Introduction: Suppose I have a rigid body with the inertia matrix in the initial position $I_0$ . If the body fixed coordinate axis rotate with matrix $R$ then it's inertia matrix will be $I_B = R^T I_0 R$. Suppose the eigenvectors of $I_0$ are $\vec{v}_1, \vec{v}_2, \vec{v}_3$. Being principal axis they are orthgonal. Then $R^T v_1$ is an eigenvector of $I_B$.

Question: if a torque $T_c$ parallel with $R^T v_1$is applied at the center of mass, will the body have an angular velocity also parallel with $R^T v_1$ ? That is: will $\omega \times T_c = 0$ knowing that $I_B T_c = \lambda_1 T_c$ ?

I think the answer is yes, but I am unable to prove it ... I know that $T_c = \omega \times (I_B \omega) + I_B \dot{\omega}$ I tried to show $\omega \times T_c = 0$ but $\omega \times T_C = \omega (\omega^T I_B \omega) - I_B\omega (\omega^T \omega) + \omega \times I_B\omega$. I do not know how to proceed ...

Of course, if $\omega$ is a principal axis, then $T_c = I_B \dot{\omega}$ is also an eigenvector of $I_B$, but I am interested if the converse is true ...

  • 1
    $\begingroup$ What are your thoughts on this? $\endgroup$
    – JMac
    Feb 21, 2017 at 21:52
  • $\begingroup$ I would appreciate a positive feedback, logic and math. Please if possible give that ... $\endgroup$
    – C Marius
    Feb 21, 2017 at 22:00
  • $\begingroup$ Is $T_c$ in same coordinates as $I_0$ or $I_B$? $\endgroup$ Feb 22, 2017 at 15:29
  • $\begingroup$ It is easy to show that when the rotation axis is along a principal direction then the angular momentum is parallel to said axis. And torque is the rate of change of angular momentum. $\endgroup$ Feb 22, 2017 at 19:44
  • $\begingroup$ $T_C$ is in the same coordinate as $I_B$ ... I just realised that my notations are a little confusing ... Yes the torque is the rate of change of angular momentum, but what I know is that $\dot{L}$ is a principal axis. Does this always mean that $L$ is also a principal axis? If so, why? $\endgroup$
    – C Marius
    Feb 22, 2017 at 21:15

1 Answer 1


The the equations of motion on the body frame are

$$ \vec{T}_B = \mathrm{I}_B \dot{\vec{\omega}}_B + \vec{\omega}_B \times \mathrm{I}_B \vec{\omega}_B $$

or in component form

$$ \begin{pmatrix} T_1 \\ T_2 \\ T_2 \end{pmatrix} = \begin{vmatrix} I_{1} & 0 & 0 \\ 0 & I_{2} & 0 \\ 0 & 0 & I_{3} \end{vmatrix} \begin{pmatrix} \dot{\omega}_1 \\ \dot{\omega}_2 \\\dot{\omega}_2 \end{pmatrix} + \begin{vmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{vmatrix} \begin{vmatrix} I_{1} & 0 & 0 \\ 0 & I_{2} & 0 \\ 0 & 0 & I_{3} \end{vmatrix} \begin{pmatrix} {\omega}_1 \\ {\omega}_2 \\ {\omega}_2 \end{pmatrix} $$

where $1$, $2$ and $3$ are the principal directions. I think your question is what happens if a torque is applied along a principal direction? The problem is that only angular acceleration depends on torque and angular velocity is usually defined in direction by the kinematics. So you can't ask If I apply a torque, what will the speed be?

In any case suppose the general case where $T_1 \neq 0$ and $T_2 = T_3 = 0$

$$\begin{aligned} T_1 & = I_1 \dot{\omega}_1 + (I_3-I_2) \omega_2 \omega_3 \\ 0 & = I_2 \dot{\omega}_2 + (I_2-I_3) \omega_1 \omega_3 \\ 0 & = I_3 \dot{\omega}_3 + (I_2-I_1) \omega_1 \omega_2 \end{aligned} $$

So you are trying to understand what motions obey the above equations of motion. Assuming the general case of $I_1 \neq I_2 \neq I_3$ you see that the torque along $1$ does not affect the angular acceleration along $2$ and $3$. So a torque applied along a principal axis, will accelerate the body also along the principal axis only if the body is already rotating along the same axis already. Only when $\omega_2 = \omega_3 =0$ you decouple the system to

$$\begin{aligned} T_1 & = I_1 \dot{\omega}_1\\ \dot{\omega}_2 & = 0\\ \dot{\omega}_3 & = 0 \end{aligned} $$

  • $\begingroup$ First of all: thank you so much for your answer! I will use the same meaning of $R$ as you did (I usually use $R^T $ in the sens you seem to use $R$), so: I think the rotation being about $\omega$ axis it is true that $\omega_c = R\cdot \omega_c = R^T \cdot \omega_c$ and the same for $\dot{\omega}_c$ but yes this will also yield $R T_B = R (\omega_B \times (I_B \omega_B) + I_B \dot{\omega}_B)$ and yes if $\omega_C = \omega_B$ is a principal axis then $T_B = I_B \dot{\omega}_B$. What I am not able to prove to my self, is your last proposition: "In addition if the torque ..." why is this? Why ? $\endgroup$
    – C Marius
    Feb 22, 2017 at 21:05
  • $\begingroup$ If $\dot{\vec{\omega}}$ is along principal axis then $\vec{T} = \mathrm{I} \dot{\vec{\omega}} = \lambda \dot{\vec{\omega}}$ which is obviously parallel to $\dot{\vec{\omega}}$ $\endgroup$ Feb 23, 2017 at 2:02
  • $\begingroup$ Also, it is common convention to represent the rotation matrix $\mathrm{R}$ as in $(\mbox{local}) \rightarrow (\mbox{global})$ transformation. That is, it transforms local vectors (body coordinates) to global vectors (world coordinates). $\endgroup$ Feb 23, 2017 at 2:05
  • $\begingroup$ Yes, if assuming $\omega$ is a principal axis then $T$ is also a principal axis, that I wrote inside the question, but I am interested in the converse ... How can I prove that if $T$ is along a principal axis, then also $\omega$ is ... $\endgroup$
    – C Marius
    Feb 23, 2017 at 8:38
  • $\begingroup$ Because if $T$ and $\dot{\omega}$ are related with a scalar value. By definition they are parallel. $\endgroup$ Feb 23, 2017 at 13:23

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