---
title: "Inertia force, inertia torque and D'Alembert's principle"
subject: "Theory of Mechanism and Machine I"
chapter: 8
key: "semester_6/theory_of_mechanism_and_machine_i:centrifugal-inertia-force-dalembert-principle"
source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/centrifugal-inertia-force-dalembert-principle
site: BCT Engineering
---

# Inertia force, inertia torque and D'Alembert's principle

Theory of Mechanism and Machine I · Chapter 8: Force Analysis of Mechanisms

- Asked in: year not given, 3+3 marks
- Video: [Enginering Mechanics Dynamics D'Alembert Principle 1](https://www.youtube.com/watch?v=UJb-Mdd4imk) — Mechanical Engineering by Ashish Purohit
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/centrifugal-inertia-force-dalembert-principle
- Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-8

## Question

(a) Define centrifugal force, inertia force and inertia torque, and state D'Alembert's principle. A 5 kg mass is attached to a shaft at a radius of 0.4 m and rotates at 300 rpm; find the centrifugal force. (b) A link of mass 12 kg and moment of inertia 0.8 kg m^2 about its centre of mass G has, at an instant, an acceleration of 25 m/s^2 of G and an angular acceleration of 40 rad/s^2 anticlockwise. Find the inertia force and inertia couple and the single equivalent inertia force, with its distance from G.

## Short answer

Centrifugal force = m ω² r outward; inertia force = –m a_G; inertia torque = –I_G α; D’Alembert’s principle adds them to external forces for static equilibrium. (a) 1973.9 N. (b) Inertia force 300 N opposite a_G, inertia couple 32 N·m clockwise, equivalent force shifted 106.7 mm from G.

## Answer

#### (a) Definitions
- **Centrifugal force**: the outward radial force on a body in circular motion, $F_c = m\omega^2 r$. It is the reaction to the centripetal force. For analysis it is taken as an inertia force.
- **Inertia force**: the fictitious force equal to $m a_G$ acting at the centre of mass G, opposite to the acceleration of G: $\vec F_I = -m\vec a_G$.
- **Inertia torque (couple)**: equal to $I_G\alpha$, opposite to the angular acceleration $\alpha$: $C_I = -I_G\alpha$.
- **D'Alembert's principle**: if the inertia force and inertia torque are added to the actual external forces on a body, the body is in dynamic equilibrium. Then $\sum \vec F + \vec F_I = 0$ and $\sum M + C_I = 0$, so the methods of statics can be used for dynamic problems.

**Numerical.**

$$
\omega = \frac{2\pi \times 300}{60} = 31.416\ \text{rad/s}
$$

$$
F_c = m\omega^2r = 5 \times 986.96 \times 0.4 = 1973.9\ \text{N}
$$

#### (b) Link
Inertia force at G:

$$
F_I = m a_G = 12 \times 25 = 300\ \text{N}
$$

directed opposite to the acceleration of G.

Inertia couple:

$$
C_I = I_G\alpha = 0.8 \times 40 = 32\ \text{N m}
$$

acting clockwise (opposite to $\alpha$).

**Single equivalent force.** The force $F_I$ and couple $C_I$ can be replaced by the same force $F_I$ acting along a line parallel to its original line, shifted by

$$
h = \frac{C_I}{F_I} = \frac{32}{300} = 0.1067\ \text{m} = 106.7\ \text{mm}
$$

from G, on the side that gives a moment about G in the same sense as $C_I$ (clockwise).

```text
         h
    ----------|-> F_I (shifted line)
   G (x)  <- C_I clockwise
```

**Answer: (a) $F_c = 1973.9$ N; (b) $F_I = 300$ N, $C_I = 32$ N m (clockwise), shifted force at $h = 106.7$ mm from G.**

## Check yourself

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

1. What does D'Alembert's principle state?
   A. Action and reaction are equal and opposite
   B. Energy is conserved in a closed system
   C. Adding inertia force and inertia torque to the external loads puts the body in dynamic equilibrium
   D. Momentum is conserved during impact
   Answer: C. With inertia force and torque added, the sums of forces and moments are zero, so statics can be used. (Key point: D'Alembert's principle)

2. The inertia force on a body is:
   A. m a_G acting in the same direction as a_G
   B. I_G x alpha
   C. m a_G opposite to the acceleration of G, acting at G
   D. m w^2 r at the pivot
   Answer: C. F_I = -m a_G, acting at the centre of mass opposite to its acceleration. (Key point: Inertia force direction)

3. Centrifugal force for 5 kg at 0.4 m radius and 300 rpm is:
   A. 1973.9 N
   B. 197.4 N
   C. 986.96 N
   D. 62.8 N
   Answer: A. F = m w^2 r = 5 x 986.96 x 0.4 = 1973.9 N. (Key point: Centrifugal force)

4. For the link with I_G = 0.8 kg m^2 and alpha = 40 rad/s^2 anticlockwise, the inertia couple is:
   A. 300 N m anticlockwise
   B. 32 N m clockwise
   C. 32 N m anticlockwise
   D. 0.02 N m clockwise
   Answer: B. C_I = I_G alpha = 32 N m, acting opposite to alpha, so clockwise. (Key point: Inertia couple)

5. How far from G does the single equivalent inertia force act?
   A. h = F_I x C_I = 9600 m
   B. h = C_I x m = 384 m
   C. h = F_I/C_I = 9.4 m
   D. h = C_I/F_I = 32/300 = 106.7 mm
   Answer: D. The offset is h = C_I/F_I, on the side giving a moment about G in the sense of C_I. (Key point: Equivalent single force)

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