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Kinetic energy and work of rigid body

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.

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.

Kinetic energy of a rigid body in plane motion

Let the body have mass mm, angular velocity ω\omega, and let the mass centre G have velocity vG\mathbf{v}_G. A particle of mass dmdm at position ρ\boldsymbol\rho relative to G has velocity vG+ω×ρ\mathbf{v}_G + \boldsymbol\omega\times\boldsymbol\rho (velocity of G plus velocity relative to G, magnitude ωρ\omega\rho).

T=∫12∣vG+ω×ρ∣2 dm=12vG2∫dm+vG⋅(ω×∫ρ dm)+12ω2∫ρ2 dmT = \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, ∫ρ dm=0\int\boldsymbol\rho\,dm = 0, so the middle term vanishes, and ∫ρ2 dm=IG\int\rho^2\,dm = I_G:

T=12mvG2+12IG ω2T = \tfrac12 m v_G^2 + \tfrac12 I_G\,\omega^2

Special cases: pure translation T=12mv2T = \tfrac12mv^2; rotation about a fixed point O, T=12IOω2T = \tfrac12 I_O\omega^2; rolling without slipping, T=12ICω2T = \tfrac12 I_C\omega^2 about the contact point.

Principle of work and energy for a rigid body

T1+U1→2=T2T_1 + U_{1\to2} = T_2

where U1→2U_{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 F\mathbf{F}: U=∫F⋅drU = \int \mathbf{F}\cdot d\mathbf{r} at its point of application; weight: U=−WΔhU = -W\Delta h; spring: U=−12k(x22−x12)U = -\tfrac12 k(x_2^2 - x_1^2).
  • Couple MM: U=∫M dθU = \int M\,d\theta, which is MΔθM\Delta\theta for constant MM.

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 −μN s-\mu N\,s (energy lost as heat).

✓ Answer checked by Claude Opus 5.5 (high effort)

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