# In a rigid body rotating in 3D around the origin, is the norm of the linear velocity of each particle constant?

I was trying to prove that kinetic energy is constant in a rigid body undergoing some arbitrary rotation around the origin in a 3D space, and I ended up proving a stronger statement: that the norm of the linear velocity of any one particle is constant.

Is this true?

UPDATE:

I'm assuming the total linear momentum of the body is zero, and therefore its center of mass stays at the origin.

Here is my proof:

Given that the rigid body has some angular velocity $$\boldsymbol\omega$$, which can also be represented by the skew-symmetric matrix $$\bf{W}$$, and a particle of the body with position $$\bf{x}$$, $$\bf{v}$$, and acceleration $$\bf{a}$$, I have:

$$\begin{array}{l} {\boldsymbol\omega} \times {\bf{x}} = {\bf{Wx}} \\ {\bf{\dot x}} = {\bf{Wx}} = {\bf{v}} \\ {\bf{\ddot x}} = {\bf{W\dot x}} = {\bf{WWx}} = {\bf{a}} \\ \frac{d}{{dt}}\frac{1}{2}{\bf{v}} \cdot {\bf{v}} = {\bf{v}} \cdot {\bf{a}} = {\bf{x}}^{\bf{T}} {\bf{W}}^{\bf{T}} {\bf{W\dot x}} = {\bf{x}}^{\bf{T}} {\bf{W}}^{\bf{T}} {\bf{WWx}} \\ {\bf{W}}^{\bf{T}} {\bf{W}} = {\boldsymbol\omega}^{\bf{T}} {\boldsymbol\omega\bf{I}} - {\boldsymbol\omega\boldsymbol\omega}^{\bf{T}} \to {\bf{W}}^{\bf{T}} {\bf{WW}} = \left( {{\boldsymbol\omega}^{\bf{T}} {\boldsymbol\omega\bf{I}} - {\boldsymbol\omega\boldsymbol\omega}^{\bf{T}} } \right){\bf{W}} \\ {\bf{x}}^{\bf{T}} {\bf{W}}^{\bf{T}} {\bf{WWx}} = {\bf{x}}^{\bf{T}} \left( {{\boldsymbol\omega}^{\bf{T}} {\boldsymbol\omega\bf{W}} - {\boldsymbol\omega\boldsymbol\omega}^{\bf{T}} {\bf{W}}} \right){\bf{x}} = \left\| {\boldsymbol\omega} \right\|^2{\bf{x}}^{\bf{T}} {\bf{Wx}} - {\bf{x}}^{\bf{T}} {\boldsymbol\omega\boldsymbol\omega}^{\bf{T}} {\bf{Wx}} = \\ = \left( {{\boldsymbol\omega} \cdot {\boldsymbol\omega}} \right)\left( {{\bf{x}} \cdot {\bf{v}}} \right) - \left( {{\bf{x}} \cdot {\boldsymbol\omega}} \right)\left( {{\boldsymbol\omega} \cdot {\bf{v}}} \right) = \left( {{\boldsymbol\omega} \cdot {\boldsymbol\omega}} \right)0 - \left( {{\bf{x}} \cdot {\boldsymbol\omega}} \right)0 = 0 \to \left\| {\bf{v}} \right\| = {\rm{const}} \\ \end{array}$$

where I use the fact that the particle's linear velocity is orthogonal to both its position as well as its angular velocity.

• If the vector angular velocity changes, this is not true.
– LPZ
Mar 23, 2023 at 11:54
• @lpz Thanks for your comment! I've updated my question to include my proof, could I please ask you to have a look and tell me where the error is? Thanks! :)
– Gabi
Mar 23, 2023 at 12:34
• @lps Ohhh I think I've assumed that angular acceleration is constant over time, but it's not, thanks! :)
– Gabi
Mar 23, 2023 at 12:42

In general, you have the velocity field of solid body can be written at every time: $$v= v_0+\omega\times r$$ with $$v_0,\omega$$ depending a priori with time. In particular, acceleration is: $$\dot v = a_0+\dot\omega\times r+\omega\times v$$ with $$a_0 = \dot v_0$$.
From this: \begin{align} \frac{d}{dt}\frac{1}{2}v^2 &= v\cdot \dot v \\ &= v\cdot (a_0+\dot\omega\times r) \end{align} which is in general not zero either when $$a_0\neq 0$$ or $$\dot\omega\neq0$$. Since you were implicitly assuming that they were zero, you do indeed obtain conservation of velocity. This is actually obvious, since the motion is uniform rotation.