Example: The Foucault Pendulum

A heavy weight on a long cable is suspended from a mount that is free to swing in any direction with very small friction. This is a Foucault pendulum. Its equilibrium position is along the line of a plumb bob, in the direction of effective gravity. With the long cable, the oscillations are small, so the velocity is horizontal and we do everything only to first order in ${r\over\ell}$.

Its natural to put the origin of coordinates at the equilibrium point, the z axis up, the y axis north, and the x axis east. The motion is in the $xy$ plane.

$$\vec{F}_{eff}=m\vec{g}+\vec{T}-2m\vec{\omega}\times\vec{v}$$

$$\vec{a}=\vec{g}+{\vec{T}\over m}-2\vec{\omega}\times\vec{v}$$

$$T\approx T\left(\hat{e}_z-{\vec{r}\over \ell}\right)$$

$$T=mg$$

$$\vec{a}=-\vec{g}+\vec{g}-{g\over\ell}\vec{r}-2\vec{\omega}\times\vec{v}$$

$$\vec{\omega}=\omega\left(\cos\lambda\hat{e}_y+\sin\lambda\hat{e}_z\right)$$

$$\ddot{\vec{r}}=-2\vec{\omega}\times\vec{v}-{g\over\ell}\vec{r}$$

$$\ddot{\vec{r}}=-2\omega\left(-\cos\lambda v_x\hat{e}_z+\sin\lambda v_x\hat{e}_y-\sin\lambda v_y\hat{e}_x\right) -{g\over\ell}\vec{r}$$

$$\ddot{\vec{r}}=-2\omega\sin\lambda\left(v_x\hat{e}_y-v_y\hat{e}_x\right) -{g\over\ell}\vec{r}$$

$$\ddot{x}= -{g\over\ell}x+2\omega\sin\lambda\dot{y}$$

$$\ddot{y}= -{g\over\ell}y-2\omega\sin\lambda\dot{x}$$

$$q=x+iy$$

We add the $x$ equation plus $i$ times the $y$ equation.

$$\ddot{q}+{g\over\ell}q=-2i\omega\sin\lambda\dot{q}=-2i\omega_z\dot{q}$$

$$q=q_0e^{iat}$$

$$-a^2+{g\over\ell}=-2i\omega_z(ia)$$

$$a^2+2\omega_za-{g\over\ell}=0$$

$$a={-2\omega_z\pm\sqrt{4\omega_z^2+4g/\ell}\over 2}$$

$$a=-\omega_z\pm\sqrt{\omega_z^2+g/\ell}$$

$$g/\ell >> \omega_z^2$$

$$a=-\omega_z\pm\sqrt{g/\ell}$$

$$q(t)=q_0e^{-i\omega_z t}\left[A e^{i\sqrt{g/\ell}t} + B e^{-i\sqrt{g/\ell}t} \right]$$

Pick the initial conditions on the pendulum to give cosine for the sum of exponentials.

$$x+iy=q=q_0e^{-i\omega_z t}\cos\left(\sqrt{g\over\ell}t\right)$$

$$x=x_0\cos(\omega_z t)\cos\left(\sqrt{g\over\ell}t\right)$$

$$y=-x_0\sin(\omega_z t)\cos\left(\sqrt{g\over\ell}t\right)$$

The pendulum precesses with an angular frequency $\omega_z=\omega\sin\lambda$.