- #1
garethpriede
- 6
- 0
Hi, hopefully someone can check my logic.
I have a lever with a mass on top which is rising towards the vertical on a frictionless pivot. The length of the lever can change. (In reality this is a robot rocking onto a foot and straightening its leg). The intention is to bring the lever to a halt at the vertical by continuously controlling the length.
I've been trying to use the Langrangian, but I'm not sure if I've captured the variables correctly.
x is the length of the leg, theta is the angle from the vertical (so 0 is upright). (I hope my latex paste comes out ok)
L=T-V \\
L=\left[\frac{1}{2}m\omega^{2}+\frac{1}{2}m\dot{x}^{2}\right]-mgxcos(\theta) \\
L=\frac{1}{2}m(x^{2}\dot{\theta}^{2}+\dot{x}^{2})-mgxcos(\theta)\\
\frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}}=\frac{\partial L}{\partial\theta}\\
m\dot{x}^{2}\ddot{\theta}=mgxsin(\theta)\\
\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}=\frac{\partial L}{\partial x}\\
m\ddot{x}=mx\dot{\theta}^{2}+mgsin(\theta)
What worries me is the first partial derivative, something seems to be missing. Do I need to account for the energy involved in extending the leg?
Then if you don't mind a mini part 2, if the pivot has friction, can I still use the Lagrangian (I assume not, but perhaps it's just an extra coefficient somewhere)
I have a lever with a mass on top which is rising towards the vertical on a frictionless pivot. The length of the lever can change. (In reality this is a robot rocking onto a foot and straightening its leg). The intention is to bring the lever to a halt at the vertical by continuously controlling the length.
I've been trying to use the Langrangian, but I'm not sure if I've captured the variables correctly.
x is the length of the leg, theta is the angle from the vertical (so 0 is upright). (I hope my latex paste comes out ok)
L=T-V \\
L=\left[\frac{1}{2}m\omega^{2}+\frac{1}{2}m\dot{x}^{2}\right]-mgxcos(\theta) \\
L=\frac{1}{2}m(x^{2}\dot{\theta}^{2}+\dot{x}^{2})-mgxcos(\theta)\\
\frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}}=\frac{\partial L}{\partial\theta}\\
m\dot{x}^{2}\ddot{\theta}=mgxsin(\theta)\\
\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}=\frac{\partial L}{\partial x}\\
m\ddot{x}=mx\dot{\theta}^{2}+mgsin(\theta)
What worries me is the first partial derivative, something seems to be missing. Do I need to account for the energy involved in extending the leg?
Then if you don't mind a mini part 2, if the pivot has friction, can I still use the Lagrangian (I assume not, but perhaps it's just an extra coefficient somewhere)