Physical Pendulum and Torsion
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Direct answer
Any rigid body pivoted at a point away from its centre of mass swings as a physical pendulum with period T = 2π√(I/mgl), where I is the moment of inertia about the pivot, m the mass and l the distance from pivot to centre of mass. Writing I = m(k² + l²), with k the radius of gyration about the centre of mass, gives an equivalent simple-pendulum length L = (k² + l²)/l, so the body keeps time with a simple pendulum of that length. A torsion pendulum — a disc or bar hung from a wire that twists — obeys T = 2π√(I/C) with C the torsional constant of the suspension, and this angular simple harmonic motion is the working principle of the Cavendish balance.
What you must remember
- Physical pendulum: T = 2π√(I/mgl), all quantities about or from the pivot; the moment of inertia must include the parallel-axis term, never just I about the centre of mass.
- Equivalent length: L = I/ml = (k² + l²)/l; the period equals that of a simple pendulum of length L, and L > l always, so a rigid body swings slower than a bob hung at its centre of mass.
- Minimum period: T is smallest when the pivot sits at l = k from the centre of mass; T_min = 2π√(2k/g), and T → ∞ both as l → 0 and as l → ∞, so every other pivot distance gives the same period twice.
- Kater's reversible pendulum: when the periods about the two pivots are equal, g = 4π²L/T² with L the distance between the pivots — the classical laboratory route to g that does not need I at all.
- Torsion pendulum: restoring torque τ = -Cθ, so T = 2π√(I/C); for a solid cylindrical wire of radius r and length L, C = πGr⁴/(2L), so stiffness grows as the fourth power of the wire radius.
- Standard inertias: disc I = MR²/2, ring I = MR², uniform rod about its centre I = ML²/12 — most numericals collapse once these are paired with the parallel-axis theorem.
- Seconds pendulum convention: T = 2 s (one second each swing), needing l ≈ 0.99 m at g = 9.8 m/s².
Where JEE questions bite
The recurring trap is substituting I about the centre of mass into 2π√(I/mgl); the formula demands the pivot value, and examiners build wrong options precisely around the parallel-axis omission. A second trap: the question "at what distance should a uniform disc be pivoted for minimum period" expects l = k = R/√2, testing whether you can convert the general condition into a body-specific number. Advanced-level papers also like the symmetry fact — a period achieved at one pivot distance recurs at a second, conjugate point. The seconds pendulum length (about 0.99 m) appears in assertion-reason format against g = 10 m/s² simplification, so keep both values in mind.
Frequently asked questions
When does a physical pendulum have the same period as a simple pendulum of length l?
Only when the pivot distance equals the equivalent length, which forces k = 0; a true rigid body always has k > 0, so its equivalent length exceeds l and it swings slower.
How does Kater's pendulum measure g without knowing the moment of inertia?
When the periods about its two pivots are adjusted to be equal, the pivot separation itself becomes the equivalent length, giving g = 4π²L/T² directly.
What supplies the restoring effect in a torsion pendulum?
The elastic resistance of the twisted suspension wire, exerting torque τ = -Cθ; gravity plays no role, so the torsion pendulum oscillates even inside an orbiting satellite.
How does the torsional constant of a wire depend on its geometry?
For a solid wire C = πGr⁴/(2L), so doubling the radius makes the wire sixteen times stiffer and cuts the period four-fold.
Why does the period of a physical pendulum diverge for pivots very close to the centre of mass?
As l → 0 the restoring torque mgl sinθ shrinks to zero while the inertia stays finite, so the equivalent length I/ml blows up and T grows without bound.