- #1

Trenthan

- 54

- 0

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

q

I

'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]

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 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

q

_{1}and q_{2}are generalised coordinatesI

_{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

[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 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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