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Ring pendulum minimum time period

Short answer

Moment of inertia about pivot: I = M(R²+x²). Period T = 2π√((R²+x²)/(gx)). Minimizing T² wrt x gives d(T²)/dx = 0 → -R²/x²+1=0 → x = R. Second derivative positive, so this is minimum. Minimum period: T_min = 2π√(2R/g).

A heavy circular ring of radius R oscillates in a vertical plane about a horizontal axis at a distance x from the center. Show that the time period is minimum when x = R.

A ring of mass M and radius R swings about a horizontal axis at distance x from its centre G (the axis is parallel to the ring's axis).

        O  pivot
        |  x
        |
      .-G-.   ring, radius R
     (     )
      '---'

Moment of inertia

  1. About the axis through G (perpendicular to the plane of the ring): I_G = MR², so k² = R².
  2. About the pivot (parallel-axis theorem): I = MR² + Mx².

Time period

For small oscillations, torque = −Mgx θ, so I θ'' = −Mgx θ:

T = 2π √(I/(Mgx)) = 2π √((R² + x²)/(g x))

Minimum

T² = (4π²/g)(R²/x + x). Set d(T²)/dx = 0:

−R²/x² + 1 = 0   →   x² = R²   →   x = R

d²(T²)/dx² = 2R²/x³ > 0, so this is a minimum.

Minimum period: T_min = 2π√(2R/g). It occurs when the axis is at the rim of the ring (x = R), i.e. when the distance from the centre equals the radius of gyration k = R.

✓ Answer checked by Claude Opus 5.5 (high effort)

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