Past exam of the physics B course of the University of Cambridge / 2016 / Paper 1 / Section C / C11

For a body-fixed mass element at position $\mathbf{r}$ relative to the fixed tip, $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$. ThereforeThe inertia tensor is real symmetric. Its orthogonal eigenvectors define principal axes and its eigenvalues are principal moments of inertia. angular momentum is parallel to angular velocity when rotation is along a principal axis, but generally not for arbitrary rotation.

Write $\mathbf{n}={\hat{\mathbf{e}}}_3$ and $s=I_3{\omega }_3$. Axial symmetry gives $\mathbf{L}=I_1\mathbf{\omega }+(I_3-I_1){\omega }_3\mathbf{n}$. Since $\dot{\mathbf{n}}=\mathbf{\omega }\times \mathbf{n}$, the vector triple-product identity gives $\mathbf{n}\times \dot{\mathbf{n}}=\mathbf{\omega }-{\omega }_3\mathbf{n}$. HenceDifferentiate in the inertial frame, using constant $s$ and $\dot{\mathbf{n}}\times \dot{\mathbf{n}}=0$. The gravitational torque about the fixed tip is $(\ell \mathbf{n})\times (-mg\hat{\mathbf{z}})$, yielding the symmetry-axis equation of a heavy symmetric top:The tip reaction exerts no moment about that tip.

For steady precession, $\mathbf{n}=(\sin \theta \cos {\Omega }_{P}t,\sin \theta \sin {\Omega }_{P}t,\cos \theta )$. At azimuth zero, put $s_{\theta }=\sin \theta$ and $c_{\theta }=\cos \theta$. Then $\dot{\mathbf{n}}=(0,{\Omega }_P{s}_{\theta },0)$, $\ddot{\mathbf{n}}=(-{\Omega }_P^2{s}_{\theta },0,0)$, and the $y$ component gives $s{\Omega }_P{s}_{\theta }-I_1{\Omega }_P^2{s}_{\theta }{c}_{\theta }=mg\ell s_{\theta }$. For a nonvertical axis this becomesReal steady branches require a nonnegative discriminant. In particular, for a top pointing above the horizontal there is a spin threshold for these fixed-angle solutions. At $\cos \theta =0$ use the quadratic's linear limit ${\Omega }_P=mg\ell /s$, rather than dividing by zero. For a vertical axis, azimuth does not describe a physical conical motion and the division by $\sin \theta$ is inappropriate.

For rapid positive spin $s$ with $s^2\gg 4mg\ell I_1\cos \theta$, the slow and fast steady precession roots areThe approximation assumes a nonsingular fixed inclination. For $g=0$, the quadratic instead gives ${\Omega }_P=0$ or ${\Omega }_P=s/(I_1\cos \theta )$. The zero branch has a stationary symmetry axis and pure axial spin. On the other branch angular momentum is constant and parallel or antiparallel to the chosen $z$ axis; $L_z=I_1{\Omega }_P$ and $s=L_z\cos \theta$. This is torque-free precession in space, not the different rate at which transverse components rotate in body coordinates.