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Euler’s equations of motion for a rigid body can be interpreted as a rewriting of Newton’s second law for rotations in a rotating frame. They basically tell us the sum of the torques equals the rate of change of the body’s angular momentum, In the rotating frame. Do we then not need to take into account inertial forces when computing the torques in rotating coordinates?

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Do we then not need to take into account inertial forces when computing the torques in rotating coordinates?

No, but there is an inertial torque you have to worry about.

From the perspective of an inertial frame, the rotational analog of Newton's second law for rotation about the center of mass is $$\frac{d\boldsymbol L}{dt} = \sum_i \boldsymbol \tau_{\text{ext},i}\tag{1}$$ where $\boldsymbol L$ is the object's angular momentum with respect to inertial, $\boldsymbol \tau_{\text{ext},i}$ is the $i^\text{th}$ external torque, and the differentiation is from the perspective of the inertial frame. Note that this pertains to non-rigid objects as well as rigid bodies.

The relationship between the time derivatives of any vector quantity $\boldsymbol q$ from the perspectives of co-located inertial and rotating frames is $$\left(\frac{d\boldsymbol q}{dt}\right)_\text{inertial} = \left(\frac{d\boldsymbol q}{dt}\right)_\text{rotating} + \boldsymbol \Omega \times \boldsymbol q\tag{2}$$ where $\boldsymbol\Omega$ is the frame rotation rate with respect to inertial.

For a rigid body, the body's angular momentum with respect to inertial but expressed in body-fixed coordinates is $\boldsymbol L = \mathbf I\,\boldsymbol \omega$ where $\mathbf I$ is the body's moment of inertia tensor and $\boldsymbol \omega$ is the body's rotation rate with respect to inertial but expressed in body-fixed coordinates. Since a rigid body's inertia tensor is constant in the body-fixed frame, we have $$\left(\frac{d\boldsymbol L}{dt}\right)_\text{body-fixed} = \frac{d(\mathbf I \boldsymbol \omega)}{dt} = \mathbf I \frac{d\boldsymbol\omega}{dt}\tag{3}$$

Combining equations (1), (2), and (3) yields $$\mathbf I \frac{d\boldsymbol\omega}{dt} = \sum_i \boldsymbol \tau_{\text{ext},i} - \boldsymbol \omega\times(\mathbf I \times \boldsymbol \omega)\tag{4}$$

This is Euler's equations of motion for a rigid body. No inertial forces come into play. However, the term $-\boldsymbol \omega\times(\mathbf I \times \boldsymbol \omega)$ is essentially an inertial torque. Just as inertial forces vanish in inertial frames, so does this inertial torque.

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  • $\begingroup$ You very clearly state that L is the inertial system torque, and (3) merely expresses it in body coordinates for use in (1) which applies to the inertial frame hence no fictitious force torques are present. Many explanations do not clearly point this out. Some explanations leave the impression that the torque IN the non-inertial system is used in the right-hand side of (3). Thanks for the clarity! $\endgroup$
    – John Darby
    Jul 15, 2022 at 21:52
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Euler's equations for a rigid body express the change in angular momentum in the inertial frame $\vec L$ with respect to the non-inertial rotating frame coordinates. See relationships (2) and (3) and the associated discussion in the answer by @David Hammen. You use relationship (3) in David's relationship (1) which is in the inertial system, where no fictitious torques are present.

For example, see the explanations in Goldstein, Classical Mechanics or in Symon, Mechanics. This was confusing to me in my first Mechanics physics course.

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