Euler's equations

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This page discusses rigid body dynamics. For other uses, see Euler function (disambiguation).

In physics, Euler's equations describe the rotation of a rigid body in a frame of reference fixed in the rotating body

\begin{matrix} I_1\dot{\omega}_{1}+(I_3-I_2)\omega_2\omega_3 &=& M_{1}\\ I_2\dot{\omega}_{2}+(I_1-I_3)\omega_3\omega_1 &=& M_{2}\\ I_3\dot{\omega}_{3}+(I_2-I_1)\omega_1\omega_2 &=& M_{3} \end{matrix}

where Mk are the applied torques, Ik are the principal moments of inertia and ωk are the components of the angular velocity vector \boldsymbol\omega along the principal axes.

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[edit] Motivation and derivation

In absolute space, i.e., in a "space-fixed" frame of reference, the time derivative of angular momentum equals the applied torque

\frac{d\mathbf{L}}{dt} \ \stackrel{\mathrm{def}}{=}\  \frac{d}{dt} \left( \mathbf{I} \cdot \boldsymbol\omega \right) = \mathbf{M}

where \mathbf{I} is the moment of inertia tensor. Although this law is universally true, it is not always helpful in solving for the motion of a general rotating rigid body, since both \mathbf{I} and \boldsymbol\omega can change during the motion.

Therefore, we change to a coordinate frame fixed in the rotating body, and chosen so that its axes are aligned with the principal axes of the moment of inertia tensor. In this frame, at least the moment of inertia tensor is constant (and diagonal), which simplifies calculations. As described in the moment of inertia, the angular momentum vector \mathbf{L} can be written

\mathbf{L} \ \stackrel{\mathrm{def}}{=}\   L_{1}\mathbf{e}_{1} + L_{2}\mathbf{e}_{2} + L_{3}\mathbf{e}_{3} =  I_{1}\omega_{1}\mathbf{e}_{1} + I_{2}\omega_{2}\mathbf{e}_{2} + I_{3}\omega_{3}\mathbf{e}_{3}

where the Ik are the principal moments of inertia, the \mathbf{e}_{k} are unit vectors in the directions of the principal axes, and the ωk are the components of the angular velocity vector along the principal axes. In a rotating reference frame, the time derivative must be replaced with

\left(\frac{d\mathbf{L}}{dt}\right)_\mathrm{rot}+ \boldsymbol\omega\times\mathbf{L}=\mathbf{M}

where the rot subscript indicates that it is taken in the rotating reference frame. Substituting L_{k} \ \stackrel{\mathrm{def}}{=}\  I_{k}\omega_{k}, taking the cross product and using the fact that the principal moments do not change with time, we arrive at the Euler equations

\begin{matrix} I_1\dot{\omega}_{1}+(I_3-I_2)\omega_2\omega_3&=&M_{1}\\ I_2\dot{\omega}_{2}+(I_1-I_3)\omega_3\omega_1&=&M_{2}\\ I_3\dot{\omega}_{3}+(I_2-I_1)\omega_1\omega_2&=&M_{3} \end{matrix}

[edit] Torque-free solutions

For the RHSs equal to zero there are non-trivial solutions: torque-free precession. Notice that if I is constant (because the inertia tensor is the identity, because we work in the intrinsecal frame, or because the torque is driving the rotation around the same axis \mathbf{\hat{n}} so that I is not changing) then we may write

\mathbf{M} \ \stackrel{\mathrm{def}}{=}\  I \frac{d\omega}{dt}\mathbf{\hat{n}} =  I \alpha \mathbf{\hat{n}}

where

α is called the angular acceleration (or rotational acceleration) about the rotation axis \mathbf{\hat{n}}.

But if I is not constant in the external reference frame (ie. the body is moving and its inertia tensor is not the identity) then we cannot take the I outside the derivate. In this cases we will have torque-free precession, in such a way that I(t) and w(t) change together so that their derivative is zero.

[edit] Generalizations

It is also possible to use these equations if the axes in which \left(\frac{d\mathbf{L}}{dt}\right)_\mathrm{relative} is described are not connected to the body. \boldsymbol\omega should then be replaced with the rotation of the axes instead of the rotation of the body. It is, however, still required that the chosen axes are still principal axes of inertia! This form of the Euler equations is handy for rotation-symmetric objects that allow some of the principal axes of rotation to be chosen freely.

[edit] See also

[edit] References