Consider a thin ring of mass M and radius R. (Here thin means that it is a line.) The linear mass density of the ring is ρ=M/(2πR). The moment of inertia I₁ for spinning about an axis perpendicular to the plane of the ring which passes through the ring center is MR². There is another spin of a ring which is like the flipping of a coin. This is a rotation about an axis which is a diameter of the ring. The moment of inertia for this type of spinning is given by

I₂ = ∫(R·cos(θ))²ρRdθ = ρR³∫cos²(θ)dθ

where the integration is from 0 to 2π. The integral of cos²(θ) from 0 to π is ½ so the integral from 0 to 2π is 1 and hence

I₂ = ρR³ = (2πRρ)R²/(2π) = (1/(2π))MR².

This flipping rotation is of particular interest for the structure of nuclei. Experimental measurement indicate that that nuclei have at least approximately spherical shapes. If a circular band of nucleons rotates in this flipping fashion the dynamic appearance of the nucleus would be that of a sphere.

A spin can also take place perpendicular to the diameter considered above. The moment of inertia about that axis is same as that of I₂. It unnecessarily complicates the analysis and will not be included now.

The angular momentum of the ring spinning at an angular rate of ω about the axis perpendicular to its plane is L=MR(Rω)=MR²ω. This means that

ω = L/(MR²)
and hence
½I₁ω² = ½MR²(L²/(MR²)² = L²/(2MR²)

For the spin about a diameter the angular momentum Λ is found as follows.

Λ = ∫ (R·cos(θ))(R·cos(θ)Ω)ρdθ = R²ρΩ∫cos²(θ)dθ
with the integration being from 0 to 2π
which reduces to
Λ = R²ρΩ = R²(M/(2πR)Ω = MRΩ/(2π)
and hence
Ω = 2πΛ/(MR)

Thus

½I₂Ω² = ½((1/(2π))MR²)(2πΛ/(MR))² = Λ²/(4πMR)

The kinetic energy of the spinning ring is then

K = L²/(2MR²) + Λ²/(4πMR)

and there is no potential energy. This means that the ring will continue spinning at the rates ω and Ω indefinitely. These rates can have any real values. This is the classical solution.

The Quantum Mechanical Analysis

To get the quantum mechanical solution the energy is expressed as

E = ½I₁(dθ/dt)² + ½I₂(dφ/dt)²

The momentum associated with θ is (∂E/∂θ) which is I₁(dθ/dt). Let this be denoted as p_θ. Thus (dθ/dt)=p_θ/I₁. Likewise the momentum associated with φ is p_φ=(dφ/dφ)/I₂. Therefore the Hamiltonian function for the the spinning ring is

H = p_θ/(2I₁) + p_φ/(2I₂)

and the Hamiltonian operator is

H^ = −h²(∂²/∂θ²) −h²(∂²/∂φ²)
and the time-independent
Schrödinger equation is
−h²(∂²ψ/∂θ²) −h²(∂²ψ/∂φ²) = Eψ

If it is assumed that the wave function ψ is of the form Θ(θ)Φ(φ) then above equation becomes

−h²Θ"(θ)Φ(φ) −h²Θ(θ)Φ"(φ) = EΘ"(θ)Φ(φ)
which upon division
by Θ(θ)Φ(φ) gives
−h²Θ"/Θ −h²Φ"/Φ = E

This equation can be expressed as

−Θ"/Θ = Φ"/Φ + E/h²

The LHS is independent of φ and the RHS independent of θ. Therefore their common value must be a constant, say k². This means that

Θ"(θ) + k²Θ(θ) = 0

The solution is

Θ(θ) = A·cos(k(θ+θ₀)

where A and θ₀ are constants. By proper choice of the coordinate system θ₀ can be made equal to zero. Then k(2π) must be an integral multiple of 2π. Therefore k must be an integer.

The other equation is

Φ"/Φ + E/h² = k²
or, equivalently
Φ" + (E/h² − k²)Φ = 0

From the previous case it is found that the coefficient of Φ must be a squared integer; i.e.,

(E/h² − k²) = q²
or, equivalently
E/h² = k² + q²

Thus the energy of the spinning ring is quantized such that it is equal to a multiple of of the sum of squared integers. The wave function then has the form

ψ(θ, φ) = cos(kθ)cos(qφ)
for 0≤θ≤2π
and 0≤φ≤2π

This means that the probability density function P(θ, φ) is given by

P(θ, φ) = cos²(kθ)cos²(qφ)

When a second rotation axis is included the quantization condition for energy involves the sum of the squares of three integers.