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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 ...

C Marius
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1 Answers1

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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

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

John Alexiou
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