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General Atwood system with moment of inertia

  1. Feb 3, 2012 #1
    1. The problem statement, all variables and given/known data

    img1622.png

    I have taken this image from http://farside.ph.utexas.edu/teaching/336k/lectures/node79.html

    Hello, I am currently doing a project on Lagrangian mechanics. As part of my project, it was advised that I should redo each of the problems done using the Lagrangian method in the Newtonian format.

    So essentially I am requesting help on determining the acceleration of the system


    2. Relevant equations

    [itex]F=m\ddot{x}[/itex]

    [itex]τ=Iα[/itex]

    [itex]τ=FR[/itex]

    3. The attempt at a solution

    I am getting increasingly confused especially when the signs don't match :(
    (a = radius of the pulley, not acceleration)
    (acceleration is denoted as [itex]\ddot{x}[/itex]

    [itex]m_{1}\ddot{x_{1}}=T_{1}-m_{1}g[/itex]
    [itex]m_{2}\ddot{x_{2}}=T_{2}-m_{2}g[/itex]

    [itex]τ=Iα[/itex]
    [itex]τ=(T_{1}-T_{2})a[/itex]

    Tangential acceleration of the pulley is the same as the acceleration of the mass, assuming that there is absolute grip of the rope to the pulley.

    [itex]\ddot{x}=αa[/itex]

    However we can see that

    [itex]\ddot{x_{1}}=-\ddot{x_{2}}[/itex]

    Therefore,

    [itex]m_{1}\ddot{x_{1}}=T_{1}-m_{1}g[/itex]
    [itex]m_{2}\ddot{x_{1}}=m_{2}g-T_{2}[/itex]

    and,

    [itex]α=(T_{1}-T_{2})a/I[/itex]
    [itex]\ddot{x}=(T_{1}-T_{2})a^{2}/I[/itex]

    Adding the first 2 equations

    [itex]m_{1}\ddot{x_{1}}+m_{2}\ddot{x_{1}}=T_{1}-T_{2}+m_{2}g-m_{1}g[/itex]

    Replacing the Tensions

    [itex]m_{1}\ddot{x_{1}}+m_{2}\ddot{x_{1}}=\ddot{x}I/a^{2}+m_{2}g-m_{1}g[/itex]

    Finally rearranging it

    [itex](m_{1}+m_{2}-I/a^{2})\ddot{x_{1}}=(m_{2}-m_{1})g[/itex]

    [itex]\ddot{x_{1}}=\frac{(m_{2}-m_{1})g}{(m_{1}+m_{2}-I/a^{2})}[/itex]

    The signs are everywhere and completely different to the derivation obtained through the Lagrangian method. :I Can anyone explain what my issue is?

    EDIT: Included the solution obtained via Lagrangian method

    [itex]\ddot{x_{1}}=\frac{(m_{1}-m{2})g}{m_{1}+m_{2}+I/a^{2})}[/itex]
     
  2. jcsd
  3. Feb 3, 2012 #2
    The minus sign difference in the numerator is not a problem but the minus sign in the detonator of your solution is a problem, the inertia of the pulley should have a plus sign, its inertia tends to reduce the acceleration of the system.
     
  4. Feb 6, 2012 #3
    Thanks for your reply Spinnor,

    I agree the problem is within the negative moment of Inertia in the final acceleration equation.

    However, I can't seem to find a reason to justify it. (Doing so would somehow need to make the initial F=T-mg equation negative or somehow justify T_2 - T_1 instead of T_1 - T_2 for the angular acc. equation)
     
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