# Foucault pendulum coordinate representation

I have a homework question that boils down to showing that plane of oscillation for a Foucault pendulum rotates 2$$\pi\cos(\theta)$$ in a day, where $$\theta$$ is the colatitude angle. I have seen an approach on Wikipedia: https://en.wikipedia.org/wiki/Foucault_pendulum#Precession_as_a_form_of_parallel_transport

However, I don't seem to get why does the pendulum in Wikipedia's case has an x and y acceleration. Below is the picture of the system I have used.

$\theta$ is the colatitude angle.">

The red line is the pendulum, x-axis points into the plane.

Now I have set my plane of oscillations to be in the yz-plane so

$$\vec{R} = (0,R\sin(\phi),R\cos(\phi)).$$

Next projecting $$\vec{\omega}$$ onto the axes gives

$$\vec{\omega} = (0,\omega\sin(\phi),\omega\cos(\phi))\implies \vec{v} = (0,\omega\dot{\phi}\cos(\phi),-\omega\dot{\phi}\sin(\phi)).$$

Now the Coriolis acceleration is given by $$\vec{a}_c = -2m(\vec{w}\times\vec{v})= \omega R \phi(\cos(\theta+\phi),0,0).$$ This is where I differ from the wikipedia article. After this I go on to to use the small angle approxtimation to get $$\phi(t) = A\cos(\sqrt{g/R}t)$$. Which gives a bad expression. My goal was to find the equations of motion for the x coordinate and then look at $$\vec{R}$$ with that x coordintate to see if I could see some sort rotation.

In conclusion, my problem is that I don't get why there is an $$v_x$$ and $$v_y$$ component in the Wikipedia article.

• Spot checking $\omega \times v$, using the $\omega$ and the $v$ vectors, the cross product is incorrect since both vectors have a $0$ is the $\hat i$ position - which implies the cross product has a $0$ in the $\hat i$ and $j$ positions. The only component which would survive cross product is $\hat k$ component. – Cinaed Simson Dec 10 '19 at 21:11
• The above comment is wrong - I should have calculated it. At this point in the calculation, a trig approximation won't generate a $\hat j$ component. So there's something wrong with your $v$. Don't set $v=d\omega/dt$. Instead, write $v=(v_{e},v_{n},v_{u})$ as in the "en.wikipedia.org/wiki/Coriolis_force" article. Then take limiting cases and see it makes sense, e.g., project it onto the $xy$ plane and set $\phi=0$ (the equator). – Cinaed Simson Dec 10 '19 at 22:49