# Physical Pendulum and Torsion

> Physical and torsion pendulums in JEE Physics: T = 2π√(I/mgl), equivalent length, Kater's pendulum, torsion constant C and radius of gyration results.

- Canonical URL: https://prepelephant.com/topics/jee/physics/physical-pendulum-and-torsion
- Exam / course: JEE · Subject: Physics
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- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Physical Pendulum and Torsion", PrepElephant, https://prepelephant.com/topics/jee/physics/physical-pendulum-and-torsion

## 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.

## Worked example: the bar pendulum

Take a uniform metre-scale-like bar of length 1 m pivoted at a hole drilled 0.4 m from its centre. About the centre, k² = L²/12 = 1/12 = 0.0833 m², so about the pivot I = m(0.0833 + 0.16) = 0.2433m kg m². The equivalent length is I/ml = 0.2433/0.4 = 0.608 m, and the period follows at once: T = 2π√(0.608/9.8) ≈ 1.56 s. Notice the routine: parallel axis, divide by ml, insert into 2π√(L/g). Every physical pendulum numerical in JEE is this three-step chain, and most lost marks come from skipping the parallel-axis term.

Now slide the pivot. Differentiating (k² + l²)/l shows the equivalent length is minimum at l = k = 1/√12 ≈ 0.289 m, where L = 2k and T_min = 2π√(0.577/9.8) ≈ 1.53 s. The curve is flat near its minimum — why a bar pendulum keeps time over a range of pivot holes, and why Kater's pendulum with two adjustable bobs could locate the conjugate points where both periods match.

## 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.