Physlib

Physlib.ClassicalMechanics.RigidBody.AngularMomentum

Angular momentum of a rigid body

For a rigid body rotating with angular velocity `ω` about its reference point, each body point at position `r` moves with velocity `ω × r`, so the body's angular momentum about that point is `L = ∫ r × (ω × r) dm`. Expanding the double cross product, `r × (ω × r) = |r|² ω − (r · ω) r`, shows that `L` is linear in `ω` with matrix the inertia tensor: `L = I ω`.

References

2 declarations

definition

Angular momentum L\mathbf{L} of a rigid body with angular velocity ω\boldsymbol{\omega}

Given a 3D rigid body RR and its angular velocity ω∈R3\boldsymbol{\omega} \in \mathbb{R}^3, the angular momentum L∈R3\mathbf{L} \in \mathbb{R}^3 is defined. Its ii-th component is obtained by applying the body's mass distribution functional ρ\rho to the scalar field fi(r)=(r×(ω×r))if_i(\mathbf{r}) = (\mathbf{r} \times (\boldsymbol{\omega} \times \mathbf{r}))_i, where r\mathbf{r} denotes the position of a point in the body. This corresponds to the integral L=∫r×(ω×r) dm\mathbf{L} = \int \mathbf{r} \times (\boldsymbol{\omega} \times \mathbf{r}) \, dm.

theorem

L=Iω\mathbf{L} = \mathbf{I} \boldsymbol{\omega} for a Rigid Body

For a three-dimensional rigid body RR rotating with angular velocity ω∈R3\boldsymbol{\omega} \in \mathbb{R}^3, the angular momentum L\mathbf{L} is equal to the matrix-vector product of the body's inertia tensor I\mathbf{I} and the angular velocity vector ω\boldsymbol{\omega}: L=Iω\mathbf{L} = \mathbf{I} \boldsymbol{\omega}