So I have been trying to derive the equations of motion of the inverted physical pendulum in a cart, but I seem to be confused about the derivation of its Kinetic Energy. I know this physical system is very popular and while I have searched and searched I couldn't find an answer to my question anywhere.
So I divided the kinetic energy into the cart's and pendulum's:
$$ T = T_C + T_P $$
The cart's one is pretty straight forward $T_C = 1/2 M \dot{x}^2$, where I am denoting $x$ the horizontal coordinate of the cart's point mass.
My trouble is now with the pendulum's Kinetic Energy. I would assume I would have to sum the translational energy of the pivot point $T_{pivot}=1/2 m \dot{x}^2$ to the rotational energy of the pendulum $T_{rot} = 1/2 I \dot{\theta}^2$, where $I$ is the moment of inertia of the pendulum with respect to the pivot point (Note: the angle $\theta$ i chose is with respect to the upper vertical, unlike in the image up there).
With this I got:
$$ \mathcal{L} = \frac{1}{2}(M+m) \dot{x}^2 + \frac{1}{2} I \dot{\theta}^2 – mgl\cos\theta $$
And therefore the equations of motion:
$$ (M+m) \ddot{x} = F(t) $$
$$ I \ddot{\theta} – mgl \sin\theta = 0 $$
These equations, though, seem too simple compared to the equations I have seen out there for this problem.
I would really appreciate if someone could point out my mistakes.
Best Answer
I had the same question and after reading some definitions, I've got the answer: The kinetic energy of a rigid body which has planar motion is always
$$T=T_{Gtranslate}+T_{rotate/G}$$
or
$$T=1/2Mv^2_G+1/2l_G\omega^2$$
where $G$ is the center of mass. So in this pendulum you have to calculate $v_G=\sqrt{\dot{x_G}^2+\dot{y_G}^2}$ and $\omega=\dot{\theta}$ and. Then the kinetic energy will be
$$T=\frac{1}{2}M(\dot{x_G}^2+\dot{y_G}^2)+\frac{1}{2}I_G\dot{\theta}^2 + T_{cart}$$
There is a paper from MIT 2.003SC course which has the same solution: http://bit.ly/PendulumonACart