Complex Motion – Dimitri Karagiannis

Below is an animation of a rigid body experiencing complex motion in a plane. It is changing its orientation, and there are no fixed points, the path traced by any given point is not circular.

Note that complex motion is any simultaneous combination of pure translation and pure rotation. If we trace the paths of two points, we can see that they are neither parallel (as they would be for pure translation), nor circular (as they would be for pure rotation).

Note the path of point A is the same path traced out by the pure translation example, and also observe that the path of B relative to A is the same circular back-and-forth motion of the pure rotation example. This is made more clear by attaching the “camera” to point A, so that as A moves, the viewing window moves with it, making it appear fixed. Note how B moves in pure rotation with respect to A. Note the similarity of the relative position vector $\vec{v}_{B/A}$ in this animation, and the position vector $\vec{v}_B$ from the pure rotation example.

See the animation below as well, it is identical to the one above, but does not trace the absolute paths of the two points. Because we cannot see the environment moving in the background, it appears as if A is fixed, and the rigid body is moving in pure rotation around it.

If we display the velocity vectors of A and B, it is clear that, as always, the velocity of each point is tangent to each point’s respective path. Because each point follows a different path, naturally, the two points have different velocities at any given time.

The relative velocity equation, however, must hold, no matter the type of motion. Therefore,

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

The following animation shows the absolute velocity of points A and B, and also shows the velocity of B from A, tacked onto point B. To show that the relative velocity equation is valid, these three vectors are also plotted off to the side, where it is clear that the velocity of B is the sum of the velocity of A plus the velocity of B with respect to A.

Notice that the velocity vector, $\vec{v}_{B/A}$ , is always perpendicular to the vector $\vec{r}_{B/A}$ (not shown, but it is the position vector starting at A and pointing at B). This must always be the case, because the velocity of B with respect to A is the velocity of B from the perspective of A, as if A were a fixed point. And, if A were a fixed point, then B would be in pure rotation around it, which would mean it would have a circular path around A, which means that its velocity would be tangent to that perpendicular path, which is perpendicular to the vector $\vec{r}_{B/A}$ . This is a critically important point! To repeat: the relative velocity vector $\vec{v}_{B/A}$ must always be perpendicular to the relative position vector $\vec{r}_{B/A}$ .

This is a critical point to remember; to make it clearer, see the animation below, which shows the vector $\vec{r}_{B/A}$ , so that it is clear that the relative position vector $\vec{v}_{B/A}$ is always perpendicular to it.

Instantaneous Center

Remember that complex motion is a simultaneous combination of pure translation and pure rotation. Only rigid bodies in pure rotation have a fixed “center” point, around which all points move with a circular path. However, all rigid bodies experiencing complex motion have an “instantaneous” center point, around which all points move with a circular path. (Only bodies moving in Pure Translation have no instantaneous center.)

This fact leads us to define the concept of the Instantaneous Center of Zero Velocity, alternatively it could be called the Instantaneous Center of Rotation, or simply just the Instantaneous Center, (IC). If it the location of the can be identified, the process of problem solving can be greatly simplified. The reason for this is because the velocity equation for a body in pure rotation can be used with respect to the IC:

$\vec{v}_A=\vec{\omega}\times\vec{r}_{A/IC}$ .

Or, in words, the velocity of any point is equal to the cross product of the angular velocity vector of the rigid body and the position vector going to that point (A) from the instantaneous center (IC).

Recalling that the result of a cross product between two vectors is another vector in the direction perpendicular to the two vectors, $\vec{v}_A$ must be perpendicular to the plane created by $\vec{\omega}$ and $\vec{r}_{A/IC}$ . If the body is rotating in the 2D plane, $\vec{\omega}$ points out of the 2D plane, and therefore we can say more simply that, $\vec{v}_A$ must be perpendicular to $\vec{r}_{A/IC}$ . This must also be true of any other point on the rigid body, i.e. $\vec{v}_B$ must be perpendicular to $\vec{r}_{B/IC}$ , $\vec{v}_C$ must be perpendicular to $\vec{r}_{C/IC}$ , etc.

This fact makes locating the instantaneous center very simple if one happens to know merely the direction of the velocity of two points on a rigid body. One needs only to draw a perpendicular line to each of those two known velocity directions — the instantaneous center must be at their intersection. See the animation below:

Note that for velocity in the plane, the scalar equation will also hold:

$v_{A}=\omega r_{A/IC}$ ,

which will also be true for every other point, i.e. $v_{B}=\omega r_{B/IC}$ , $v_{C}=\omega r_{C/IC}$ , etc. From this we can observe that the location of the instantaneous center can also be determined if one knows the velocity of only one point, but also knows the angular velocity. This method was reviewed in class and in the handouts given in class, along with a method for finding the IC when the known velocity vectors of two points are parallel (but not equal).

This fact also allows us to make one more observation, which is that, rearranging the equation to solve for $r_{A/IC}$ (the distance between the point and the IC), we have

$r_{A/IC} = \frac{v_{A}}{\omega}$ ,

which means that the smaller the absolute velocity of a point is, or the larger the angular velocity is, the closer the IC will be to the point of interest. That can be observed from the animation above — the IC approaches much closer to the rigid body whenever the angular velocity $\omega$ is relatively large.

The path traced by the instantaneous center is completely determined by the motion of the rigid body. The shape of the path is less interesting than being able to determine the location of the IC at any given time, or, ideally, if you are able to analytically solve for the location of the IC at every point in time. The path traced out by the instantaneous center for this rigid body is shown below. Note that viewing window has been zoomed out so that more of the path can be seen, and therefore the rigid body appears much smaller.