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Atwood's machine with two connected discs

  1. Sep 3, 2016 #1
    1. The problem statement, all variables and given/known data
    The system looks like this:
    image.jpg
    I have two discs which are connected.
    Disc 1 has ##R_1##(radius) and ##M_1##(mass)
    Disc 2 has ##R_2## and ##M_2##
    ## R_2 > R_1 ##
    ## M_2 > M_1 ##

    on both discs weights are attached on opposite sides.
    On smaller ##m_1## is pulling and on bigger ##m_2##

    ##m_2 > m_1##

    i need to calculate angular acceleration and accelerations of both weights


    2. Relevant equations

    ## F = m * a ##
    ## \alpha = \frac{a}{R} ##
    ## M = I * \alpha ##
    ## Q = m * g ##

    3. The attempt at a solution

    The force that would cause acceleration of system is equal to :

    ## F = Q_2 - Q_1 ##
    ## F = m_2*g - m_1*g ##

    The force that will cause tangential acceleration of discs would be equal to :

    ## I * \alpha = M ##
    ## \frac{1}{2} m * R^2 * \frac{a}{R} = F * R ##
    ## \frac{1}{2}*m * a = F ##



    Now we know that Disc 2 and ##m_2## will have the same accelerations and Disc 1 and ##m_1## will have the same accelerations. We also now that Disc 1 and Disc 2 also have same angular acceleration. So from that i thought i can write this:

    ## a_1*(\frac{1}{2} * M_1 + m_1) + a_2(\frac{1}{2}*M_2 + m_2) = F ## (force that cause acceleration)

    ## a = \alpha * R##

    ##\alpha * R_1 *(\frac{1}{2} * M_1 + m_1) + \alpha * R_2 *(\frac{1}{2}*M_2 + m_2) = F##

    and i have everything to get an alpha but its wrong

    if somebody solve this on different way, could please explain why is my approach wrong
     
    Last edited: Sep 3, 2016
  2. jcsd
  3. Sep 3, 2016 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Can you explain how you derived this?

    I suggest you set up three force equations: one for each hanging mass and one for the disk. You can combine those equations, adding what you know about the relationships between the angular and linear accelerations.
     
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