0
$\begingroup$

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 ...

$\endgroup$
  • 1
    $\begingroup$ What are your thoughts on this? $\endgroup$ – JMac Feb 21 '17 at 21:52
  • $\begingroup$ I would appreciate a positive feedback, logic and math. Please if possible give that ... $\endgroup$ – C Marius Feb 21 '17 at 22:00
  • $\begingroup$ Is $T_c$ in same coordinates as $I_0$ or $I_B$? $\endgroup$ – ja72 Feb 22 '17 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. $\endgroup$ – ja72 Feb 22 '17 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 '17 at 21:15
2
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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 '17 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 $\dot{\vec{\omega}}$ $\endgroup$ – ja72 Feb 23 '17 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). $\endgroup$ – ja72 Feb 23 '17 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 '17 at 8:38
  • $\begingroup$ Because if $T$ and $\dot{\omega}$ are related with a scalar value. By definition they are parallel. $\endgroup$ – ja72 Feb 23 '17 at 13:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.