Pure Translation – Dimitri Karagiannis

See the video below for a rectangular rigid body in pure translation. Note that it is not moving in a straight line (rectilinear), but along a curved path (curvilinear). Nonetheless, the rigid body experiences pure translation because its orientation does not change.

Let us label two points on this rigid body, A and B, at the bottom left and top right corner, respectively. If we also draw the vector $\vec{r}_{B/A}$ , we can observe two things about this vector.

Its magnitude never changes. This is necessarily true for two points on a rigid body, no matter what type of motion it has. Why? Because, by definition, a rigid body does not change shape, which means the distance between any two points must always be constant. Therefore the magnitude of the vector connecting two points is always constant.
Its direction never changes. This is a result of the fact that it is in pure translation. If it was not in pure translation, the direction of the vector $\vec{r}_{B/A}$ would be changing continuously, because the orientation of the rigid body would be changing continuously. The angle $\theta$ in the video below could be used to measure the angular position (orientation) of the rigid body. Once again, note that this angle does not change.

We can also notice from this animation another key feature of pure translation: the paths of any two points are parallel. Whether those paths be rectilinear (straight) or curvilinear (curved), they are parallel.

Another observation can be made if we draw the velocity vectors for each point. Remember, the relative position equation:

$\vec{r}_B = \vec{r}_A + \vec{r}_{B/A}$

For rigid bodies in pure translation, we have already observed that $\vec{r}_{B/A} = \vec{constant}$ , and therefore $\vec{v}_{B/A}=\vec{0}$ , and therefore the relative velocity equation is:

$\vec{v}_{B} = \vec{v}_{A}$

This is visualized in the animation below. Note also that the velocities are always tangent to the path.

Furthermore, the acceleration at every point should also be the same for rigid bodies in pure translation. Taking the derivative of the relative position equation of a rigid body in pure translation, we get:

$\vec{a}_B = \vec{a}_A$

See the animation below, and also observe that the acceleration has components in the tangential and normal directions at all times.

A complete animation, showing all of the above information, is also provided below:

_{All animations created by Dr. Dimitri Karagiannis}