Problem source

A horizontal spindle rotates freely in a fixed bearing. 
Three light rods are each attached by one end to the spindle
so that they rotate in a vertical plane.
A particle of mass $m$ is fixed to the other   end 
of each of the three rods.
The rods 
 have lengths $a$, $b$ and $c$, with  $a > b > c\,$ 
and the angle between any pair of rods is
$\frac23 \pi$.
The angle between the rod of length $a$ and the vertical  is
$\theta$, 
as shown in the diagram.
\vspace*{-0.1in}
\begin{center}
\psset{xunit=0.45cm,yunit=0.45cm,algebraic=true,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25}
\begin{pspicture*}(-6.49,-3.44)(6.06,9.38)
\psline[linestyle=dashed,dash=1pt 1pt](0,9.15)(0,-3.05)
\psline(0,2.51)(2.41,8.3)
\psline(0,2.51)(3.89,-0.76)
\psline(0,2.51)(-4.39,0.87)
\parametricplot{-2.7855695569416454}{1.176155335856138}{1*1.77*cos(t)+0*1.77*sin(t)+0|0*1.77*cos(t)+1*1.77*sin(t)+2.51}
\parametricplot{1.1701030633139027}{1.5707963267948966}{1*2.47*cos(t)+0*2.47*sin(t)+0|0*2.47*cos(t)+1*2.47*sin(t)+2.54}
\rput[tl](0.08,4.53){$\theta$}
\rput[tl](0.59,3.24){$\frac{2}{3}\pi$}
\rput[tl](-0.46,2.08){$\frac{2}{3}\pi$}
\rput[tl](1.56,6.08){$a$}
\rput[tl](2.57,1.3){$b$}
\rput[tl](-2.8,2.31){$c$}
\begin{scriptsize}
\psdots[dotsize=6pt 0,dotstyle=*](2.41,8.3)
\psdots[dotsize=6pt 0,dotstyle=*](3.89,-0.76)
\psdots[dotsize=6pt 0,dotstyle=*](-4.39,0.87)
\end{scriptsize}
\end{pspicture*}
\end{center}

 
Find an expression for the energy of the system and 
show that, if the system is in equilibrium, then 
\[ 
\tan \theta = -\frac{(b-c) \sqrt{3}}{2a-b-c}\;. 
\] 
Deduce that there are exactly two equilibrium positions 
and determine
which of the two equilibrium positions is stable. 
 
Show that, for the system to make complete revolutions, 
it must pass through its position of stable equilibrium 
with an angular velocity of at least 
\[ 
\sqrt{\frac{4gR}{a^2+b^2+c^2}} \, , \]
where $2R^2 =  (a-b)^2+(b-c)^2 +(c-a)^2 \;$.