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Dynamics of rotational motion: the yo-yo

  1. Apr 1, 2006 #1
    the question is about that:
    Ayo-yo is made from two uniform disks, each with mass m and radius R,
    connected by a light axle of radius b. A light, thin string is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration and angular acceleration of the yo-yo and the tension in the string.

    the question from me is that , what is the use of radius b of axle?
    the question did not give me the mass of axle, and i can't find the
    moment of inertia of the axle..........
    can anyone tell me?
     
  2. jcsd
  3. Apr 1, 2006 #2

    Galileo

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    I think you can ignore the mass/moment of inertia of the axle compared to those of the uniform disks since they gave information other than that it's light.

    The radius b is important, since that will relate the falling velocity of the yo-yo to the rotational velocity of the disks.
     
  4. Apr 1, 2006 #3
    how the b be so important?
     
  5. Apr 1, 2006 #4

    Galileo

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    Did you read my post? "The radius b is important, since that will relate the falling velocity of the yo-yo to the rotational velocity of the disks"
    Now you have to find this relation, assuming the yo-yo unwinds without slipping.
    If the yo-yo falls a distance [itex]2\pi b[/itex], how many revolutions will the disks have made?
     
  6. Apr 1, 2006 #5
    javascript:;/2PI?
     
  7. Apr 1, 2006 #6
    2*pi*b/2*pi?
     
  8. Apr 1, 2006 #7
    i am sorry , i am making the mistake
     
  9. Apr 1, 2006 #8
    the axle and the disks are connect together?
    is the disks also make 2*PI*b revolutions?
     
  10. Apr 1, 2006 #9

    Galileo

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    No. Imagine the disks are turning. How much rope is unwound per disk revolution?

    EDIT: Yeah, the disks and axle are rigidly connected ofcourse, so they have the same angular velocity.
     
    Last edited: Apr 1, 2006
  11. Apr 1, 2006 #10
    2*PI*b
    can i write the formula like this?
    2mg-T=2ma---------first formula
    Tb=m*(R^2)*(a/b)--------second formula
    a-acceleration T-tension
    then solve those equations,
    is that right?
     
  12. Apr 1, 2006 #11

    Galileo

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    Yes, that looks correct.
     
  13. Apr 1, 2006 #12
    but i can't get the answer
    my director have the answer like that:
    a=(b^2/b^2+R^2)g α=a/b T=(2MR^2/b^2+R^2)g
    the answers have confused me so much.......
     
  14. Apr 1, 2006 #13

    Galileo

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    Yeah, there's a factor of two missing somewhere...
    If you take your second equation:
    Tb=m(R^2)(a/b)
    and replace it with
    Tb=2m(R^2)(a/b)
    you get the same answer. But I think your equation is correct. The total moment of inertia is 2(1/2mR^2)=mR^2 not 2mR^2. I don't see the error.
     
  15. Apr 1, 2006 #14
    exactly, i think he give me the wrong answer
    thank you so much~
     
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