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Circular disk as physical pendulum

Short answer

Period when pivoted at rim: T = 2π√((3/2)MR²/(MgR)) = 2π√(3R/(2g)) ≈ 0.873 s for R=0.126 m. For a pivot at distance r from centre givng same period, solv r + R²/(2r) = 3R/2 → r = R/2 = 6.3 cm.

A uniform circular disk whose radius R is 12.6 cm is suspended as a physical pendulum from a point on its rim. (a) What is its period? (b) At what radial distance r < R is there a pivot point that gives the same period?

Given: R = 12.6 cm = 0.126 m, g = 9.8 m/s². The disc is pivoted at a point on its rim, with the axis perpendicular to the disc.

(a) Period

  1. Moment of inertia about the centre: I_G = ½MR², so k² = R²/2.
  2. About the rim (parallel-axis theorem): I = ½MR² + MR² = (3/2)MR².
  3. Distance from the pivot to the centre: l = R. For a physical pendulum T = 2π√(I/(Mgl)):
T = 2π √( (3/2) M R² / (M g R) ) = 2π √(3R/2g)
T = 2π √(3 × 0.126 / (2 × 9.8)) = 2π √(0.01929) = 2π × 0.1389

Answer (a): T ≈ 0.873 s.

(b) Pivot at distance r from the centre giving the same period

For a pivot at distance r from the centre, the equivalent simple-pendulum length is L = r + k²/r. It must equal that of the rim pivot, L = 3R/2:

r + R²/(2r) = 3R/2
2r² − 3R r + R² = 0
(2r − R)(r − R) = 0     →   r = R/2  or  r = R

The required point with r < R is r = R/2 = 0.063 m.

Answer (b): r = R/2 = 6.3 cm from the centre. This agrees with interchangeability: the centre of oscillation lies at k²/R = R/2 from the centre, on the side opposite the pivot.

✓ Answer checked by Claude Opus 5.5 (high effort)

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