# Euler force for pendulum

Hello I have a question related to the Euler force. Why is this force never considered for a simple pendulum?

As far as I understand, Euler force is given by (assume I would consider the 2d pendulum in a 3D space, that the quantities are vectors) $$$$\boldsymbol{\dot{\omega}} \times \mathbf{r}$$$$ This means for the force to vanish, $$\boldsymbol{\dot{\omega}} = 0$$, or $$\mathbf{r} = 0$$, or the vectors must point in the same direction. I do not see why one of these conditions is satisfied.

• What do you think $r$ and $\omega$ are here? If you think it is the length of the pendulum and the angular velocity of the pendulum respectively as you look at it from an inertial reference frame, then that is where your mistake lies. You will probably find your answer by truly understanding what $r$ and $\omega$ represent in this equation. – Aaron Stevens Jun 19 at 17:52
• I think you are entirely right. This is what I am searching for. – Q.stion Jun 19 at 19:14

It is an inertial pseudo-force arising when you are in a non galilean frame. $$\omega$$ is then the radial velocity of your reference frame relative to a galilean frame. Here it is not the case.
• I have a question to this. In the standard textbooks when deriving all these forces, $\omega$ is the same in the centifugal force and euler force. So it is the same compensated by the centripetal force. So it is there, right? – Q.stion Jun 19 at 19:16
• Centrifugal force is radial and proportional to $\omega$; Euler force is tangential and proportional to time derivative of $\omega$. – Matt Jun 19 at 20:40