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 Incoming links: Hoop with a suspended counterweight

Use the angle orientation in the PDF:

M

has coordinates

(R sin θ, - R cos θ)

relative to the hoop centre. The wrapped string segment has length

R (π /2 - θ)

θ

increases, it releases length

Rθ

, so the suspended mass moves down by

Rθ

. Both speeds have magnitude

R ∣ \dot{θ} ∣

T = \frac{1}{2} (M + m) R^{2} \dot{θ}^{2}, V = - M g R cos θ - m g Rθ + constant,

and

L = T - V

(M + m) R \ddot{θ} = g (m - M sin θ) .

At equilibrium,

θ_{0} = arcsin (m / M), 0 < θ_{0} < π /2.

The positive curvature

V^{''} (θ_{0}) = M g R cos θ_{0}

establishes stability. Put

θ = θ_{0} + δ

and linearize. The equilibrium force cancels, leaving

\ddot{δ} + ω^{2} δ = 0, ω^{2} = \frac{M g c o s θ _{0}}{( M + m ) R} = \frac{g}{R} \frac{M - m}{M + m} .

Hence

ω = (\frac{M - m}{M + m})^{1/4} \frac{g}{R} .

For

M ≫ m

θ_{0} ≃ m / M

and

ω \to g / R

R

. The linearization assumes an amplitude small enough to remain inside the stated angular range and a taut string throughout.