0
$\begingroup$

This question already has an answer here:

I have I have some objects assembled like this :

enter image description here

The inertia tensor would be :

$$I=I_1+I_2+I_3-m_1 \,\tilde{r}_{01}\,\tilde{r}_{01}-m_3\,\,\tilde{r}_{03}\,\tilde{r}_{03}$$

Where :

$$\tilde{r}_{01}=\begin{bmatrix} 0 & -z & 0 \\ z & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

and $$\tilde{r}_{03}=\begin{bmatrix} 0 & +z & 0 \\ -z & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

According to this.

Detailed notations here.

But what is the inertia if -

  • I rotate the planar object on the very top(everything with it) around $Z$-axis
  • Just rotate the rod and sphere (except the planar object), just like a pendulum. Let angle with $X$-axis is $\alpha$ and angle with $Y$-axis is $\beta$
  • Do both together
$\endgroup$

marked as duplicate by G. Smith, Jon Custer, Aaron Stevens, Kyle Kanos, ZeroTheHero Oct 13 at 15:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @G.Smith maybe if you take some time, you can see I asked that question. And that has an accepted answer, Why would I ask the same question ? Maybe it has something new. And yes it does, some calculation about angles . Can you help with it, please ? $\endgroup$ – Maifee Ul Asad Sep 30 at 17:22
  • $\begingroup$ The intertia tensor about which point? The yellow circle in your figure? $\endgroup$ – SuperCiocia Sep 30 at 17:30
  • $\begingroup$ About the dot you can see in the planar object. @SuperCiocia $\endgroup$ – Maifee Ul Asad Sep 30 at 17:43
  • 1
    $\begingroup$ You can rotate the inertia tensor like this $I_{i}\rightarrow S^{T}I_{i}S$ S is orthonormal rotation matrix , fBut to add the inertia tensor after rotation , the coordinate systems must be parallel $\endgroup$ – Eli Sep 30 at 18:13
1
$\begingroup$

The inertia tensor obeys the congruent transformation from the local coordinates to the world coordinates.

If you have a 3×3 rotation matrix $\mathbf{R}$ then you have

$$ \mathbf{I}_{\rm world} = \mathbf{R} \, \mathbf{I}_{\rm body} \mathbf{R}^\top $$

So the combined inertia would be

$$ \mathbf{I} = \mathbf{R} \, \left(\mathbf{I}_1 +\mathbf{I}_2 + \mathbf{I}_3 \right) \mathbf{R}^\top - m_1 \overline{\boldsymbol{r}}_1 \overline{\boldsymbol{r}}_1 - m_3 \overline{\boldsymbol{r}}_3 \overline{\boldsymbol{r}}_3$$

You must make sure the vectors $\boldsymbol{r}_n$ point to the center of mass in the world coordinate system after the body is rotated.

See also this answer to a similar question.

$\endgroup$

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