$$ A wheel of radius \(a\), with its mass concentrated on the rim, is rolling with velocity \(v\) round a circle of radius \(R(\gg a)\), maintaining a constant inclination \(\alpha\) to the vertical. Show that \(v=a \omega=R \Omega\), where \(\omega\) is the angular velocity of the wheel about its axis, and \(\Omega(\ll \omega)\) is the precessional angular velocity of the axis. Use the momentum equation to find the horizontal and vertical components of the force at the point of contact. Then show from the angular momentum equation about the centre of mass that \(R=2 v^{2} / g \tan \alpha .\) Evaluate \(R\) for \(v=5 \mathrm{~ms}^{-1}\) and \(\alpha=30^{\circ}\).

Define P as the point of contact between the wheel and the circle and C as the center of mass of the wheel. Let M be the midpoint of the line segment connecting P and C. Since the angle α is maintained constant as the wheel rolls around the circle, \(\triangle PMC\) is always an isosceles right triangle. From the definition of the angular velocities ω and Ω, we know that \(v = a\omega\) and \(v = R\Omega\). Therefore, \(a\omega = R\Omega\).

In order to calculate the horizontal and vertical components of the force, we can analyze the forces acting on the wheel and write the momentum equations. Let Fx and Fy be the horizontal and vertical components of the force at point P. The forces acting on the wheel are its weight mg acting downward from point C and the force F at point P. The momentum equations are: \(\begin{aligned} m \frac{\mathrm{d} v}{\mathrm{d} t} &=F_{x} \\ m g-F_{y} &=m \frac{v^{2}}{R} \end{aligned}\) From the above equations, we get: \(\begin{aligned} F_{x} &=m \frac{\mathrm{d} v}{\mathrm{d} t} \\ F_{y} &=m g - m \frac{v^{2}}{R} \end{aligned}\)

Now, we will write the angular momentum equation about the center of mass C. The moment of inertia of the wheel about its axis is \(I=\frac{1}{2}ma^2\). The angular momentum equation about point C is given by: \(m a^{2} \frac{\mathrm{d} \omega}{\mathrm{d} t} =R \sin \alpha \cdot F_{x}-R \cos \alpha \cdot F_{y}\) Now, using the relations obtained in step 2, we have: \(\frac{1}{2}ma^{2} \frac{\mathrm{d} \omega}{\mathrm{d} t} =R \sin \alpha \cdot m \frac{\mathrm{d} v}{\mathrm{d} t}-R \cos \alpha \cdot\left(m g - m \frac{v^{2}}{R}\right)\) Since \(a\omega = R\Omega\), differentiating both sides with respect to time, we get \(a\frac{\mathrm{d} \omega}{\mathrm{d} t} = R\frac{\mathrm{d} \Omega}{\mathrm{d} t}\). Notice that \(\frac{\mathrm{d} v}{\mathrm{d} t} = a\frac{\mathrm{d} \omega}{\mathrm{d} t}\). Canceling terms and rearranging the equation, we get: \(R = \frac{2v^2}{g \tan \alpha}\)

We are given \(v=5 \mathrm{~ms}^{-1}\) and \(\alpha=30^{\circ}\). Plugging these values into the equation for R, we have: \(R = \frac{2(5)^2}{(9.81)(\tan 30^{\circ})} \approx 29.3 \thinspace \mathrm{m}\) Hence, the radius R of the circular path is approximately 29.3 meters.