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.
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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 \(2R^2 = (a-b)^2+(b-c)^2 +(c-a)^2 \;\).