The velocity is then \[ \mathbf{v}(t) = \mathbf{\dot{r}}(t) = [\dot{r}\cos\theta\sin\alpha - \omega r\sin\theta\sin\alpha, \dot{r}\sin\theta\sin\alpha + \omega r\cos\theta \sin\alpha, \dot{r}\cos\alpha]. \]

We can then compute the square \[ |\mathbf{v}|^2 = \dot{r}^2 + (r\omega \sin\alpha)^2. \]

The Lagrangian is then given by \[ \mathcal{L} = T - V = \frac{1}{2}m \left[ \dot{r}^2 + (r\omega\sin\alpha)^2\right] - mgr\cos\alpha. \]

By Hamilton's principle, we now seek to minimize the integral of the Lagrangian. Notice that $\pd{\mathcal{L}}{t} = 0$, so we can apply Beltrami's identity, \[ \dot{r}\pd{\mathcal{L}}{\dot{r}} - \mathcal{L} = \text{constant}. \]

This gives \[ \frac{m}{2} \dot{r}^2 - \frac{m}{2}(\omega\sin\alpha)^2 r^2 + m(g\cos\alpha) r = H, \]

where $H$ is constant.

Now there is a nice way to simplify this problem through the process of *non-dimensionalisation*. Effectively, we seek to scale the distance, $r$ and the time $t$ so that the problem is simplified. We thus set
\[
r = [r]\overline{r} \quad \text{and} \quad t = [t]\overline{t},
\]

If we do this and simplify, we obtain \[ \overline{\dot{r}}^2 - (\omega\sin\alpha)^2[t]^2 \overline{r}^2 + (2g\cos\alpha)\frac{[t]^2}{[r]} \overline{r} = \frac{2H}{m} \frac{[t]^2}{[r]^2}. \]

Now the trick is to choose \[ [t] = \frac{1}{\omega \sin\alpha} \quad \text{and} \quad [r] = 2g\cos\alpha[t]^2 = \frac{2g\cos\alpha}{\omega^2\sin^2\alpha}, \]

so that the coefficients are simplified. This leaves \[ \overline{\dot{r}}^2 - \overline{r}^2 + \overline{r} = \overline{H}, \]

where $\overline{H}$ is a non-dimensional constant. We now seek to solve the problem in which the particle is launched from the origin at speed $r(t) = \lambda g\cot\alpha/omega$. We can verify that in terms of the non-dimensional quantities \[ \overline{\dot{r}} = \frac{[t]}{[r]} \frac{\lambda g}{\omega} \cot\alpha = \frac{\lambda}{2}. \]

We now apply this initial speed and initial location $\overline{r}(0) = 0$. This gives the resultant problem \[ \overline{\dot{r}}^2 - \overline{r}^2 + \overline{r} = \frac{\lambda}{2}, \]

The maximum height will be achieved at the point where $\dot{r} = 0$. Thus we seek \[ \overline{r}^2 - \overline{r} + \frac{\lambda}{2} = 0 \Rightarrow \left(\overline{r}-\frac{1}{2}\right)^2 + \frac{(\lambda^2-1)}{4} = 0. \]

Therefore, when $\lambda > 1$, there is no point where $\overline{\dot{r}} = 0$, whereas as soon as $\lambda \leq 1$, there is a point where the particle achieves a max. Now consider the case when $\lambda = 1$. We seek to solve \[ \overline{\dot{r}}^2 - \overline{r}^2 + r = \frac{1}{4}, \]

and so \[ \overline{\dot{r}} = \pm\left(r - \frac{1}{2}\right). \]

Since we require that $\overline{r} < 1/2$, we take the negative sign. Solving the equation and applying $\overline{r}(0) = 0$, then gives \[ \overline{r}(\overline{t}) = \frac{1}{2} \left( 1 - e^{-\overline{t}}\right). \]

We can now return to non-dimensionalized values to give \[ r(t) = \frac{g\cos\alpha}{\omega^2\sin^2\alpha} \left(1 - e^{-\omega\sin\alpha t}\right). \]