Show that the equation of motion for a particle on Norton's Dome is $\frac{d^2 r}{dt^2}=r^{1/2}$

Question

A particle sits at the top of a dome, whose height drops away from the centre, with a drop given by $$h=\frac{2r^{3/2}}{3g}$$ where $g$ is the acceleration due to gravity, and $r$ is a coordinate measured radially along the surface from the peak. At $t = 0$ the particle is at rest at the top of the dome.

Show that the equation of motion is $$\frac{d^2 r}{dt^2}=r^{1/2}\tag{1}$$

Resolving the gravitational acceleration along the surface, $g \cos\theta$, where $\theta$ is the angle between the surface and the vertical. A sketch shows that $dh = \cos \theta \,dr$. Hence $$\cos \theta = \frac{dh}{dr}=\frac{r^{1/2}}{g}$$

I am unable to proceed any further. Could someone please explain to me how to obtain equation $(1)$?

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Note: Please do not migrate this to Physics.SE (it's already there). What I am asking here is purely about the mathematics behind equation $(1)$. Thanks.

Answer 1

Consider conservation of energy, which gives $$\frac 12 m\dot{r}^2=mgh$$

Now differentiate this with respect to time so you have $$m\dot{r}\ddot{r}=mg\frac{dh}{dt}$$

You also have $$\frac{dh}{dt}=\frac{r^{\frac 12}\dot{r}}{g}$$ and the result follows immediately

Answer 2

Here is a physics proof: $$\frac{d^2 r}{dt^2}$$ is the acceleration over path $r$. Denote as a.

It is easy to show that $a = g sin(\gamma) = g cos(\theta) = g \frac{r^{1/2}}{g} = r^{1/2}$