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Bar vs torsional pendulum; minimum period

Short answer

Both are angular SHM with isochronous small oscillations. Bar pendulum uses gravitational torque (T=2π√((k²+l²)/gl), torsional uses elastic torque (T=2π√(I/C)). Minimum T occurs when d(T²)/dl = 0 → l = k, so radius of gyration about CG equals distance from suspension to CG.

Point out the similarities and dissimilarities between the oscillations of a bar pendulum and a torsional pendulum. Show that the radius of gyration is equal to the distance from the center of suspension to the center of gravity of a compound pendulum, when the time period is minimum.

Similarities

  • In both, the restoring torque is proportional to the angular displacement, so the motion is angular SHM: I d²θ/dt² = −(constant) θ.
  • Both have T = 2π√(I/restoring torque per unit angle).
  • Both are isochronous: the period does not depend on amplitude (for small amplitude).
  • Both exchange kinetic and potential energy with constant total energy (undamped case).

Dissimilarities

Bar pendulumTorsional pendulum
Axis horizontal; body swings in a vertical planeAxis vertical (wire); body twists in a horizontal plane
Restoring torque from gravity, mglRestoring torque from elasticity of wire, C
T = 2π√((k² + l²)/gl)T = 2π√(I/C)
Depends on g and sinθ ≈ θ (small angles)Independent of g; limit is elastic limit of the wire
Used to find gUsed to find η and I

Minimum period of a compound pendulum

With G the centre of gravity, k the radius of gyration about G and l the distance of the suspension point from G:

T = 2π √((k² + l²)/(g l)) ,   T² = (4π²/g)(k²/l + l)

For minimum T: d(T²)/dl = (4π²/g)(1 − k²/l²) = 0, so l² = k², l = k.

d²(T²)/dl² = (4π²/g)(2k²/l³) > 0, so T is a minimum.

Thus when the period is minimum, k = l: the radius of gyration (about G) equals the distance from the centre of suspension to the centre of gravity, and T_min = 2π√(2k/g).

✓ Answer checked by Claude Opus 5.5 (high effort)

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