working on some exam practice questions, and just needed a bit of help to check my answer to the last part of this question:
A particle of unit mass is projected around the inner surface of an upright circular cone with its vertex down. The equation of the conical surface is $x^2 + y^2 = \frac{z^2}{a^2}$
(i) Find the Lagrangian for the particle in terms of the cylindrical coordinates $r$ and $\theta$
(ii) Find two integrals of the motion.
(iii) If the angular momentum is $h = \sqrt{2ga}$ and the energy is $E = 2ga$, show that $(1 + a^2)\dot{r^2} + 2ga(r − 2 +\frac{1}{r^2} ) = 0 $
(iv) Using the factorisation $(x^3 − 2x^2 + 1) = (x − 1)(x^2 − x − > 1)$, or otherwise, show that the particle moves between two fixed limits in $r$ and find those limits.
I am comfortable that I have down the first 3 parts correctly and just wanted to check part (iv):
The turning points in the radius occur at $\dot{r}=0$,
so from the part (iii), I have:
$$\frac{2ga(r-1)(r^2-r-1)}{r^2} = 0 $$
$$\implies (r-1)(r^2-r-1) = 0 $$
So we get that $r=1$ and $r = \frac{1\pm \sqrt{5}}{2}$, but since $r > 0$, I just take the positive.
Would this be correct?
Thanks so much.