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?


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.

  • 2
    $\begingroup$ If the vector angular velocity changes, this is not true. $\endgroup$
    – LPZ
    Mar 23, 2023 at 11:54
  • $\begingroup$ @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! :) $\endgroup$
    – Gabi
    Mar 23, 2023 at 12:34
  • $\begingroup$ @lps Ohhh I think I've assumed that angular acceleration is constant over time, but it's not, thanks! :) $\endgroup$
    – Gabi
    Mar 23, 2023 at 12:42

1 Answer 1


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.

Hope this helps.

  • $\begingroup$ Thanks! I didn't mean to assume constant angular acceleration, just constant angular momentum. I'll try to redo the proof and see what I can get, if I can't do it I'll post another question, hopefully you or someone else might be able to help me :) $\endgroup$
    – Gabi
    Mar 23, 2023 at 19:22

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