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Homework Help: Atwood's Pulley

  1. May 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Equation 1: m1g - T1 = m1a (for first mass)
    Equation 2: T2 - m2g = m2a (for second mass)
    Equation 3: (T1-T2)R = IA (for pulley; A = rotational acceleration)

    What are the steps I must take to get the final equation that describes the motion of the system:
    (m1-m2) g = (m1+m2+I/R^2)a ??

    Thank you


    2. Relevant equations



    3. The attempt at a solution

    I can figure out a = g(m1-m2)/(m1+m2) BUT i dont know how to factor in equation 3?
     
  2. jcsd
  3. May 21, 2010 #2

    Are you trying to find the linear acceleration of the system?
     
  4. May 21, 2010 #3

    Doc Al

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    Staff: Mentor

    That's only true for a massless pulley.
    Hint: Express the rotational acceleration A in terms of a. Then you'd have 3 equations and 3 unknowns.
     
  5. May 21, 2010 #4
    No, I'm trying to show how to derive the last equation from the first 3.

    Doc Al:
    I know that A =a/R, which would give me (T1-T2)R = I (a/R) , I'm just not sure how to combine that with the two equations I combined for the masses: a = g(m1-m2)/(m1+m2)
     
  6. May 22, 2010 #5

    Doc Al

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    Staff: Mentor

    Good. That becomes your revised third equation.
    Careful: You did not get this equation by combining the first two equations! The first two equations have three unknowns--a, T1, T2--there's no way to eliminate the two tensions using just the first two equations. (You would get this equation if you combined the two equations for an Atwood's machine with a massless pulley. But those equations are different, since there is only a single tension to worry about.)

    To combine the first two equations, you just add them. To combine all three, just add all three!
     
  7. May 22, 2010 #6
    Ok, so I was wrong by assuming a massless pulley therefore making that first equation I solved for "a" wrong.

    Now I have:
    Equation 1: m1g - T1 = m1a (for first mass)
    Equation 2: T2 - m2g = m2a (for second mass)
    Equation 3: (T1-T2)R = I(a/R) (for pulley)

    Do I set 1 and 2 equal to zero, and then make them equal to eachother so I can isolate T1 and T2 to make (T1-T2). Then do I sub (T1-T2) into equation 3?

    Im confused
     
  8. May 22, 2010 #7

    Doc Al

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    Staff: Mentor

    I don't really know what you mean by setting them equal to each other or setting them equal to zero.

    In any case, it's much easier than all that.

    Why don't you do what I suggested in the last sentence of my previous post?
     
  9. May 22, 2010 #8
    Ok I think Im starting to get it, I got this far:

    Adding equations 1 and 2: m1g + (-m2g) - T1 + T2 = m1a + m2a
    Adding in equation 3: g (m1-m2)+ (T1-T2) = a (m1+m2) + I (a/R^2)

    g(m1-m2) = a (m1+m2+I/R^2)

    Everything look right?
     
  10. May 22, 2010 #9

    Doc Al

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    Good.
    I forgot to mention that you needed to divide both sides of equation 3 by R to give units of force. So equation 3 becomes: (T1-T2) = I(a/R^2)

    Adding that, you get:
    g (m1-m2) - T1 +T2 + (T1-T2) = a (m1+m2) + I (a/R^2)

    Which gives you your final result:
    Looks good!
     
  11. May 22, 2010 #10
    Thanks so much for your help Doc Al! Very appreciated
     
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