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Classical Physics - Pulley Problem

  1. Aug 29, 2007 #1
    A massless string is placed over a massless pulley, and each end is wound around and fastened to a vertical hoop. The hoops have masses M1 and M2 and radii R1 and R2. The apparatus is placed in a uniform gravitation field g and released with each end of the string aligned along the field.

    I have to show that the tension is T = gM1M2/(M1+M2)

    I can sort of solve the problem by just saying that (M1M2)/(M1+M2) is the reduced mass. Then all we have to do is say the tension is balanced with the weight, so that T = Mg = gM1M2/(M1+M2). But then doesn't this imply that the hoop isn't moving? I think I'm missing something here... It is also important to realize that both hoops are rolling down the string so the acceleration of one hoop downward is not necessarily the acceleration of the other one upward... Any help / hints would be appreciated... thanks
     
  2. jcsd
  3. Aug 29, 2007 #2
    <statement retracted>
     
    Last edited: Aug 29, 2007
  4. Aug 29, 2007 #3
    <statement retracted>
     
    Last edited: Aug 29, 2007
  5. Aug 29, 2007 #4

    Dick

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    Try this. Each hoop has a torque on it of T times the appropriate radius. This determines the angular acceleration alpha of each hoop. Let s be the amount of string as a function of time. Then I claim s''=a1+a2=R1*alpha1+R2*alpha2 where the a's are the linear acceleration of each hoop. (The sum of the center of mass accelerations determines the second derivative of the length of string but the amount is also determined by the angular accelerations). Put this together with the usual force balance diagram for the vertical forces on each hoop and you will get what you want.
     
  6. Aug 29, 2007 #5

    Dick

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    That's the result if the hoops don't spin.
     
  7. Aug 29, 2007 #6
    I see.
    <edits>
     
  8. Aug 29, 2007 #7
    ok I'm here... about to pull my hair out
     
  9. Aug 29, 2007 #8

    Dick

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    Can you work out a1 and a2 from linear force balance? Can you work out alpha1 and alpha2 from moment of inertia and torque? Let's get started... Do you accept the equality of the two expressions for s''? Does it seem right?
     
  10. Aug 29, 2007 #9
    yes to all
     
  11. Aug 29, 2007 #10
    so just using T - M1g = M1A1
    and T - M2g = M2A2
     
  12. Aug 29, 2007 #11
    but these accelerations are center of mass acceleration for each hoop
     
  13. Aug 29, 2007 #12
    and the equations for s'' make perfect sense
     
  14. Aug 29, 2007 #13

    Dick

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    My convention was to label positive acceleration down. Suggest you do the same if you want s''=a1+a2. Now the rotational part.
     
  15. Aug 29, 2007 #14

    Dick

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    Center of mass acceleration is also related to the rates of string being played out. Think about it.
     
  16. Aug 29, 2007 #15
    and Torque1 = Inertia_hoop1*alpha1 = Tension*R1
    Torque2 = Inertia_hoop2*alpha2 = Tension*R2
     
  17. Aug 29, 2007 #16

    Dick

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    Doing great! Put in the I's, solve for the alphas and put into the s'' equation along with the a's.
     
  18. Aug 29, 2007 #17
    hmm having trouble relating alphas to Acom
     
  19. Aug 29, 2007 #18
    oh just A = alpha*r right?
     
  20. Aug 29, 2007 #19
    hmm that's tangential though not center of mass accel
     
  21. Aug 29, 2007 #20
    and Inertia1 = (1/2)MR1^2
    Inertia2 = (1/2)MR2^2
     
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