---
title: "Kinetic energy and work of rigid body"
subject: "Engineering Mechanics"
chapter: 7
key: "semester_3/engineering_mechanics:kinetic-energy-rigid-body-work-energy"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/kinetic-energy-rigid-body-work-energy
site: BCT Engineering
---

# Kinetic energy and work of rigid body

Engineering Mechanics · Chapter 7: Plane Motion of Rigid Bodies: Work and Energy Method

- Asked in: year not given, 6 marks
- Video: [Rigid Bodies Work and Energy Dynamics (Learn to solve any question)](https://www.youtube.com/watch?v=AMrKLzNoTWA) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/kinetic-energy-rigid-body-work-energy
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-7

## Question

Derive the expression for the kinetic energy of a rigid body in plane motion. State the principle of work and energy for a rigid body and explain which forces do no work on a rolling body, and how the work of a couple is calculated.

## Short answer

Kinetic energy of a rigid body in plane motion is T = ½ m v_G² + ½ I_G ω². The work-energy principle states T₁ + U₁→₂ = T₂. Forces doing no work on a rolling body include friction (contact point at rest) and normal reaction. Work of a couple is ∫ M dθ.

## Answer

#### Kinetic energy of a rigid body in plane motion
Let the body have mass $m$, angular velocity $\omega$, and let the mass centre G have velocity $\mathbf{v}_G$. A particle of mass $dm$ at position $\boldsymbol\rho$ relative to G has velocity $\mathbf{v}_G + \boldsymbol\omega\times\boldsymbol\rho$ (velocity of G plus velocity relative to G, magnitude $\omega\rho$).

$$
T = \int \tfrac12 |\mathbf{v}_G + \boldsymbol\omega\times\boldsymbol\rho|^2\,dm
= \tfrac12 v_G^2\int dm + \mathbf{v}_G\cdot\left(\boldsymbol\omega\times\int\boldsymbol\rho\,dm\right) + \tfrac12\omega^2\int\rho^2\,dm
$$

Since G is the mass centre, $\int\boldsymbol\rho\,dm = 0$, so the middle term vanishes, and $\int\rho^2\,dm = I_G$:

$$
T = \tfrac12 m v_G^2 + \tfrac12 I_G\,\omega^2
$$

Special cases: pure translation $T = \tfrac12mv^2$; rotation about a fixed point O, $T = \tfrac12 I_O\omega^2$; rolling without slipping, $T = \tfrac12 I_C\omega^2$ about the contact point.

#### Principle of work and energy for a rigid body

$$
T_1 + U_{1\to2} = T_2
$$

where $U_{1\to2}$ is the work of all external forces and couples (and internal ones that do work, such as springs or friction between connected parts) between positions 1 and 2. For a system of connected bodies, add the kinetic energies of all bodies.

#### Work of forces and couples
- Force $\mathbf{F}$: $U = \int \mathbf{F}\cdot d\mathbf{r}$ at its point of application; weight: $U = -W\Delta h$; spring: $U = -\tfrac12 k(x_2^2 - x_1^2)$.
- Couple $M$: $U = \int M\,d\theta$, which is $M\Delta\theta$ for constant $M$.

#### Forces that do no work
- Reactions at smooth fixed pins and smooth surfaces (perpendicular to motion).
- The friction force on a body **rolling without slipping**, because its point of application (the contact point) is instantaneously at rest, so no energy is dissipated. The normal reaction on a rolling body also does no work.
- Internal forces between particles of a rigid body (equal, opposite, no relative displacement).

If the body slips, the friction force does work $-\mu N\,s$ (energy lost as heat).

## Check yourself

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

1. What is the kinetic energy of a rigid body in plane motion?
   A. (1/2) m v_G^2 only
   B. (1/2) m v_G^2 + (1/2) I_G omega^2
   C. (1/2) I_G omega^2 only
   D. m v_G^2 + I_G omega^2
   Answer: B. Translation of G plus rotation about G. (Key point: Kinetic energy expression)

2. Why does the cross term in the derivation vanish?
   A. v_G is zero
   B. The integral of rho dm is zero because G is the mass centre
   C. omega is perpendicular to v_G
   D. The particles have equal speed
   Answer: B. Position relative to the mass centre integrates to zero over the mass. (Key point: Cross term vanishes)

3. Why does friction do no work on a body rolling without slipping?
   A. Friction is zero
   B. Its point of application (the contact point) is instantaneously at rest
   C. Friction is perpendicular to the motion
   D. Work done by friction is stored as potential energy
   Answer: B. No displacement of the contact point means no work and no dissipation. (Key point: Friction in rolling)

4. How is the work of a couple calculated?
   A. U = M × distance moved by G
   B. U = integral of M d(theta), or M × delta theta for constant M
   C. U = M × omega
   D. U = M / theta
   Answer: B. A couple does work through the angle the body rotates. (Key point: Work of a couple)

## Related in Engineering Mechanics

- [Plane motion and relative velocity](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Aplane-motion-types-relative-velocity) — Engineering Mechanics, chapter 5
- [Work-energy principle, power and efficiency](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Awork-energy-principle-power-efficiency) — Engineering Mechanics, chapter 3
- [Equations of motion for plane motion](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Aequations-plane-motion-rigid-body) — Engineering Mechanics, chapter 6
- [Virtual work and its principle](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Avirtual-work-principle-definition) — Engineering Mechanics, chapter 1
- [Rolling cylinder attached to a spring](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arolling-cylinder-spring-energy) — Engineering Mechanics, chapter 7
- [Lagrange's equations from D'Alembert's principle](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Alagrange-equations-derivation-dalembert) — Engineering Mechanics, chapter 9

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