# Tension of a Spinning Ring

fun with physics

What is the tension in a spinning ring, for example Bank's orbitals or a Niven Ring?

Let us break the ring into N sections. The angle (in radians) subtended by each section, dθ, is 2π / N. Suppose the radius of the ring is R and the mass-per-unit-length of the ring is λ; then the mass of a section is λ R dθ and the mass of the whole ring M = 2π_R_λ.

Ignoring gravity, there are two forces pulling on each small section: tension pulling to the left and tension pulling to the right. Because the ring is circularly symmetric, the magnitude of the two tensions are equal. The directions of the two forces are dθ from being opposite.

If at some moment in time, the left tension, Tl, is pointing in the - direction, then the right tension will be

Tr = T sin(dθ) ŷ + T cos(dθ)

where T is the magnitude of either tension.

As N gets large: dθ gets small, cos(dθ) → 1, and sin(dθ) → dθ. So:

Tl = -T

Tr = +T + Tŷ

Consequently, the centripetal force on the small section is _T_dθ. From Newton's second law:

F = m a

Tl + Tr = m a

Tŷ = m a

Let a = |a|, the centripetal acceleration.

T dθ = (λ R dθ) a

T = λ R a

Part B: Plug in some numbers.

Let S be the specific strength, which is the maximum tension divided by density: T / λ.

SR a

RS / a

Let P be the period of revolution. Since centrepetal acceleration, a = v² / R.

a = v² / R

a = (2π R / P)² / R

a = 4π² R / _P_²

a _P_² = 4π² R

a _P_² ≤ 4π² (S / a)

_P_² ≤ 4π² S / _a_²

P ≤ 2π √(S) / a

Assume our desired centripetal acceleration is an Earthly 10 m/s².

If the ring is made of solid steel, S = 154000 N·m/kg (specific strength). With a = 10 m/s², we find that the maximum radius for a ring of steel is 15 kilometers. Period of revolution = 243 seconds = 4 minutes. If you assume that half of the mass of your ring is structural material and the other half is nonstructural, then you get a diameter of 15 km. Multiply the period by 1/√2.