# Rotation of rigid bodies about a fixed axis

The concept of rotation of rigid bodies about an axis has confused me for some time now. When we say rigid,3-dimensional body is rotating about an axis with an angular velocity $$\Omega$$, how do we know that all particles on rigid body are exhibiting circular motion? Also why do the circular paths of each of the particles, lie in the plane perpendicular to the axis of rotation?

• Are you looking for a calculus based answer or a non-calculus based answer? The calculus based answer is short and sweet, but won't be very helpful if you were looking for a non-calculus based answer. Commented Jul 30 at 15:04

Because that's what rigid means. If the particles were moving in different directions relative to each other, the body wouldn't be rigid.

We know it because we have defined it to be so - we are discussing a rigid, rotating body. If all the particles are not moving in a circular pattern around a common axis in a plane perpendicular to the axis, either the body is not rigid (since particles are moving with respect to one another), or the body isn't rotating (since particles are not moving in a circle around an axis). A body with particles moving in non-circular paths relative to one another and not perpendicular to some axis cannot be described as a rigid, rotating body.

arbitrary point $$~p~$$ has the coordinates in the rotating system

$$\mathbf P= \left[ \begin {array}{ccc} \cos \left( \psi \right) &-\sin \left( \psi \right) &0\\ \sin \left( \psi \right) &\cos \left( \psi \right) &0\\ 0&0&1\end {array} \right] \, \left[ \begin {array}{c} x\\ y\\ z\end {array} \right] = \left[ \begin {array}{c} \cos \left( \psi \left( t \right) \right) x -\sin \left( \psi \left( t \right) \right) y\\ \sin \left( \psi \left( t \right) \right) x+\cos \left( \psi \left( t \right) \right) y\\z\end {array} \right]$$

where $$~x~,y~,z~$$ are constant

to proof that the trajectory is circular, we can analyze the distance from the origin at any time t. The trajectory will describe a circle if this distance is constant for all t.

$$~P_z~$$ is constant and $$~P_x^2+P_y^2=x^2+y^2~$$ also constant . Thus, the overall motion is circular with a fixed height z, in 3D space.

the angular velocity is $$~\Omega=\frac {d}{dt}\,\psi(t)~$$

how do we know that all particles on rigid body are exhibiting circular motion?

A circle and a sphere is the only shapes that maintain a constant radius from any fixed point. Since all distances between all particles must be maintained, it is intuitive that each orbit must be circular (if planar, see below).

Also why do the circular paths of each of the particles, lie in the plane perpendicular to the axis of rotation?

This is a bit more tricky to explain, although it is an excellent question to ask. Consider a particle attached to the rotating body and the plane that goes through this particle and is perpendicular to the axis of rotation.

We resolve the velocity of the particle into in-plane and out-of-plane components and note that all the out of plane components are common among all particles (or all are zero), and that the in plane components change with location. For a pure rotation about an axis all the particles have out of plane velocity components that are zero. Visually, the particles move in plane that is perpendicular to the axis of rotation.