1. ## Mechanics Problem

I did
$\displaystyle \tan \theta = \frac{r}{h}$ where r is the radius
so $\displaystyle h \tan \theta = r$

$\displaystyle T \cos \theta = mg$ where T is the tension
$\displaystyle T = \frac{mg}{\cos \theta}$
using the standard circular motion result
$\displaystyle T \sin \ theta = m \omega ^2 r$
$\displaystyle mg \tan \theta = m \omega ^2 h \tan \theta$

giving $\displaystyle \omega^2 = \frac{g}{h}$

where did i go wrong ?

2. Originally Posted by bobak
I did
$\displaystyle \tan \theta = \frac{r}{h}$ where r is the radius
so $\displaystyle h \tan \theta = r$

$\displaystyle T \cos \theta = mg$ where T is the tension
$\displaystyle T = \frac{mg}{\cos \theta}$
using the standard circular motion result
$\displaystyle T \sin \ theta = m \omega ^2 r$
$\displaystyle mg \tan \theta = m \omega ^2 h \tan \theta$

giving $\displaystyle \omega^2 = \frac{g}{h}$

where did i go wrong ?
The only thing wrong is that there are two horizontal forces on the ring: not just a component Tsin(θ) from the RA section of the string but also the tension T in the RB section of the string.

3. Silly me. Thank a lot Opalg.