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advanced24 min studyLesson 154 in Physics

Euler equations for rigid bodies

Derive rotational equations in principal body axes and use torque, energy, and angular momentum as independent checks.

Before you begin
On this page

The question for today

What you’ll understand

  • Derive the gyroscopic term from a moving body basis.
  • Calculate instantaneous angular acceleration in principal axes.
  • Check torque-free energy and inertial angular-momentum conservation.

Newton’s rotational law is simplest in an inertial frame: torque changes angular momentum. The inertia tensor is simplest in a principal body frame: it is constant and diagonal. Euler’s equations combine those two advantages by paying careful attention to the rotating basis. The additional cross product is essential bookkeeping for a moving frame, not an extra applied torque.

Combine a fixed tensor with a rotating basis

Take moments about the center of mass of a free body, or about a fixed inertial point for a supported body. These choices avoid the additional transport terms associated with an arbitrary accelerating origin. In body components, the inertia tensor is constant under the rigid-body assumption, so the component derivative of angular momentum is the inertia tensor times the component derivative of angular velocity.

The inertial derivative also includes the turning body basis. Insert the rotating-frame derivative rule into the torque law and express every quantity in the same body frame. The resulting vector equation includes angular velocity crossed with angular momentum. Even when the external torque vanishes, angular-momentum components in the body frame can change because the frame itself keeps rotating.

Iω˙b+ωb×(Iωb)=τbI\dot{\boldsymbol\omega}_b+\boldsymbol\omega_b\times(I\boldsymbol\omega_b)=\boldsymbol\tau_b
The dot means a derivative of body-frame components. The equation assumes a constant body inertia tensor and a center-of-mass origin or a fixed inertial pivot.↔ Scroll wide equations horizontally.

Read the component signs physically

In principal axes the tensor multiplication becomes three independent scale factors, but the cross product still couples the angular-velocity components. Write the first cross-product component explicitly before simplifying. It equals the second angular velocity times the third momentum minus the third angular velocity times the second momentum. Factoring it gives the difference of the third and second principal moments.

The cyclic structure supplies the remaining equations without guessing signs. For a spherical inertia tensor all principal moments match, and every coupling term vanishes. For rotation initially along one principal axis, the cross product also vanishes at that instant. Neither statement justifies discarding the coupling for a generic asymmetric rotation containing several nonzero components.

I1ω˙1+(I3−I2)ω2ω3=τ1,I2ω˙2+(I1−I3)ω3ω1=τ2,I3ω˙3+(I2−I1)ω1ω2=τ3I_1\dot\omega_1+(I_3-I_2)\omega_2\omega_3=\tau_1,\quad I_2\dot\omega_2+(I_1-I_3)\omega_3\omega_1=\tau_2,\quad I_3\dot\omega_3+(I_2-I_1)\omega_1\omega_2=\tau_3
These equations use a right-handed principal body frame. A torque-free system sets all three applied torque components to zero, not all three angular accelerations.↔ Scroll wide equations horizontally.

Conservation laws test the calculation

Dot the vector equation with angular velocity. The cross product makes no contribution because it is perpendicular to angular velocity, leaving the mechanical power as torque dotted with angular velocity. Therefore rotational energy is constant for an ideal torque-free rigid body. Separately, the inertial angular-momentum vector stays fixed; its magnitude is consequently constant in every orthonormal frame.

These are useful independent checks on numerical integration, but conserving two scalar quantities alone does not guarantee an accurate orientation history. The vector lab provides a component exercise: predict how changing a vector angle alters the signs of its projections, then verify. Use those signs when checking products in Euler’s equations. The diagram does not integrate the coupled equations or display physical torque.

Work it through

A torque-free body with changing angular velocity

Let the principal moments be (2, 3, 4) kilogram square meters and the initial body angular velocity be (1, 2, 3) radians per second. With zero external torque, find the instantaneous body angular acceleration and verify the instantaneous energy derivative.

  1. The first equation gives omega_1-dot = -(4 - 3) times 2 times 3 divided by 2, which is -3 radians per second squared.

  2. The second equation gives omega_2-dot = -(2 - 4) times 3 times 1 divided by 3, which is 2. The third gives omega_3-dot = -(3 - 2) times 1 times 2 divided by 4, which is -0.5.

  3. Evaluate the energy derivative as the sum of I_i omega_i omega_i-dot. The terms are -6, 12, and -6 watts; their sum is zero despite the three nonzero component derivatives.

The result

The instantaneous body angular acceleration is (-3, 2, -0.5). Constant energy and inertial angular momentum coexist with changing body angular-velocity components.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Check component signs before coupling them

Predict the projection signs of a vector at 30 and 150 degrees, then verify using the angle control. Use this sign discipline when evaluating cross-product components by hand.

Check component signs before coupling themPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA projection diagram supports component arithmetic. It is not an Euler-equation solver and does not compute applied or gyroscopic torque.

Pause & reason

Check your understanding

For a torque-free asymmetric rigid body, which statement must hold?

Your turn

Try it yourself

A spherical rigid body has principal moments all equal to 5 kilogram square meters and body angular velocity initially (1, -2, 3). With no external torque, determine its body angular acceleration and rotational energy.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

All differences between principal moments vanish. Compute energy from the sum of the squared components.

Compare with the solution

Every component derivative is zero. The squared angular-speed magnitude is one plus four plus nine, giving fourteen, so the energy is one half times five times fourteen, or thirty-five joules. This simplified evolution follows from isotropic inertia; it cannot be transferred to a body with unequal principal moments.

Keep the idea

Take it with you

  • Euler’s gyroscopic term accounts for the derivative of a rotating basis.
  • Zero torque does not generally mean zero body angular acceleration.
  • Energy and inertial angular momentum provide complementary checks on rotational dynamics.

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