Pure Rotation – Dimitri Karagiannis

The animation below shows a rigid body in pure rotation.

Notice that every point on this rigid body moves in a circular path around a single point, which has no motion at all. This is, by definition, necessary to have pure rotation. This can be made clearer by tracing the paths of two arbitrary points, and identifying the point which experiences no motion.

If we also draw vectors $\vec{r}_A , \vec{r}_B , \vec{r}_{B/A}$ , we can clearly see that the relative position relationship is valid,

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

Note that as the rigid body moves, the relative position vector $\vec{r}_{B/A}$ does not change its magnitude. This is a requirement of rigid bodies, because, by definition of a rigid body, it does not change its shape. Therefore, the distance between two points on a rigid body cannot change.

This vector does, however, very clearly change its direction. It is changing continuously, because the orientation of the rigid body is changing continuously. One valid measure of the orientation, or the angular position of the rigid body, is shown by the variable $\theta$ in the animation. Note that the angular position can be measured from any fixed line to any vector between two points on the rigid body. In this case, $\theta$ is the angle between the fixed horizontal and the position vector $\vec{r}_A$ , or $\vec{r}_{A/O}$ .

The velocity of any point on the particle is, necessarily, in the direction tangent to the path. Because the body is in pure rotation, the path of every point is circular, with the fixed point as its center. Therefore the velocity is always tangent to this circular path. This can be seen in the animation below.

Note that, unlike rigid bodies in pure translation, the velocity vectors at different points are not the same. They have different magnitudes, and different directions. The further from the fixed point a point is, the larger the magnitude of the linear velocity at that point will be. Therefore, as you can see from the animation, the linear velocity at point A, $\vec{v}_A$ has a larger magnitude than the linear velocity at point B, $\vec{v}_B$ , because point A is further from the fixed point O than point B is.

Please note that there is a difference between Linear Velocity, $\vec{v}$ , and Angular Velocity $\vec{\omega}$ . In general, the linear velocity is different for every single point on a rigid body (as can be seen above). The angular velocity, however, is the rate of change of the angular position, $\vec{\omega} =\frac{d}{dt}\vec{\theta}$ , and is therefore not specific to any point, but to the body as a whole. Also note that for planar problems (2D), there is only an angular position in one direction, $\vec{\theta} = 0\hat{i} + 0\hat{j} + \theta\hat{k}$ , and hence it can just be considered as a scalar: the angular position is simply a scalar value $\theta$ . Therefore the angular velocity will also simply be a scalar value, $\omega = \frac{d}{dt} \theta$ .

However, if you want to use the cross product to calculate the linear velocity via the equation:

$\vec{v} =\frac{d\vec{r}}{dt} = \vec{\omega} \times \vec{r}$ ,

you need to be able to turn the scalar angular velocity back into a vector (i.e., $\vec{\omega} = 0\hat{i} + 0\hat{j} + \omega \hat{k}$ ).

Finally, we can also consider the linear and angular acceleration of the rigid body. The linear acceleration of points A and B are shown in the animation below:

Note that the total linear acceleration is a combination of its two components: linear acceleration in the normal direction (sometimes called centripetal acceleration, points toward the center of rotation, $\vec{a}_n = \vec{\omega}\times (\vec{\omega}\times \vec{r})$ ; magnitude in 2D: $a_n= \frac{v^2}{r} = r \dot{\theta}^2 = v \dot{\theta} = r \omega^2= v\omega$ ), and linear acceleration in the tangential direction (points tangent to the path in either direction, $\vec{a}_t = \vec{\alpha}\times \vec{r}$ ; magnitude in 2D: $a_t= \dot{v} = r\ddot{\theta} = r\alpha$ .)

Remember that there is a difference between Linear Acceleration, $\vec{a}=\frac{d\vec{v}}{dt}$ , which is different at every point on a rigid body, and Angular Acceleration, $\vec{\alpha}$ , which is a measure of the rate of change of the angular velocity for the entire rigid body. In 2D planar motion, it has a component in only one direction, $\vec{\alpha} = 0\hat{i}+0\hat{j}+\alpha\hat{k}$ , and can therefore be considered as a scalar value, $\alpha = \frac{d\omega}{dt}$ . However, you need the angular acceleration as a vector in order to use the following equation to calculate linear acceleration:

$\vec{a} = \vec{\alpha}\times \vec{r} + \vec{\omega}\times (\vec{\omega}\times \vec{r})$

_{All animations created by Dr. Dimitri Karagiannis}