---
title: "Flywheel size from turning moment areas"
subject: "Theory of Mechanism and Machine II"
chapter: 2
key: "semester_7/theory_of_mechanism_and_machine_ii:flywheel-energy-areas-numerical"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/flywheel-energy-areas-numerical
site: BCT Engineering
---

# Flywheel size from turning moment areas

Theory of Mechanism and Machine II · Chapter 2: Turning Moment Diagram and Flywheel

- Asked in: year not given, 6 marks
- Video: [Dynamics of Machinery Lectures| Flywheel Problems| Turning moment diagram for multi cylinder engines](https://www.youtube.com/watch?v=98AFawBzH7k) — Michael Thomas Rex F
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/flywheel-energy-areas-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-2

## Question

The areas (in kJ) between the turning moment curve and the mean torque line of an engine, taken in order from the start of the cycle, are $+4.8$, $-5.2$, $+5.0$, $-5.7$, $+5.9$ and $-4.8$. The mean speed is 240 rpm and the coefficient of fluctuation of speed is 0.03. Find the maximum fluctuation of energy and the mass of the flywheel rim if its mean radius is 0.9 m. Neglect the arms and hub.

## Short answer

By cumulatively adding energy areas, maximum energy is 4.8 kJ, minimum -1.1 kJ, so ΔE = 5.9 kJ. With ω=25.13 rad/s and Cs=0.03, required moment of inertia I=311.4 kg·m²; flywheel rim mass = I/R² = 384 kg.

## Answer

**Data:** mean speed $N = 240$ rpm, $C_s = 0.03$, rim mean radius $R = 0.9$ m.

#### Energy at each point
Take the energy at A as zero and add the areas in order (positive area = gain of kinetic energy, negative = loss).

| Point | Cumulative energy (kJ) |
|---|---|
| A | +0.0 |
| B | +4.8 |
| C | -0.4 |
| D | +4.6 |
| E | -1.1 |
| F | +4.8 |
| G | +0.0 |

Maximum energy = 4.8 kJ, minimum energy = -1.1 kJ.

$$
\Delta E = E_{max} - E_{min} = 4.8 - (-1.1) = 5.9\ \text{kJ} = 5900\ \text{J}
$$

#### Moment of inertia
$$
\omega = \frac{2\pi \times 240}{60} = 25.13\ \text{rad/s}
$$

$$
\Delta E = I\omega^2 C_s \Rightarrow I = \frac{\Delta E}{\omega^2 C_s} = \frac{5900}{(25.13)^2 \times 0.03} = 311.4\ \text{kg·m}^2
$$

#### Mass of the rim
Treating the rim as a thin ring, $I = mR^2$:

$$
m = \frac{I}{R^2} = \frac{311.4}{0.9^2} = 384\ \text{kg}
$$

**Answer:** $\Delta E = 5.9$ kJ; mass of rim $\approx 384$ kg.

## Check yourself

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

1. How is the maximum fluctuation of energy found from the list of areas?
   A. Add all positive areas
   B. Take cumulative energy from a starting point and subtract the minimum from the maximum
   C. Take the largest single area
   D. Average the positive and negative areas
   Answer: B. Add areas in order; Delta E = E_max - E_min of the cumulative values. (Key point: Method for energy fluctuation)

2. With cumulative energies 0, +4.8, -0.4, +4.6, -1.1, +4.8, 0 kJ, the maximum fluctuation of energy is:
   A. 4.8 kJ
   B. 5.9 kJ
   C. 5.2 kJ
   D. 3.7 kJ
   Answer: B. E_max = 4.8 kJ and E_min = -1.1 kJ, so Delta E = 5.9 kJ. (Key point: Result of Delta E)

3. Which relation gives the required moment of inertia of the flywheel?
   A. I = Delta E / (w C_s)
   B. I = Delta E / (w^2 C_s)
   C. I = Delta E x w^2 x C_s
   D. I = 2 Delta E / w^2
   Answer: B. Delta E = I w^2 C_s, so I = Delta E / (w^2 C_s) = 5900/(25.13^2 x 0.03) = 311.4 kg.m^2. (Key point: Moment of inertia)

4. Treating the rim as a thin ring of mean radius R, the mass is:
   A. m = I R^2
   B. m = I / R^2
   C. m = 2I / R^2
   D. m = I / R
   Answer: B. I = m R^2 so m = 311.4/0.81 = 384 kg. (Key point: Rim mass)

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