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There is something that has been bugging me for a while.

Suppose I have a mass $m$ attached to a rod of length $l=1$ and with no mass and that the axes that we choose are not the inertial axis.

Suppose everything is initially static.

Then, suppose we apply a moment about the x axis.

Considering the formulas, I will have a angular acceleration about the $y$ axis $\alpha_y=M_x/I_{xy}$ with $I_{xy}=m* cos(10°) sin(10°)$

But from the drawing it seems that $m$ would turn about the $x$ axis rather than about the $y$ axis, no? Where am I mistaking?

enter image description here

Edit :

So if I want a rotation about the x-axis, I should apply a moment about the x and y axis or the x and z axis? Intuitively, I would say x and z axis but since I have the equation $M_y=I_{yx} \alpha_x$ it seems the physic says that I should apply a tork about the x and y axis. Which one is correct?

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1 Answer 1

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Your angular acceleration will be about the same axis as the moment. Due to $$\vec M=I\vec\alpha$$ This is similar to linear acceleration being in the same direction as the force, due to $$\vec F =m\vec a$$

These are vector relationships, and since $I$ (as well as $m$) are not vectors, these equations will cause the acceleration vector direction to be to be the same. I believe your subscripts are simply wrong.

Answer to comment:

@roi_saumon It should be like this, right: $$\begin{bmatrix} I_{xx} & I_{xy} & 0 \\ I_{yx} & I_{yy} & 0 \\ 0 & 0 & I_{zz} \end{bmatrix}$$ Multiplying the $\alpha$ vector: $$\begin{bmatrix} M_{x} \\ 0 \\ 0\end{bmatrix}=\begin{bmatrix} I_{xx} & I_{xy} & 0 \\ I_{yx} & I_{yy} & 0 \\ 0 & 0 & I_{zz} \end{bmatrix}\cdot \begin{bmatrix} \alpha_{x} \\ \alpha_{y} \\ \alpha_z \end{bmatrix}=\begin{bmatrix} \alpha_{x}I_{xx}+ \alpha_{y}I_{xy}\\ \alpha_{x}I_{yx}+ \alpha_{y}I_{yy} \\ \alpha_zI_{zz} \end{bmatrix}$$

Here we see that $\alpha_z=0$ and $\alpha_{x}I_{yx}+ \alpha_{y}I_{yy}=0$. These must be true for the matrix equation to be fulfilled. From the latter we get:

$$\alpha_{x}I_{yx}+ \alpha_{y}I_{yy}=0\quad\Leftrightarrow \quad\alpha_{y}=-\alpha_x\frac{ I_{yx}}{I_{yy}}$$

Inserting this in the matrix equation, we have:

$$\begin{bmatrix} M_{x} \\ 0 \\ 0\end{bmatrix}=\begin{bmatrix} \alpha_{x}I_{xx} -\alpha_x\frac{ I_{yx}}{I_{yy}}I_{xy}\\ 0 \\ 0 \end{bmatrix}=\begin{bmatrix} \alpha_{x}\left(I_{xx} -\frac{ I_{yx}^2}{I_{yy}}\right)\\ 0 \\ 0 \end{bmatrix}$$

Now we are back to the $\vec M=I\vec\alpha$ situation. There is only $\alpha_x$ an $M_x$ present in this. Maybe the moment-of-inertia can be shorted down even further.

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  • $\begingroup$ But if the inertia tensor in my case is $ \begin{bmatrix} 0 & I_{xy} & 0 \\ I_{yx} & 0 & 0 \\ 0 & 0 & I_{zz} \end{bmatrix} $ I would have : $ \begin{bmatrix} M_x \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix} I_{xy} \alpha_y \\ I_{yx} \alpha_x \\I_{zz} \alpha_y \end{bmatrix} $ which shoes that $\alpha_y=\frac{M_x}{I_{xy}}$ no? $\endgroup$
    – roi_saumon
    Commented Nov 10, 2017 at 19:56
  • $\begingroup$ Can somebody explain me the issue here? Why the upper reasoning is not correct? $\endgroup$
    – roi_saumon
    Commented Nov 12, 2017 at 23:11
  • $\begingroup$ @roi_saumon Why are most of the elements in the inertia tensor zero? $\endgroup$
    – Steeven
    Commented Nov 13, 2017 at 6:21
  • $\begingroup$ $@$Steeven True, there should be $I_{xx}$ and $I_{zz}$ as well $\endgroup$
    – roi_saumon
    Commented Nov 13, 2017 at 16:31
  • $\begingroup$ @roi_saumon Those $I_{xx}$ and $I_{yy}$ do make quite a difference. See the edit to the answer. Only $\alpha_x$ is present. $\endgroup$
    – Steeven
    Commented Nov 15, 2017 at 7:23

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