# Cart Pole kinetic energy

As explained in [1], the kinetic energy of a Cart Pole is:

$$\frac{1}{2} (M+m)\dot x^2 + \frac{1}{2} m L^2 \dot \theta^2 - m L cos(\theta) \dot \theta \dot x$$

Where $$m$$ is the mass at the tip of the pole, $$M$$ the mass of the cart, $$L$$ the length of the pole, $$x$$ the position of the cart and $$\theta$$ the angle of the pole.

I understand the first term as being the linear kinetic energy, and the second as being the rotational kinetic energy, but can you give me an intuition on the last term ($$- m L cos(\theta) \dot \theta \dot x$$) ?

• The position vector of the pendulum mass is$\vec r = \begin{pmatrix}x_p\\y_p\end{pmatrix} = \begin{pmatrix}-L \sin\theta + x\\L\cos\theta\end{pmatrix},$
– Eli
May 10, 2022 at 6:40
• Thus $T=\dfrac{1}{2}m\overrightarrow{v}\cdot \overrightarrow{v}+\frac 12M\dot x^2$
– Eli
May 10, 2022 at 6:50

Let us start with an example first. Consider a pendulum with constantly accelerated support (instead of the support being a degree of freedom of the system). The position of said pendulum can be parameterized via $$\vec r = \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}-L \sin\theta + \frac 1 2 at^2\\L\cos\theta\end{pmatrix},$$ so the velocity vector is given via $$\dot{\vec r} = \begin{pmatrix}-L \cos\theta\ \dot\theta + at\\ -L\sin\theta\ \dot\theta\end{pmatrix}.$$ The kinetic energy is hence given via $$T = \frac 1 2 m \dot{\vec r} ^2 = \frac 1 2 m(L^2\dot\theta^2 - 2 at\, L\ cos(\theta)\,\dot\theta + a^2 t^2)\ .$$ Using the Lagrange formalism to obtain the equations of motion the last term does not contribute, since it is a total derivative. The middle term resembles the term you are wondering about and will result in the fictitious force expected in the equations of motion, due to the accelerated frame of reference. For an interpretation closer to the concept of energy, this term quantifies how energy gets pumped into the system by the external forcing/acceleration.