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Derivatives and Partial Derivatives

  1. Sep 30, 2010 #1
    ey guys

    Generally i just do these without thinking, however i was checking some work today with a friend and he is adament i did my derivative wrong....

    If i can double check with you

    Well firstly
    'c' is simply a constant
    q1 and q2 are generalised coordinates
    IZG1 is simply the moment of inertia
    'T' being Kinetic energy

    For those who dont notice what this is. Its apart of Lagrange's equation during the formation of the Equation of Motion

    EDIT**, i should mention that q1 and q2 are both functions of time


    T = \frac{1}{2}m_{1}c^{2}\dot{q_{1}}^2 + \frac{1}{2}I_{ZG1}\dot{q_{1}}^{2} + \frac{1}{2}m_{2}(\dot{q_{2}}^{2}+\dot{q_{1}}^{2}q_{2}^{2})+\frac{1}{2}I_{ZG2}\dot{q_{1}}^{2}



    \\ \frac{\partial T}{\partial \dot{q_{1}}} = m_{1}c^{2}\dot{q_{1}} + I_{ZG1}\dot{q_{1}}+ m_{2}q_{2}^{2}\dot{q_{1}}+ I_{ZG2}\dot{q_{1}}


    His version
    \frac{d}{dt}\left ( \frac{\partial T}{\partial \dot{q_{1}}} \right ) = m_{1}c^{2}\ddot{q_{1}}+I_{ZG1}\ddot{q_{1}} + m_{2}\ddot{q_{1}}q_{2}^{2} + 2m_{2}q_{2}\dot{q_{1}} + I_{ZG2}\ddot{q_{1}}


    Mine(below this), so unless im mistaken it should have
    in the third term????

    \frac{d}{dt}\left ( \frac{\partial T}{\partial \dot{q_{1}}} \right ) = m_{1}c^{2}\ddot{q_{1}}+I_{ZG1}\ddot{q_{1}} + m_{2}\ddot{q_{1}}q_{2}^{2} + 2m_{2}q_{2}\dot{q_{2}}\dot{q_{1}} + I_{ZG2}\ddot{q_{1}}

    Now if someone can clarify which version is correct big help. Ive gone back to my text book however i dont have anything similar. Everything is linear so no 'squared' stuff

    Cheers Trent, and thanks in advance
    Last edited: Sep 30, 2010
  2. jcsd
  3. Sep 30, 2010 #2
    Your version is the correct one.
  4. Sep 30, 2010 #3
    wohhh sorry i didnt make it clear

    His version is the top

    Mine is the lower one****

    So you reffering to the top or bot

    Soz lolz
  5. Sep 30, 2010 #4
    Bottom. I guess you could have guessed this answer!
  6. Mar 23, 2011 #5
    i agree with..the bottom equation is correct..
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