---
title: "Flywheel energy relation and rim sizing"
subject: "Theory of Mechanism and Machine II"
chapter: 2
key: "semester_7/theory_of_mechanism_and_machine_ii:flywheel-energy-fluctuation-derivation"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/flywheel-energy-fluctuation-derivation
site: BCT Engineering
---

# Flywheel energy relation and rim sizing

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

- Asked in: year not given, 5 marks
- Video: [Flywheels](https://www.youtube.com/watch?v=eamfxa1gGZw) — Youngstone Videos
- 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-fluctuation-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-2

## Question

Derive the relation $\Delta E = I\omega^2 C_s$ for a flywheel. Why is most of the mass of a flywheel placed in the rim? Compare a rimmed flywheel with a solid disc flywheel of the same mass and outer radius.

## Short answer

ΔE = I ω (ω₁ - ω₂) = I ω² C_s since mean ω = (ω₁+ω₂)/2 and ω₁-ω₂ = C_s ω. Most mass is placed in the rim to maximize radius of gyration for given inertia, reducing weight. A thin rim stores twice the energy of a solid disc of equal mass and outer radius.

## Answer

#### Derivation
Let the flywheel have moment of inertia $I$, maximum speed $\omega_1$, minimum speed $\omega_2$ and mean speed $\omega = \dfrac{\omega_1+\omega_2}{2}$.

Maximum kinetic energy: $E_{1} = \tfrac12 I\omega_1^2$. Minimum kinetic energy: $E_{2} = \tfrac12 I\omega_2^2$.

$$
\begin{aligned}
\Delta E &= E_1 - E_2 = \tfrac12 I(\omega_1^2-\omega_2^2) \\
&= \tfrac12 I(\omega_1-\omega_2)(\omega_1+\omega_2) \\
&= I\,\omega\,(\omega_1-\omega_2)
\end{aligned}
$$

since $\omega_1+\omega_2 = 2\omega$. Using $C_s = \dfrac{\omega_1-\omega_2}{\omega}$, we get $\omega_1-\omega_2 = C_s\omega$, so

$$
\Delta E = I\omega^2 C_s = 2E\,C_s
$$

where $E = \tfrac12 I\omega^2$ is the mean kinetic energy.

#### Why the mass is in the rim
For a required $\Delta E$, speed and $C_s$, the value of $I$ is fixed. Since $I = mk^2$, a large radius of gyration $k$ gives the same $I$ with less mass. So the material is placed at the rim, which keeps the weight and cost low. The limit is the rim stress $\sigma = \rho v^2$, which restricts the rim speed.

#### Rimmed ring vs solid disc
| Point | Thin rim | Solid disc |
|---|---|---|
| Moment of inertia | $mR^2$ | $\tfrac12 mR^2$ |
| Energy stored (same $m$, $R$, $\omega$) | $\tfrac12 mR^2\omega^2$ | $\tfrac14 mR^2\omega^2$ |
| Mass for same $I$ | smaller | twice as much |

For the same mass and outer radius, the rim stores about twice the energy of a solid disc.

## Check yourself

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

1. In deriving Delta E = I w^2 C_s, the step that uses the mean speed is:
   A. w1 + w2 = 2w so Delta E = I w (w1 - w2)
   B. w1 - w2 = 2w so Delta E = I w (w1 + w2)
   C. w1 x w2 = w^2 so Delta E = I w^2
   D. w1 = w2 = w
   Answer: A. Delta E = 1/2 I (w1-w2)(w1+w2), and w1+w2 = 2w gives I w (w1-w2); with w1-w2 = C_s w this gives I w^2 C_s. (Key point: Derivation step)

2. Why is most of a flywheel's mass placed in the rim?
   A. A large radius of gyration gives the required I with less mass
   B. It reduces the required Delta E
   C. It lowers the rim stress
   D. It makes C_s larger
   Answer: A. Since I = m k^2, a large k gives the same I with less mass, weight and cost. (Key point: Why the rim carries the mass)

3. What limits how large the rim speed can be?
   A. Rim stress sigma = rho v^2
   B. The coefficient of steadiness
   C. Mean torque
   D. Bearing diameter
   Answer: A. Centrifugal stress in the rim is sigma = rho v^2, so permissible stress restricts the rim speed. (Key point: Limit on rim)

4. For the same mass, outer radius and speed, how does a thin rim compare to a solid disc?
   A. Stores equal energy
   B. Stores half the energy
   C. Stores about twice the energy
   D. Stores four times the energy
   Answer: C. Rim I = m R^2, disc I = 1/2 m R^2, so the rim stores twice the kinetic energy. (Key point: Ring vs disc)

## Related in Theory of Mechanism and Machine II

- [Flywheel size from turning moment areas](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Aflywheel-energy-areas-numerical) — Theory of Mechanism and Machine II, chapter 2
- [Flywheel for a punching machine](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Apunching-machine-flywheel-design) — Theory of Mechanism and Machine II, chapter 2
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