---
title: "Flywheel for a punching machine"
subject: "Theory of Mechanism and Machine II"
chapter: 2
key: "semester_7/theory_of_mechanism_and_machine_ii:punching-machine-flywheel-design"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/punching-machine-flywheel-design
site: BCT Engineering
---

# Flywheel for a punching machine

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

- Asked in: year not given, 8 marks
- Video: [Problem on Flywheel used in Punching Machine](https://www.youtube.com/watch?v=aJLuP3sCZKA) — Neeraj Gautam
- 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/punching-machine-flywheel-design
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-2

## Question

A punching machine punches one 30 mm diameter hole in a 10 mm thick plate in every revolution of the flywheel shaft. The ultimate shear strength of the plate is 350 MPa. The punch has a stroke equal to the plate thickness, the punching takes place during $20\%$ of the revolution, and the shaft runs at a mean speed of 300 rpm. The motor torque is constant and the speed must not vary by more than $\pm 1.5\%$ of the mean speed. Find the power of the motor and the mass of a flywheel of mean rim radius 0.6 m. Neglect friction and the mass of arms and hub.

## Short answer

Maximum punching force 329.9 kN, energy per hole = 1649 J. Motor power = (energy × rpm/60) = 8.25 kW. Flywheel supplies 1319 J during punching; with ω=31.42 rad/s and Cs=0.03, I=44.56 kg·m², rim mass = 123.8 kg.

## Answer

**Data:** $d = 0.03$ m, $t = 0.01$ m, $\tau_u = 350$ MPa, $N = 300$ rpm, $R = 0.6$ m, $C_s = 0.03$ (since $\pm1.5\%$ means a total variation of $3\%$).

#### Energy for one punching
Maximum punching force:

$$
F = \pi d t\,\tau_u = \pi \times 0.03 \times 0.01 \times 350\times10^{6} = 329867\ \text{N}
$$

The force falls from $F$ to zero over the stroke, so the average force is $F/2$:

$$
E = \frac{F}{2}\times t = \frac{329867}{2}\times 0.01 = 1649\ \text{J}
$$

#### Motor power
The motor supplies this energy uniformly over every revolution (one revolution takes $60/300 = 0.2$ s):

$$
P = E \times \frac{N}{60} = 1649 \times 5 = 8247\ \text{W} = 8.25\ \text{kW}
$$

#### Fluctuation of energy
During the punching (20% of a revolution) the motor supplies only $0.2E$:

$$
E_{motor} = 0.2 \times 1649 = 330\ \text{J}
$$

The flywheel must supply the rest:

$$
\Delta E = E - E_{motor} = 1649 - 330 = 1319\ \text{J}
$$

#### Flywheel
$$
\omega = \frac{2\pi \times 300}{60} = 31.42\ \text{rad/s}
$$

$$
I = \frac{\Delta E}{\omega^2 C_s} = \frac{1319}{(31.42)^2 \times 0.03} = 44.56\ \text{kg·m}^2
$$

$$
m = \frac{I}{R^2} = \frac{44.56}{0.6^2} = 123.8\ \text{kg}
$$

**Answer:** Motor power $\approx 8.25$ kW; flywheel rim mass $\approx 123.8$ kg.

## Check yourself

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

1. How is the energy needed for one punching found?
   A. Maximum force x thickness
   B. Half of maximum force x plate thickness (force falls linearly to zero)
   C. Maximum force x hole diameter
   D. Maximum force x 20% of thickness
   Answer: B. The average force over the stroke is F/2, so E = (F/2) x t = 1649 J. (Key point: Energy per punch)

2. The motor torque is constant, so the motor power equals:
   A. Energy per punch x punches per second
   B. Energy per punch / punching time
   C. Energy per punch x 0.2
   D. Energy per punch x angular speed
   Answer: A. The motor supplies the energy of one punch evenly over each revolution: P = 1649 x 5 = 8.25 kW. (Key point: Motor power)

3. Why is Delta E smaller than the punching energy E?
   A. Friction absorbs part of E
   B. The motor supplies 20% of E during the punching, so the flywheel supplies the rest
   C. The plate returns energy
   D. The flywheel supplies only 20% of E
   Answer: B. During 20% of a revolution the motor gives 0.2E = 330 J, so Delta E = 1649 - 330 = 1319 J. (Key point: Fluctuation of energy)

4. A speed variation of plus or minus 1.5% of mean speed corresponds to C_s of:
   A. 0.015
   B. 0.03
   C. 0.0075
   D. 0.3
   Answer: B. The total variation is 3%, so C_s = 0.03. (Key point: C_s for plus/minus 1.5%)

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