is 'some kinematics'
     
  14. Mar 23, 2013 #13

    Doc Al

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  15. Mar 23, 2013 #14
    so if i am correct this is how u do it

    alpha=3

    v=v0+at
    0=15+3t
    -5=t
     
  16. Mar 23, 2013 #15

    Doc Al

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    If alpha = 3 rad/s^2, then that would be correct. (For one thing, it's slowing down so alpha would be negative.)

    You first need to calculate alpha.
     
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