2. Torsional Pendulum — Rigidity Modulus and Moment of Inertia by the Two-Body Method¶

Module

Rotational Dynamics · Module 1

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Aim¶

To determine the rigidity (shear) modulus of the material of a suspension wire and the moment of inertia of a circular disc, using the two-body method of a torsional pendulum.

Theoretical Background¶

A torsional pendulum consists of a heavy circular disc suspended from a rigid support by a thin wire fixed rigidly at both ends. When the disc is rotated in its own horizontal plane through a small angle and released, the wire, twisted through that angle, exerts a restoring torque proportional to the twist (Hooke's law in shear), so the disc executes angular simple harmonic motion.

The proportionality constant relating restoring torque to twist angle — the torsional rigidity C of the wire — depends on the wire's rigidity (shear) modulus η, its length l and, very sensitively, its radius r (as the fourth power), since C = πηr⁴/2l. A thin wire is therefore very much 'softer' torsionally than a thick one of the same material and length.

The period of oscillation depends on both C and the disc's own moment of inertia I about the wire's axis: T = 2π√(I/C). Since I is generally not known beforehand for an arbitrary disc, the two-body method sidesteps this by placing two identical, equal point masses symmetrically on the disc at a known distance d from its centre. This adds a known increment 2md² to the moment of inertia without changing C, giving a second, longer period T' whose difference from the first period T lets I be eliminated and calculated purely from measured quantities.

Once I is found this way, the same period T (measured with the disc alone) yields C, and hence the rigidity modulus η of the wire — one of the two fundamental elastic constants of an isotropic solid, alongside Young's modulus.

Governing Formulas¶

Torsional constant of the wire:

\[ C = \dfrac{\pi \eta r^{4}}{2l} \]

Period with disc alone / with two added masses:

\[ T = 2\pi\sqrt{\dfrac{I}{C}}, \qquad T' = 2\pi\sqrt{\dfrac{I + 2md^{2}}{C}} \]

Two-body method — moment of inertia of the disc:

\[ I = \dfrac{2 m d^{2} T^{2}}{T'^{2} - T^{2}} \]

Rigidity modulus:

\[ \eta = \dfrac{8\pi l I}{r^{4} T^{2}} \]

Labelled apparatus layout — original schematic prepared for this guide (not a photograph). A heavy circular disc suspended by a thin wire clamped rigidly at the top, free to execute torsional oscillations in its own plane; symmetric slots allow two equal masses to be clamped at known radii for the two-body method.

Simulator Controls¶

Control	Symbol	Range	Unit
Wire length	l	30 – 100	cm
Wire radius	r	0.20 – 0.80	mm
Added-mass distance from axis	d	4 – 12	cm
Each added mass	m	50 – 250	g

Readouts:

Quantity	Symbol	Unit
Period, disc alone	T	s
Period, with two added masses	T′	s

Procedure¶

Launch the Torsional Pendulum panel; note the default wire length, radius, and the readout period T of the bare disc.
Set the added-mass distance d and mass m to their default values; note the new readout period T′ with both masses symmetrically placed.
Repeat for several different values of d (keeping l, r, m fixed) and tabulate T and T′ each time.
Use the two-body formula to calculate the disc's moment of inertia I for each trial and check its consistency (I should not depend on d).
Using I and the bare-disc period T, calculate the rigidity modulus η of the wire from the given formula, for several values of l and r.

Observation Table¶

Table: Two-body method: period with and without added symmetric masses

S. No.	d (cm)	m (g)	T (s)	T′ (s)	I = 2md²T²/(T′²−T²) (g·cm²)
1
2
3
4
5

Graph¶

Plot T′² − T² (y-axis) against d² (x-axis).

Expected trend: a straight line through the origin, since T′² − T² = (2m/C)·d².

(T′² − T²) vs d², expected to be a straight line through the origin

15×15 grid — choose your own scale per square; plot from your observation table

Calculation¶

Verify that I calculated from the two-body formula is essentially constant across different values of d — this is the key internal consistency check of the method.
Using the mean I and the bare period T, compute C = 4π²I/T², then η = 2Cl/(πr⁴).
Repeat for at least two different wire radii r and compare the two independently-determined values of η — they should agree within experimental error, since η is a material property independent of the wire's dimensions.

Result¶

Table: Rigidity modulus of the wire material

Trial	l (cm)	r (mm)	I (g·cm²)	η (dyne/cm²)
1
2
3

Maximum Permissible Error¶

η = 8πlI/(r⁴T²) depends on r to the fourth power and T squared — even a small uncertainty in the wire radius (measured with a screw gauge) dominates the total error.

Precautions¶

Add the two masses exactly symmetrically about the axis, at equal distances d, so no net translational imbalance or wobble is introduced.
Keep the twist angle small and let the disc settle into a clean, wobble-free oscillation before timing.
Measure the wire radius r with a screw gauge at several points along its length, since η depends on r⁴ and this is usually the dominant source of error.

Self-Check Quiz¶

1. Doubling the wire radius r (all else fixed) changes the torsional constant C by a factor of:

A) 2
B) 4
C) 8
D) 16

Answer: D — C ∝ r⁴, so doubling r increases C by a factor of 2⁴ = 16 — a thicker wire is dramatically stiffer torsionally.

2. In the two-body method, why is it valid to say I is independent of the added-mass distance d used in the trial?

A) Because I is a property of the disc alone and does not depend on what is temporarily added to it
B) Because d cancels algebraically in every formula
C) Because the masses are massless in the idealised model
D) It is not independent — this is a limitation of the method

Answer: A — I refers to the disc's own moment of inertia; the added masses are external, known contributions used only to create a second data point — the recovered I should be the same regardless of which d was used.

3. If a wire twice as long but of the same radius and material is used, the period T of the same disc will:

A) Increase by a factor of √2
B) Decrease by a factor of √2
C) Stay the same
D) Double

Answer: A — C ∝ 1/l, so T = 2π√(I/C) ∝ √l — doubling l increases T by √2.

Viva-Voce Questions¶

Why does the rigidity modulus depend on the fourth power of the wire radius rather than, say, the square?
What is the two-body method designed to eliminate, and how does it achieve this algebraically?
Distinguish between Young's modulus and the rigidity (shear) modulus — what kind of deformation does each describe?
Why must the two added masses be placed at exactly equal distances from the axis and be of exactly equal mass?
How would you check experimentally that the wire is genuinely obeying Hooke's law throughout the twist range you used?
What would happen to the period if the disc were replaced by a ring of the same mass and radius?