How to apply a screw motion for this case? I'm completely new to the notion of a screw motion. As far as I know, it carries out the the rotation and translation simultaneously in comparison with the homogenous transformation matrix that carries out the process separately. Take a look this picture

The homogeneous matrix is given by (i.e. I have no problem with this motion)
$$
T_1^0 = \begin{bmatrix} R & p \\ 0 & 1 \end{bmatrix}
$$
where $R_{z,45^\circ}$ and $p=[5\ 4\ 0]^T$ are the rotation matrix and translation vector, respectively. To my understanding, the screw motion selects the rotation axis at a particular point in the space. In the planar case, it points outward/inward the page in the above picture. I don't know how can I choose this point and how much I should rotate. How can I apply same motion in the above picture so that the rotated square should end up exactly in the same pose using screw motion assuming the square initially at the origin of frame 0 with zero rotation?
 A: Screw motion is best understood in terms of velocity kinematics. You are describing the position kinematics, which is fine to represent using a transformation matrix $T_1^0$. What you need is a point $\vec{p}$ and a rotation matrix $R$.
Similarly to show the kinematic relationship of a moving body as a screw motion you need the velocity of a known point $\vec{v}_P$ and the rotational velocity of the body $\vec{\omega}$.
The body is moving with screw motion in general. A screw motion is a line in space the body is rotating about, together with a parallel translation along this axis. It is described by the following properties

*

*Axis Direction from the rotational velocity vector $$ \vec{\rm dir} = \frac {\vec{\omega}}{| \vec{\omega} |} $$


*Rotation magnitude from the rotational velocity vector $$ {\rm mag} = | \vec{\omega} | $$


*Point on the axis closest to point P $$ \vec{\rm pos} = \frac{ \vec{\omega} \times \vec{v}_P } { | \vec{\omega} |^2 } $$


*Screw pitch (ratio of translation to rotation) $$ {\rm pitch} = \frac{ \vec{\omega} \cdot \vec{v}_P } { | \vec{\omega} |^2 }$$
You can completely construct the motion of the body from these four quantities

*

*Rotational velocity $$ \vec{\omega} = ({\rm mag})\; (\vec{\rm dir})$$

*Translational velocity of point P $$ \vec{v}_P = \vec{\omega} \times (\vec{\rm pos}) + ({\rm pitch})\, \vec{\omega} $$
Note that since ${\dot R} = \vec{\omega} \times R$ the derivative of the transformation matrix is
$$ \dot{T}_1^0 = \begin{bmatrix}\vec{\omega}\times R & \vec{v}_{P}\\
0 & 0
\end{bmatrix}$$
which is how the motion kinematics relate to the transformation matrix.
A: The green marks indicate a line of material points. Start by drawing the red line between the two rectangle centers. Find the midpoint of this line at $\sqrt{41}/2$. Draw the blue perpendicular bisector. The rotation point must be somewhere along this line.

The task now is to find the point on the line where the angles of the green lines with the two purple lines coming out of the point to the rectangle centers are equal. It looks to be somewhere around here:

