= Solution
Let $\mathbf e'_1(t),\mathbf e'_2(t),\mathbf e'_3(t)$ be an <orthonormal basis> fixed in the <rotating reference frame> $S'$. A small rotation through $\boldsymbol\omega\,dt$ changes each basis vector by $\boldsymbol\omega\times\mathbf e'_j\,dt$, so its derivative in the <inertial frame> is
$$
\left(\frac{d\mathbf e'_j}{dt}\right)_S=\boldsymbol\omega\times\mathbf e'_j.
$$
Write the arbitrary <vector> as $\mathbf a=\sum_j a_j(t)\mathbf e'_j(t)$. Differentiating the components and the moving basis separately gives
$$
\left(\frac{d\mathbf a}{dt}\right)_S=\sum_j\dot a_j\mathbf e'_j+\sum_j a_j\boldsymbol\omega\times\mathbf e'_j
=\left(\frac{d\mathbf a}{dt}\right)_{S'}+\boldsymbol\omega\times\mathbf a.
$$
This proves the \b[<rotating-frame derivative formula>], including when the <angular velocity> varies with time.
Back to article page
