# Clearing Up Confusion for Torques from Fictitious Forces for Motion with Respect to the Center of Mass

The following is my attempt at explaining why torques from fictitious forces do not appear in general in the standard relationship for the change in angular momentum of a system of particles about the center of mass, and specifically in the Euler equations of motion for a rigid body.

Is this explanation reasonable and can it be improved? Responses from those with more knowledge on this topic are appreciated.

Let $$\vec L_{CM}$$ denote the angular momentum of a system of particles with respect to the center of mass, CM, of the system. As shown in physics mechanics textbooks, the change in angular momentum with respect to the center of mass is:

$$\vec M_{ext} = {d\vec L_{CM} \over dt} \tag{1}$$

$$\vec M_{ext}$$ is the total external torque in the inertial frame about the center of mass. Relationship (1) is valid even if the CM is accelerating. No torques from fictitious forces appear in relationship (1). Relationship (1) only considers translation of the CM (a point) of the system with respect to an inertial frame; relationship (1) is not the motion in a frame rotating with respect to the CM.

For a frame centered at the moving CM that is not rotating, it is true that the torque from the fictitious force does not contribute to the change in angular momentum about the CM because the fictitious force from translational acceleration only, $$-m\vec a_{CM}$$, acts at the CM. But in a non-inertial rotating frame centered at the CM, fictitious forces due to rotation can contribute to the angular momentum in that frame. [e.g., see Symon, Mechanics, 2nd edition problem 7.3] Sometimes relationship (1) is interpreted to mean torques from fictitious forces never appear for evaluations with respect to the CM, but in a frame rotating with respect to the CM torques from fictitious forces need to be considered that are not considered in relationship (1).

The Euler equations of motion for a rigid body (a special case of a system of particles) are based on relationship (1) and do not include torques from fictitious forces. Consider the rotational motion of a rigid body about its CM using relationship (1). In the space, S, frame (non-rotating), $$\vec L_{CM_S} = \bf I_{CM_S} \vec \omega$$ is the angular momentum of the body where $$\bf I_{CM_S}$$ is the inertia tensor expressed in the space frame and $$\vec \omega$$ is the angular velocity of the body about the CM. Let $$\vec M_{CM}$$ be the total external torque in the space frame about the CM. Using (1):

$$\vec M_{CM} = {d(\vec L_{CM_S}) \over dt} \tag{2}$$

But $$\bf I_{CM_S}$$ is not constant if the body rotates. We can simplify (2) using body, $$B$$, (rotating) frame coordinates.

As shown in physics mechanics textbooks, using a passive rotation, any vector $$\vec A$$ can be considered as the same vector in two frames, with common origins, one rotating with respect to the other.

$${d\vec A \over dt} = {d^*\vec A \over dt} + \vec \omega \times \vec A \tag{3}$$

The interpretation of relationship (3) is as follows. $${d\vec A \over dt}$$ is the derivative of $$\vec A$$ in one frame. $${d^*\vec A \over dt}$$ is the derivative of the same vector expressed using coordinates of a second (starred) frame. $$\vec \omega \times \vec A$$ accounts for the rotation of the starred frame with respect to the first frame.

We can apply (3) considering the space and body frames in a rigid body. Applying (3) to the space angular momentum vector in (2) we have:

$${d \vec L_{CM_S} \over dt} = {d^* \vec L_{CM_S} \over dt} + \vec \omega \times \vec L_{CM_S} \tag{4}$$

The first term on the right hand side of (4) $${d^* \vec L_{CM_S} \over dt}$$ is the derivative of the space frame angular momentum expressed using body frame coordinates, and the second term accounts for the rotation of the body frame. In the body frame coordinates the space angular momentum is $$\bf I_{CM_B}\vec \omega$$ where $$\bf I_{CM_B}$$ is constant. Therefore, we can express relationship (2) as:

$$\vec M_{CM} = \bf I_{CM_B} \dot {\vec \omega} + \vec \omega \times \bf I_{CM_B} \vec \omega \tag{5}$$

where we have expressed the time derivative of the space frame angular momentum using the body frame coordinates and in the body frame $$\bf I_{CM_B}$$ is constant. The first term on the right-hand side of (5) is not the rate of change of angular momentum in the body frame, since the angular momentum in the body frame, call it $$\vec L_{CM_B}$$ is zero; the body frame angular momentum $$\vec L_{CM_B}$$ is a different vector than the space frame angular momentum $$\vec L_{CM_S}$$. The first term is the rate of change of the space frame angular momentum, expressed using body frame coordinates. As in relationship (1), no torques from fictitious forces appear in (5). If the body axes are chosen the be the principal axes, relationship (5) provides the Euler equations of motion for a rigid body.

The Euler equations

Euler equation inertial frame at the CM \begin{align*} &\frac{d}{dt}\left(\mathbf I\,\mathbf\omega\right)=\mathbf\tau\\ &\Rightarrow\\ &\mathbf I\,\mathbf{\dot{\omega}}+\mathbf{\dot{I}}\,\mathbf\omega=\mathbf\tau\quad ,\text{with}\\ &\mathbf I=\mathbf R\,\mathbf I_B\quad ,\mathbf{\dot{I}}=\mathbf{\dot{R}}\,\mathbf I_B = \underbrace{\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right]}_{\mathbf{\omega}^\times} \,\mathbf R\,\mathbf I_B\quad\Rightarrow\\\\ &\boxed{\quad\mathbf I\,\mathbf{\dot{\omega}}+\mathbf\omega\times\,\mathbf I\,\mathbf\omega=\mathbf\tau\quad~Eq. (1)} \end{align*}

Euler equation body frame at the CM

\begin{align*} &\text{with}\\ &\mathbf I=\mathbf{R}\,\mathbf I_B\,\mathbf R^T\quad,\text{you obtain Eq. (1)}\\\\ &\mathbf{R}\,\mathbf I_B\,\underbrace{\mathbf R^T\,\mathbf{\dot{\omega}}}_{ \mathbf{\dot{\omega}}_B} +\mathbf\omega\times\, \mathbf{R}\,\mathbf I_B\,\underbrace{\mathbf R^T\,\mathbf\omega}_{ \mathbf\omega_B}=\mathbf\tau\\&\text{ multiply from the left with}~\mathbf R^T\quad,\Rightarrow\\\\ &\mathbf I_B\,\underbrace{\mathbf R^T\,\mathbf{\dot{\omega}}}_{ \mathbf{\dot{\omega}}_B} +\mathbf R^T\,\mathbf\omega\times\, \mathbf I_B\,\underbrace{\mathbf R^T\,\mathbf\omega}_{ \mathbf\omega_B}=\mathbf R^T\,\mathbf\tau\\\\ &\boxed{\quad\mathbf I_B\,\mathbf{\dot{\omega}}_B+\mathbf\omega_B\times\,\mathbf I_B\,\mathbf\omega_B=\mathbf\tau_B\quad~Eq. (2)} \end{align*}

Euler equation body frame (parallel to B frame) at body point $$~P$$ \begin{align*} &\text{with }\\ &\mathbf I_P=\mathbf I_B\underbrace{-m\, \left[ \begin {array}{ccc} 0&-z_{{{\it CP}}}&y_{{{\it CP}}} \\ z_{{{\it CP}}}&0&-x_{{{\it CP}}} \\ -y_{{{\it CP}}}&x_{{{\it CP}}}&0\end {array} \right] \, \left[ \begin {array}{ccc} 0&-z_{{{\it CP}}}&y_{{{\it CP}}} \\ z_{{{\it CP}}}&0&-x_{{{\it CP}}} \\ -y_{{{\it CP}}}&x_{{{\it CP}}}&0\end {array} \right]}_{\mathbf I_{CP}}\\ &\mathbf\omega_P=\mathbf\omega_B\\ &\mathbf\tau_P=\mathbf\tau_B+\mathbf r_{CP}\times\mathbf F_F\quad\Rightarrow\\\\ &\boxed{\quad\left(\mathbf I_B+\mathbf I_{CP}\right)\,\mathbf{\dot{\omega}}_P+\mathbf\omega_P\times\, \left(\mathbf I_B+\mathbf I_{CP}\right)\,\mathbf\omega_P=\mathbf\tau_P\quad~Eq. (3)} \end{align*}

• $$~\mathbf I~$$ Intertia tensor
• $$~\mathbf\tau~$$ External torgue
• $$~\mathbf R~$$ transformation matrix between body frame and inertial frame
• $$~\mathbf r_{CP}~$$ body fixed vector from the center of mass to point p
• $$~\mathbf F_F~$$ fictitious forces at the center of mass
• $$~$$ subscript B body frame at the center of mass
• $$~$$ subscript P body frame at body point P
• Very helpful, thanks! Especially (1) in the space frame and (3) in the body frame with P not at the CM. One question: In (1) $I$ is in the space frame and changes in time, and in (2) and (3) $I_B, I_P, and \enspace I_{CP}$ are in the body frame and constant in time; correct? Commented Jul 28, 2022 at 16:00
• @JohnDarby correct ! and P frame is parallel to B frame
– Eli
Commented Jul 28, 2022 at 16:02
• Thanks. P frame is parallel to B, I see. Your answer is very clear and concise and clears up the confusion I have with other responses to similar questions on this exchange. Commented Jul 28, 2022 at 16:04