How to find angular velocity vector of a three dimensional rigid body given velocity of three non-collinear points on the same?

Question

The position of a three-dimensional rigid body is completely defined by specifying position vectors of three non-collinear points on the body.

Similarly, one can define the motion of the rigid body by defining the velocity of the centre-of-mass and the angular velocity vector.

However, in my case, I know the linear velocity of three points on a rigid body. How can I compute the angular velocity vector from this information?

*angular velocity vector and the velocity of centre of mass

Answer 1

You have to find the vector $\boldsymbol \omega$ that satisfies the three equations $\dot{\boldsymbol x}_i=\boldsymbol\omega\times{\boldsymbol x}_i$ for $i=1,2,3\,.$ If you know the time dependent matrix $B(t)$ that generates your rotated vectors $\boldsymbol x_i(t)$ by $\boldsymbol x_i=B(t)\boldsymbol{Q}_i$ where $\boldsymbol {Q}_i$ are three linearly independent fixed vectors in the body you can just calculate the skew symmetric matrix $$ \dot B\,B^\top=\left(\begin{array}{ccc}0&-\omega_3&\omega_2\\\omega_3&0&-\omega_1\\-\omega_2&\omega_1&0\end{array}\right) $$ and read out the components of the angular velocity from that.

Answer 2

Say that we have velocities at three points, $\vec r_1$, $\vec r_2$, and $\vec r_3$, $\vec v_1$, $\vec v_2$, and $\vec v_3$ at a certian time. What is the velocity of the cemter of mass (COM), $V_c$, and rotation of rigit body about the COM, $\omega$?

Lets list the 3 equations:

\begin{align} \vec v_1 &= \vec V_c + \vec \omega \times \vec r_1;\tag{1}\\ \vec v_2 &= \vec V_c + \vec \omega \times \vec r_2;\tag{2}\\ \vec v_3 &= \vec V_c + \vec \omega \times \vec r_3;\tag{3}\\ \end{align}

The velocity of COM can be found by insert inner product of $\vec r_i$ to each equation. \begin{align} \vec r_1\cdot\vec v_1 &= \vec r_1\cdot\vec V_c + \vec r_1\cdot(\vec \omega \times \vec r_1) = V_{c1} ;\\ \vec r_2\cdot\vec v_2 &= \vec r_2\cdot\vec V_c + \vec r_2\cdot(\vec \omega \times \vec r_2) = V_{c2} ; \\ \vec r_3\cdot\vec v_3 &= \vec r_3\cdot\vec V_c + \vec r_3\cdot(\vec \omega \times \vec r_3) = V_{c3} ;\\ \end{align}

We solved $\vec V_c$ by obtained 3 components of along 3 independent directions.

We try by substrating the aboce equations to remove the velocity of the COM. \begin{align} \vec v_1 -\vec v_2 &= \vec \omega \times \left(\vec r_1 - \vec r_2\right);\\ \vec v_2 -\vec v_3 &= \vec \omega \times \left(\vec r_2 - \vec r_3 \right);\\ \vec v_3 - \vec v_1 &= \vec \omega \times \left(\vec r_3 - \vec r_1\right);\\ \end{align}

insert inner product od each position vector from the right ahnd side.

\begin{align} \left(\vec v_1 -\vec v_2\right)\cdot \vec r_1 &= \vec \omega \times \left(\vec r_1 - \vec r_2\right)\cdot \vec r_1 =\vec \omega \cdot\left( \vec r_1 - \vec r_2\right)\times \vec r_1 = \vec \omega \cdot \left( \vec r_1 \times \vec r_2\right);\\ \left(\vec v_2 -\vec v_3\right)\cdot \vec r_2 &= \vec \omega \times \left(\vec r_2 - \vec r_3 \right)\cdot \vec r_2 =\vec \omega \cdot \left(\vec r_2 - \vec r_3 \right)\times \vec r_2 =\vec \omega \cdot \left(\vec r_2 \times \vec r_3 \right);\\ \left(\vec v_3 - \vec v_1\right)\cdot \vec r_3 &= \vec \omega \times \left(\vec r_3 - \vec r_1\right)\cdot \vec r_3 = \vec \omega \cdot \left(\vec r_3 - \vec r_1\right)\times \vec r_3 = \vec \omega \cdot \left(\vec r_3 \times \vec r_1\right);\\ \end{align} The last set of 3 equations gives the components of $\vec \omega$ in 3 directions.

Answer 3

I think I got a compact solution:

Let $ \vec{v}_{21} = \vec{v_2} - \vec{v_1} $ and so on. We get two independent equations,

$ \begin{align} \vec{v}_{21} &= \vec{\omega} \times \vec{r}_{21}\\ \vec{v}_{31} &= \vec{\omega} \times \vec{r}_{31}\\ \vec{v}_{31}\times\vec{v}_{21} &= \vec{v}_{31} \times ( \vec{\omega} \times \vec{r}_{21})\\ &=(\vec{v}_{31}\cdot\vec{r}_{21})\vec{\omega} - (\vec{v}_{31}\cdot\vec{\omega})\vec{r}_{21}\\ &=(\vec{v}_{31}\cdot\vec{r}_{21})\vec{\omega}\\ \vec{\omega} &= \frac{\vec{v}_{31}\times\vec{v}_{21}}{\vec{v}_{31}\cdot\vec{r}_{21}} \end{align} $

(We know $ \vec{v}_{31}$ and $\vec{\omega}$ are orthogonal.)

Yet to find out $\vec{V}_c$ though.

@ytlu, shouldn't equations (1),(2) and (3) be

$ \vec{v}_1 = \vec{V}_c + \vec{\omega} \times (\vec{r}_1 - \vec{r}_c) $