= Solution
\b[Directions.] Choose $+x$ along the horizontal spindle from the pivot to the disk, and $+z$ vertically upward. With <lever arm> $\mathbf r=l\widehat{\mathbf x}$ and weight $\mathbf F_g=-mg\widehat{\mathbf z}$, the <torque> about the pivot is
$$
\boldsymbol\tau=\mathbf r\times\mathbf F_g=mgl\widehat{\mathbf y}.
$$
Clockwise <precession> viewed from above has $\boldsymbol\Omega=-\Omega\widehat{\mathbf z}$. The relation $\boldsymbol\tau=\boldsymbol\Omega\times\mathbf L$ therefore requires the spin <angular momentum> to point along $-x$, \b[from the disk toward the pivot]. The disk spins \b[clockwise when viewed from its outer face looking along the spindle toward the pivot]. Looking from the pivot reverses that apparent rotation. The <right-hand rule> fixes these signs unambiguously.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018-ia-physics-q8-gyroscope.png]
{title=Top-view torque and angular-momentum directions, with the disk's spin viewed from its outer face}
The drawing shows an instantaneous top view; the disk's <angular momentum> and the horizontal spindle turn clockwise together. The pivot <force> has zero <torque> about the pivot.
\b[Rate of precession.] For steady horizontal motion the spin contribution has magnitude $L=I\omega$ and turns through $d\phi=\Omega\,dt$ in time $dt$. Its change is perpendicular to itself and has magnitude $dL=L\,d\phi$. Hence
$$
|\boldsymbol\tau|=\left|\frac{d\mathbf L}{dt}\right|=\Omega I\omega=mgl,
\qquad
\boxed{\Omega=\frac{mgl}{I\omega}.}
$$
This is the usual <slow steady gyroscope precession> relation. Any constant vertical <angular momentum> associated with steady precession has zero time derivative; the horizontal spin contribution supplies the changing component here. Transient <nutation> is excluded by the prescribed horizontal steady motion.
\b[Clock design.] Convert both rotation rates to <angular frequencies>:
$$
\Omega=\frac{2\pi}{60}\,\mathrm{rad\,s^{-1}},\qquad
\omega=2\pi(300)\,\mathrm{rad\,s^{-1}}.
$$
For the <moment of inertia of a uniform disc>, $I=mR^2/2$, so the <mass> cancels:
$$
\boxed{l=\frac{\Omega R^2\omega}{2g}
=\frac{(2\pi/60)(0.10)^2(600\pi)}{2(9.8)}
=0.1007\,\mathrm m\simeq10.1\,\mathrm{cm}.}
$$
\b[Spin friction.] At fixed geometry, reducing $\omega$ increases $\Omega\propto1/\omega$. The <precession> period, $T_p=2\pi/\Omega$, therefore \b[decreases]. This trend holds while slow steady <precession> remains a suitable description; it cannot be extrapolated to a stopped disk.
\b[Counterweight.] To oppose the disk's gravitational <torque>, place the counterweight on the other side of the pivot, at $-l_c\widehat{\mathbf x}$. The net <torque> is
$$
\boldsymbol\tau=g(ml-m_cl_c)\widehat{\mathbf y}.
$$
A horizontal axle can remain fixed when this vanishes, giving
$$
\boxed{l_c=\frac{ml}{m_c}\quad\text{on the opposite side of the pivot}.}
$$
For the fixed spin direction already found, <counterbalanced gyroscope precession> has signed rate
$$
\Omega_z=-\frac{g(ml-m_cl_c)}{I\omega}.
$$
Moving the counterweight outward from the pivot first slows the clockwise motion, eliminates its driving <torque> at balance, and then reverses the steady motion to counterclockwise. A counterweight on the disk's side would add to the gravitational <torque>. These statements neglect axial spin <moment of inertia> of the counterweight and assume the stated horizontal motion, without a separately imposed precession at balance.
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