-2
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Nov 10 '17 at 19:56
  • $\begingroup$ Can somebody explain me the issue here? Why the upper reasoning is not correct? $\endgroup$ – roi_saumon Nov 12 '17 at 23:11
  • $\begingroup$ @roi_saumon Why are most of the elements in the inertia tensor zero? $\endgroup$ – Steeven Nov 13 '17 at 6:21
  • $\begingroup$ $@$Steeven True, there should be $I_{xx}$ and $I_{zz}$ as well $\endgroup$ – roi_saumon Nov 13 '17 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 Nov 15 '17 at 7:23

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.