= Solution
The PDF gives initial position $(x',y',z')=(1,0,0)$ and initial rotating-frame <velocity> zero. Use those values; the TeX erroneously places dots over the position coordinates.
For rotation about the third axis, the supplied acceleration relation and <Newton's second law> give the <equations of motion in a rotating frame>
$$
\ddot x'-2\omega\dot y'+3\omega^2x'=0,\qquad
\ddot y'+2\omega\dot x'+3\omega^2y'=0,\qquad
\ddot z'+4\omega^2z'=0.
$$
Thus $z'=0$. Set $\zeta=x'+iy'$ to combine the planar equations:
$$
\ddot\zeta+2i\omega\dot\zeta+3\omega^2\zeta=0.
$$
The two characteristic roots are $i\omega$ and $-3i\omega$. From $\zeta(0)=1$ and $\dot\zeta(0)=0$,
$$
\zeta=\frac34e^{i\omega t}+\frac14e^{-3i\omega t}.
$$
The triple-angle formulas therefore give the particularly simple <astroid motion of a harmonic oscillator in a rotating frame>
$$
\boxed{x'=\cos^3(\omega t),\qquad
y'=\sin^3(\omega t),\qquad z'=0.}
$$
The orbit satisfies $|x'|^{2/3}+|y'|^{2/3}=1$; the zero <speed> at each cusp is consistent with the smooth time-parametrized motion.
Its rotating-frame <speed> is
$$
|\boldsymbol v'|^2
=9\omega^2\sin^2(\omega t)\cos^2(\omega t)
=\frac94\omega^2\sin^2(2\omega t).
$$
Hence, for $\omega>0$, \b[the first maximum occurs at $t=\pi/(4\omega)$], and
$$
\boxed{v'_{\max}=\frac32\omega.}
$$
The maximum repeats periodically. For a signed negative angular frequency, use $\pi/(4|\omega|)$ and $3|\omega|/2$; at $\omega=0$ the particle remains at rest.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ia/paper-4-rotating-astroid.png]
{title=Astroid orbit and the first maximum <speed> in the rotating frame}
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