Past exam of the classical mechanics course of the University of Oxford / 2023 / 2 / b / iv / Solution

Now take zero torque and $\mathbf{\omega }(0)=(1,0,1)$. The axial angular velocity is ${\omega }_3=1$ and the complex transverse component obeys $\dot{z}=i\nu z$ with $z(0)=1$ and $\nu =(I_3-I_1)/I_1$. Hence the angular momentum components in the principal body frame areThe transverse projection therefore traces a circle with signed angular frequency $\nu$. Its magnitude is constant, $L=\sqrt{I_1^2+I_3^2}$.

In the inertial frame, zero torque means that $\mathbf{L}$ is fixed. Usingand the derivative of a body-fixed basis vector, the symmetry axis satisfiesThus there is uniform torque-free precession of a symmetric top about the fixed angular momentum, at a constant cone angle $\beta$ and space angular frequencyThe cone angle stays constant, with no nutation. The body-frame and space-frame frequencies of free symmetric-top precession differ: $\nu$ describes rotating components, whereas ${\Omega }_{\mathrm{space}}$ describes the symmetry axis in space. The formulas also cover $I_3=I_1$, where the angular momentum components in the principal body frame are constant.