Imagine a bowl whose shape is given in cylindrical coordinates by $z=c-\frac{1}{2g}\sin^2(r)$ for $r\in[0,\frac{\pi}{2}]$ say. Then take a ball and place it at rest anywhere in this bowl. The energy of the ball is $E = \frac{1}{2}mr^{\prime2}+mgz$, so $\frac{1}{2}r^{\prime2} = \frac{E}{m}-g(c-\frac{1}{2g}\sin^2(r)) = \frac{E}{m}-gc+\frac{1}{2}\sin^2(r)$. So if we now choose $c=\frac{E}{mg}$, we end up with $\frac{1}{2}r^{\prime2} = \frac{1}{2}\sin^2(r)$ and $r^\prime = \sin(r)$.
For this to be proper, you'd have to worry about taking the square root in the end and the point $r=0$ and dimensions working out and stuff, but you get the idea.