---
title: "Ring pendulum minimum time period"
subject: "Engineering Physics"
chapter: 1
key: "semester_1/engineering_physics:ring-pendulum-minimum-period"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/ring-pendulum-minimum-period
site: BCT Engineering
---

# Ring pendulum minimum time period

Engineering Physics · Chapter 1: Oscillation

- Asked in: 2073 Shrawan, 5 marks
- Video: [Time Period of Physical or Compound pendulum || In Nepali](https://www.youtube.com/watch?v=h2A3SuUJUJQ) — BE Crash Course
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/ring-pendulum-minimum-period
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1

## Question

A heavy circular ring of radius R oscillates in a vertical plane about a horizontal axis at a distance x from the center. Show that the time period is minimum when x = R.

## Short answer

Moment of inertia about pivot: I = M(R²+x²). Period T = 2π√((R²+x²)/(gx)). Minimizing T² wrt x gives d(T²)/dx = 0 → -R²/x²+1=0 → x = R. Second derivative positive, so this is minimum. Minimum period: T_min = 2π√(2R/g).

## Answer

A ring of mass M and radius R swings about a horizontal axis at distance x from its centre G (the axis is parallel to the ring's axis).

```text
        O  pivot
        |  x
        |
      .-G-.   ring, radius R
     (     )
      '---'
```

#### Moment of inertia
1. About the axis through G (perpendicular to the plane of the ring): I_G = MR², so k² = R².
2. About the pivot (parallel-axis theorem): I = MR² + Mx².

#### Time period
For small oscillations, torque = −Mgx θ, so I θ'' = −Mgx θ:
```text
T = 2π √(I/(Mgx)) = 2π √((R² + x²)/(g x))
```

#### Minimum
T² = (4π²/g)(R²/x + x). Set d(T²)/dx = 0:
```text
−R²/x² + 1 = 0   →   x² = R²   →   x = R
```
d²(T²)/dx² = 2R²/x³ > 0, so this is a minimum.

Minimum period: T_min = 2π√(2R/g). It occurs when the axis is at the rim of the ring (x = R), i.e. when the distance from the centre equals the radius of gyration k = R.

## Check yourself

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

1. For a ring about its centre, k^2 =?
   A. R^2/2
   B. R^2/4
   C. R^2
   D. 2R^2
   Answer: C. I_G = MR^2. (Key point: Radius of gyration)

2. About the pivot at distance x?
   A. MR^2 - Mx^2
   B. Mx^2
   C. MR^2/2 + Mx^2
   D. MR^2 + Mx^2
   Answer: D. Parallel-axis theorem. (Key point: Moment of inertia)

3. The period is?
   A. 2pi sqrt((R^2 + x^2)/(g x))
   B. 2pi sqrt(x/g)
   C. 2pi sqrt(g x/(R^2+x^2))
   D. 2pi sqrt(R/g)
   Answer: A. I/(Mgx). (Key point: Period)

4. Minimum period when?
   A. x = 0
   B. x = R, T_min = 2pi sqrt(2R/g)
   C. x = R/2
   D. x = 2R
   Answer: B. d(T^2)/dx = 0 gives x = R. (Key point: Minimum)

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