= Solution
\b[Translation and rotation.] For particles constituting a <rigid body>, the total <linear momentum> is
$$
\mathbf P=\sum_i m_i\mathbf v_i=M\mathbf V_{\rm CM},\qquad\frac{d\mathbf P}{dt}=\sum_i\mathbf F_{i,\rm ext}.
$$
Thus the total external <force> governs the motion of the <centre of mass>. About that <centre of mass>, the <angular momentum> and applied <torque> are
$$
\mathbf L_{\rm CM}=\sum_i\mathbf r_i'\times m_i\mathbf v_i',\qquad\mathbf G_{\rm CM}=\sum_i\mathbf r_i'\times\mathbf F_{i,\rm ext},\qquad\frac{d\mathbf L_{\rm CM}}{dt}=\mathbf G_{\rm CM},
$$
where primes denote relative positions and <velocities>. The <derivative> of the vector is taken relative to nonrotating axes. The <inertia tensor> relates <angular momentum> to <angular velocity>, $\mathbf L_{\rm CM}=\mathsf I\boldsymbol\omega$. For rotation about a fixed unit direction $\mathbf n$, the <moment of inertia> is $I_{\mathbf n}=\int r_\perp^2\,dm=\mathbf n\cdot\mathsf I\mathbf n$; it measures the resistance to changing that rotation and gives rotational <kinetic energy> $I_{\mathbf n}\omega^2/2$. The simpler vector relation $\mathbf L=I\boldsymbol\omega$ applies when the rotation axis is a <principal axis>, not for every orientation of an arbitrary <rigid body>. These translational and rotational balances must be applied together; a nonzero <torque> does not replace the net-<force> equation.
\b[Gyroscope precession.] A rapidly spinning <gyroscope> has a large <angular momentum> approximately along its spin axis. The <torque> component parallel to this vector changes its magnitude, while the perpendicular component turns its direction. If the vector undergoes <precession> with <angular velocity> $\boldsymbol\Omega$, then over $dt$ its change is the infinitesimal rotation $d\mathbf L=(\boldsymbol\Omega\,dt)\times\mathbf L$. Hence
$$
\boxed{\mathbf G=\frac{d\mathbf L}{dt}=\boldsymbol\Omega\times\mathbf L.}
$$
This describes direction-changing <precession> at constant $|\mathbf L|$; any spin-up contribution must be added separately. For perpendicular <torque> and <angular momentum>, the precessional component perpendicular to $\mathbf L$ is $\boldsymbol\Omega_\perp=\mathbf L\times\mathbf G/L^2$. The familiar tendency of a supported <gyroscope> to turn sideways is this vector balance, not the absence of gravitational <force>.
\b[Wheel-lift threshold without the flywheel.] Let $h=0.50\,\mathrm m$ be the <centre of mass> height, $a=1.0\,\mathrm m$ half the wheel separation, and $R_b=10\,\mathrm m$ the bend radius. The required inward <force> is $F=Mv^2/R_b$, provided by <static friction>. Let $N_o,N_i$ be the summed outer and inner wheel <normal forces>. Vertical balance gives $N_o+N_i=Mg$. Moment balance about the <centre of mass> gives
$$
a(N_o-N_i)=h\frac{Mv^2}{R_b},\qquad N_i=\frac M2\left(g-\frac{hv^2}{aR_b}\right).
$$
The inner wheels lift when $N_i=0$. Thus the <vehicle rollover threshold> is
$$
\boxed{v_{\max}=\sqrt{\frac{gaR_b}{h}}=14.01\,\mathrm{m\,s^{-1}}=50.4\,\mathrm{km\,h^{-1}}.}
$$
The stated <mass> cancels. This assumes the rigid geometry and sufficient <static friction> specified in the paper; the threshold is set by loss of contact, not by sliding.
\b[Engine flywheel.] The <moment of inertia of a uniform disc> gives
$$
I_f=\frac12M_fr_f^2=\frac12(50)(0.4)^2=4.00\,\mathrm{kg\,m^2}.
$$
Its spin <angular speed>, spin <angular momentum>, and turn <angular speed> at $40\,\mathrm{km\,h^{-1}}$ are
$$
\omega_f=4000\frac{2\pi}{60}=418.88\,\mathrm{rad\,s^{-1}},\quad L_f=I_f\omega_f=1675.52\,\mathrm{kg\,m^2\,s^{-1}},\quad\Omega=\frac{40/3.6}{10}=1.1111\,\mathrm{rad\,s^{-1}}.
$$
The transverse flywheel axis is perpendicular to the vertical turn axis, so
$$
\boxed{G=\Omega L_f=1.862\times10^3\,\mathrm{N\,m}.}
$$
To specify its direction, use right-handed axes $+x$ forward, $+y$ to the car's left, and $+z$ upward, and consider a left turn. Then $\boldsymbol\Omega=+\Omega\widehat{\mathbf z}$. Choose the flywheel <angular momentum> to the car's right, $\mathbf L_f=-L_f\widehat{\mathbf y}$. The car must apply
$$
\mathbf G_{\rm on\ flywheel}=+\Omega L_f\widehat{\mathbf x},\qquad\boxed{\mathbf G_{\rm on\ car}=-\Omega L_f\widehat{\mathbf x}}
$$
to obtain the required <precession>. This <gyroscopic roll torque in a turning vehicle> opposes the outward roll tendency in a left turn. By the <right-hand rule>, the flywheel must turn \b[counterclockwise viewed from the car's right side toward its centre], opposite to the rotation of wheels rolling forward. Reversing flywheel spin would aggravate the outward roll. Reversing the turn reverses both roll and gyroscopic <torques>, so the same flywheel spin remains helpful.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021-q6-flywheel-torque.png]
{title=<Gyroscope> vector directions for a left-turning car and the flywheel spin needed to oppose outward roll}
For comparison, the outward inertial roll moment at the stated <speed> is $Mhv^2/R_b=6173\,\mathrm{N\,m}$, and the limiting gravitational restoring moment is $Mga=9810\,\mathrm{N\,m}$. The gyroscopic reaction is about $30\%$ of the former and $19\%$ of the latter, so it is appreciable for this large flywheel. If this ideal model is extended to include the favorable gyroscopic reaction in the wheel-lift condition, it gives $Mhv^2/R_b-L_fv/R_b=Mga$, or about $56.8\,\mathrm{km\,h^{-1}}$ rather than $50.4\,\mathrm{km\,h^{-1}}$. This quantifies a useful but finite increase in the rollover threshold.
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