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Atwood machine acceleration

  1. Mar 31, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm trying to find the acceleration of the masses in an atwood machine, the pulley has a rotational inertia I. The pulley has a radius R

    Picture : http://en.wikipedia.org/wiki/File:Atwood.svg
    But I made it accelerate the other way, so the equations from newtons 2nd law are (a bit) different for this picture.


    2. Relevant equations
    [tex] m_1 \cdot a = T_1 - m_1 \cdot g[/tex]
    [tex] m_2 \cdot a = m_2 \cdot g - T_2[/tex]
    [tex]Ʃ \tau = \frac{dL}{dt}[/tex]


    3. The attempt at a solution
    I know you can do it another way, but I wan't to do this with angular momentum L(with respect to midpoint of the pulley). The pulley has a radius R.

    [tex] L= I_{pulley} \cdot \omega + (m_1 + m_2) \cdot R^2 \cdot \omega = I_{pulley} \cdot \frac{v}{R} + (m_1 +m_2)vR[/tex]

    So:
    [tex] \frac{dL}{dt} = I_{pulley} \cdot \frac{a}{R} + (m_1 +m_2)a \cdot R[/tex]
    [tex] Ʃ \tau = T_2 \cdot R - T_1 \cdot R[/tex]
    [tex] T_2 = m_2 \cdot g -m_2 \cdot a[/tex]
    [tex] T_1 = m_1 \cdot a + m_2 \cdot g [/tex]
    So
    [tex] Ʃ \tau = g \cdot R (m_2 - m_1) - a(m_2 +m_1)[/tex]
    [tex] Ʃ \tau = \frac{dL}{dt}[/tex]
    So
    [tex] g \cdot R (m_2 - m_1) - a(m_2 +m_1)R = I_{pulley} \cdot \frac{a}{R} + (m_1 +m_2)a \cdot R [/tex]

    Solve for a and i'll get:
    [tex] a=\frac{g \cdot (m_2 - m_1)}{\frac{I}{R^2}+2(m_1+m_2)}[/tex]

    The right answer should be:
    [tex] a=\frac{g \cdot (m_2 - m_1)}{\frac{I}{R^2}+m_1+m_2}[/tex]

    I notice that if i say:
    [tex] Ʃ \tau = gR(m_2-m_1) [/tex] , I do get the right answer. But this is not true is it? this would only be true if the masses are not accelerating? is the angular momentum wrong ?

    Where do I go wrong? thank you!
     
    Last edited: Mar 31, 2012
  2. jcsd
  3. Mar 31, 2012 #2

    I like Serena

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    Welcome to PF! :smile:

    I've included your picture for reference.

    200px-Atwood.svg.png


    You've treated the system as a whole.
    This means you should only look at the external forces.

    The external force applied on the left side is the weight of m1, which is m1g.
    And on the right side you have m2g.
    So the sum of all moments is:
    $$\Sigma \tau = m_1g R - m_2g R$$
    You have looked at the internal forces, but doing so, you are effectively counting things double.
     
  4. Mar 31, 2012 #3
    Oh that makes sense! So if I would not look at the whole system but just the pulley, I would get the right answer too if I say:
    [tex] Ʃ \tau = T_2R-T_1R [/tex]
    But then L would be different
    $$L= I \cdot \frac{v}{R} $$

    That would be right too?
    Thanks!
     
  5. Mar 31, 2012 #4

    I like Serena

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  6. Mar 31, 2012 #5
    thanks alot
     
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