Derivatives and Partial Derivatives

[Trenthan]

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
I_ZG1 is simply the moment of inertia
'T' being Kinetic energy

For those who don't 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

$$<br /> T = \frac{1}{2}m_{1}c^{2}\dot{q_{1}}^2 + \frac{1}{2}I_{ZG1}\dot{q_{1}}^{2} + \frac{1}{2}m_{2}(\dot{q_{2}}^{2}+\dot{q_{1}}^{2}q_{2}^{2})+\frac{1}{2}I_{ZG2}\dot{q_{1}}^{2}<br />$$


$$<br /> \\ \frac{\partial T}{\partial \dot{q_{1}}} = m_{1}c^{2}\dot{q_{1}} + I_{ZG1}\dot{q_{1}}+ m_{2}q_{2}^{2}\dot{q_{1}}+ I_{ZG2}\dot{q_{1}}<br />$$



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






Mine(below this), so unless I am mistaken it should have
$$\dot{q_{2}}$$
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. I've gone back to my textbook however i don't have anything similar. Everything is linear so no 'squared' stuff

Cheers Trent, and thanks in advance

 

[Petr Mugver]

Your version is the correct one.

 

[Trenthan]

wohhh sorry i didnt make it clear

His version is the top

Mine is the lower one****

So you referring to the top or bot

Soz lolz

 

[Petr Mugver]

Bottom. I guess you could have guessed this answer!

 

[angadaria]

i agree with..the bottom equation is correct..