- #1
Trenthan
- 50
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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 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
[tex] <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 /> [/tex]
[tex] <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 /> [/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}}<br /> [/tex]
Mine(below this), so unless I am 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. 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
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 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
[tex] <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 /> [/tex]
[tex] <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 /> [/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}}<br /> [/tex]
Mine(below this), so unless I am 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. 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
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