For any frame $S$ and vector $\mathbf{A}$, let $\left[\frac{d \mathbf{A}}{d t}\right]_{S}$ denote the derivative of $\mathbf{A}$ relative to $S$. A frame of reference $S^{\prime}$ rotates with constant angular velocity $\omega$ with respect to an inertial frame $S$ and the two frames have a common origin $O$. [You may assume that for any vector $\left.\mathbf{A},\left[\frac{d \mathbf{A}}{d t}\right]_{S}=\left[\frac{d \mathbf{A}}{d t}\right]_{S^{\prime}}+\boldsymbol{\omega} \times \mathbf{A} .\right]$

(a) If $\mathbf{r}(t)$ is the position vector of a point $P$ from $O$, show that

$$\left[\frac{d^{2} \mathbf{r}}{d t^{2}}\right]_{S}=\left[\frac{d^{2} \mathbf{r}}{d t^{2}}\right]_{S^{\prime}}+2 \boldsymbol{\omega} \times \mathbf{v}^{\prime}+\boldsymbol{\omega} \times(\boldsymbol{\omega} \times \mathbf{r})$$

where $\mathbf{v}^{\prime}=\left[\frac{d \mathbf{r}}{d t}\right]_{S^{\prime}}$ is the velocity in $S^{\prime}$.

Suppose now that $\mathbf{r}(t)$ is the position vector of a particle of mass $m$ moving under a conservative force $\mathbf{F}=-\nabla \phi$ and a force $\mathbf{G}$ that is always orthogonal to the velocity $\mathbf{v}^{\prime}$ in $S^{\prime}$. Show that the quantity

$$E=\frac{1}{2} m \mathbf{v}^{\prime} \cdot \mathbf{v}^{\prime}+\phi-\frac{m}{2}(\boldsymbol{\omega} \times \mathbf{r}) \cdot(\boldsymbol{\omega} \times \mathbf{r})$$

is a constant of the motion. [You may assume that $\nabla[(\boldsymbol{\omega} \times \mathbf{r}) \cdot(\boldsymbol{\omega} \times \mathbf{r})]=-2 \boldsymbol{\omega} \times(\boldsymbol{\omega} \times \mathbf{r})$.]

(b) A bead slides on a frictionless circular hoop of radius $a$ which is forced to rotate with constant angular speed $\omega$ about a vertical diameter. Let $\theta(t)$ denote the angle between the line from the centre of the hoop to the bead and the downward vertical. Using the results of (a), or otherwise, show that

$$\ddot{\theta}+\left(\frac{g}{a}-\omega^{2} \cos \theta\right) \sin \theta=0 .$$

Deduce that if $\omega^{2}>g / a$ there are two equilibrium positions $\theta=\theta_{0}$ off the axis of rotation, and show that these are stable equilibria.