---
title: "Moment of inertia by torsional pendulum"
subject: "Engineering Physics"
chapter: 1
key: "semester_1/engineering_physics:moment-of-inertia-torsional-pendulum"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/moment-of-inertia-torsional-pendulum
site: BCT Engineering
---

# Moment of inertia by torsional pendulum

Engineering Physics · Chapter 1: Oscillation

- Asked in: 2071 Shrawan, 5 marks
- Video: [MOMENT OF INERTIA OF AN IRREGULAR BODY USING TORSION PENDULUM || TORSIONAL PENDULUM EXPERIMENT ||](https://www.youtube.com/watch?v=GMDHLMXMz0c) — Pankaj Physics Gulati
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/moment-of-inertia-torsional-pendulum
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1

## Question

Derive a relation to find the moment of inertia of a rigid body about an axis passing through its center of gravity using the torsional pendulum.

## Short answer

Suspend the body from a wire, measure period T₁. Add a body of known moment of inertia I₀, measure new period T₂. From T₁² = 4π²I/C and T₂² = 4π²(I+I₀)/C, eliminate C to get I = I₀ T₁²/(T₂² − T₁²).

## Answer

A rigid body is hung by a wire so that the axis (through its centre of gravity) coincides with the wire. Twisting it makes it oscillate as a torsional pendulum.

```text
   /////////
      |  wire, torsional constant C
   ===O===  body, I (unknown)
```

#### Period of the torsional pendulum
Restoring torque = −Cθ, so I d²θ/dt² = −Cθ, which is angular SHM:
```text
T = 2π √(I/C)   →   I = C T²/4π²
```
C is not known directly, so use a body of known moment of inertia.

#### Method
1. Suspend the unknown body (moment of inertia I) and measure its period T₁:
```text
T₁² = 4π² I / C                       ... (1)
```
2. Place on it (or attach to the wire) a regular body, e.g. a ring or disc, with known moment of inertia I₀ about the same axis. Measure the new period T₂:
```text
T₂² = 4π² (I + I₀) / C                ... (2)
```
3. Divide (1) by (2):
```text
T₁²/T₂² = I/(I + I₀)
I T₂² = (I + I₀) T₁²
I (T₂² − T₁²) = I₀ T₁²
```
```text
I = I₀ T₁² / (T₂² − T₁²)
```
Measure T₁, T₂, and calculate I₀ from the mass and dimensions of the standard body (for a disc I₀ = ½MR²; for a ring I₀ = MR²) to obtain I of the body about the axis through its centre of gravity.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. T of torsional pendulum?
   A. 2pi sqrt(C/I)
   B. 2pi sqrt(l/g)
   C. 2pi sqrt(I/C)
   D. sqrt(I/C)
   Answer: C. Angular SHM. (Key point: Period)

2. Why add a body of known I0?
   A. To change g
   B. To remove friction
   C. To change amplitude
   D. C of the wire is unknown
   Answer: D. Ratio eliminates C. (Key point: Why a standard body)

3. T2^2 = ?
   A. 4pi^2 (I + I0)/C
   B. 4pi^2 I0/C
   C. 4pi^2 (I - I0)/C
   D. 4pi^2 I/(C I0)
   Answer: A. Moment of inertias add. (Key point: Equations)

4. I =?
   A. I0 T2^2/(T2^2 - T1^2)
   B. I0 T1^2/(T2^2 - T1^2)
   C. I0 (T2^2 - T1^2)/T1^2
   D. I0 T1/T2
   Answer: B. Dividing the two equations. (Key point: Result)

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