ey guys(adsbygoogle = window.adsbygoogle || []).push({});

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

q_{1}and q_{2}are generalised coordinates

I_{ZG1}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

[tex]

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}

[/tex]

[tex]

\\ \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}}

[/tex]

His version

[tex]

\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}}

[/tex]

Mine(below this), so unless im mistaken it should have

[tex]

\dot{q_{2}}

[/tex]

in the third term????

[tex]

\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}}

[/tex]

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

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# Derivatives and Partial Derivatives

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