= Solution
The spring length is $a+x$, its tension is $kx$, and the orbit radius is
$$
r=L+(a+x)\sin\phi_2.
$$
Vertical balance first gives
$$
kx\cos\phi_2=mg,
\qquad
\boxed{x=\frac{mg}{k\cos\phi_2}}.
$$
Radial balance is
$$
kx\sin\phi_2=m\omega^2r.
$$
Using the vertical equation,
$$
\boxed{
\omega=\left[
\frac{g\tan\phi_2}
{L+\left(a+\dfrac{mg}{k\cos\phi_2}\right)\sin\phi_2}
\right]^{1/2}}.
$$
Unlike the rigid-rod result, this depends on $m$: a larger mass stretches the spring farther and changes the radius of the orbit.
Back to article page
