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I have the following question regarding an ideal rigid body.

Firstly, is it always true without any exceptions that the angular velocity of any point about any other point on a rigid body is always the same?

If so, what happens in case of precession, in which two different angular velocities are predefined?

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  • $\begingroup$ The question is not clear. To this body you impose whatever movement you want, with whatever angular velocity you want. So, the point under consideration will move with the angular velocity v/r, where r is the distance between the two points, and v is a velocity that YOU impose. $\endgroup$
    – Sofia
    Commented Nov 14, 2014 at 17:56

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Think of rotation as a manifestation of a change in coordinate direction. Affix a coordinate system at any point on a rigid body and it is going to change direction at the same rate regardless of the point location (even at the rotation axis). This is why angular velocity is shared among the entire rigid body.

Mathematically, the rate of change of a 3×3 rotation matrix $R$ is $$ \frac{{\rm d}}{{\rm d}t}R = \vec{\omega} \times R$$ where $\vec{\omega}$ is the angular rotation vector, and $\times$ the vector cross product.

For the second part of your question a general rotation can be decomposed into three rotations about three predefined axes (like precession, nutation and spin), or by a single rotation about an arbitrary axis. In either case, the rotation is going to be about a single axis and you need three parameters to specify it exactly. Either three scalar angle speeds about three predefined axes, or two parameters for the rotation axis, and one for the rotation speed magnitude.

Note also that the general motion of a rigid body is a rotation about some axis in 3D, coupled with a parallel translation along the same axis. The rotation and translation defines a screw in 3D and given the linear and angular velocities of any point on the rigid body, the properties of the motion can be derived.

For example, said rigid body has rotational velocity $\vec{\omega}$ and linear velocity $\vec{v}_A$ at some point A with position $\vec{r}_A$ at some instant. The following properties are defined:

  1. The rotation axis direction is $$\vec{e} = \frac{\vec{\omega}}{|\vec{\omega}|}$$
  2. The rotation magnitude is $$\omega = |\vec{\omega}|$$
  3. The point C on the rotation line closest to the origin is $$\vec{r}_C = \vec{r}_A + \frac{\vec{\omega}\times\vec{v}_A}{\omega^2}$$
  4. The motion pitch (ratio of linear motion to angular motion) is $$ h = \frac{\vec{\omega} \cdot \vec{v}_A}{\omega^2}$$
  5. The linear velocity of point C is $$\vec{v}_C = h\,\vec{\omega}$$

In summary, a single rotation speed, and a linear velocity vector at some point is enough to describe the instantaneous motion of a rigid body both geometrically (rotation line in 3D, pitch and magnitude) and analytically (rigid body transformation of velocity to any other point).

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  • $\begingroup$ How does the coordinate axes change as the rigid body rotates @John Alexiou ? Please ping when replying ^^ $\endgroup$
    – Cantor Dust Drachen
    Commented Feb 24, 2021 at 18:15
  • $\begingroup$ @Buraian the rate of change (velocity vector) of each unit axis $\hat{u}$ is $$ \frac{\rm d}{{\rm d}t} \hat{u} = \vec{\omega} \times \hat{u} $$ SInce the rotation matrix $R$ is composed by the columns of the three unit axis, it also changes in a similar fashion $$ \dot{R} = \vec{\omega} \times R $$ Is this what you are asking? $\endgroup$
    – John Alexiou
    Commented Feb 24, 2021 at 18:33
  • $\begingroup$ Not exactly, it's like the coordinate axes is a mathematical construction we have made on the physical situation of how the objec rotating, my question is simply how are you attaching this coordinate system to the physical object $\endgroup$
    – Cantor Dust Drachen
    Commented Feb 24, 2021 at 18:34
  • $\begingroup$ @Buraian Take a physical body and paint on it a vector. This is one of the three vectors describing the orientation of the body. As the body moves about space you track the components of this vector as $$\hat{u} = \pmatrix{u_x \\ u_y \\ u_z}$$ as they project onto the world coordinate system. The matrix $R$ transforms vectors defined riding on the body to vectors on the common coordinate system. Three unit vectors riding on the body $\hat{u}$, $\hat{v}$ and $\hat{w}$ product the 3×3 matrix $$R = \pmatrix{u_x & v_x & w_x \\ u_y & v_y & w_y \\ u_z & v_z & w_z} $$ $\endgroup$
    – John Alexiou
    Commented Feb 24, 2021 at 18:38
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    $\begingroup$ @Buraian [see edit above]. Common coordinate system is the world coordinate system. It is a coordinate system used to take (distance) measurements on an inertial reference frame. WHen we describe forces for example, the components of the force vector are described along this common coordinate system. The same for the rest of the vectors if we hope to do vector algebra (add them etc). $\endgroup$
    – John Alexiou
    Commented Feb 24, 2021 at 18:42
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I think when a rigid body rotating about any point , the point of rotation remains fixed and all other points of the rigid body always rotate about that fixed point with same angular velocity. enter image description here

Let us suppose a disk rotating about its center with a angular velocity $\omega$ and $r$ is the radius. Now the angular velocity at point A and B must be same. If the velocity are not same then there would be a relative motion between the points A and B of the rigid body which gives rise a Shearing strain due to a tangential stress which appears due the relative motion between A and B. Hence shape of the rigid body would be changed.

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  • $\begingroup$ Isn't there a relative motion between a and b even then?Because their distance from the axis is different? $\endgroup$
    – Schwarz Kugelblitz
    Commented Sep 24, 2019 at 18:57

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