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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.

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    $\begingroup$ 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

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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.

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  • $\begingroup$ 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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