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

Take the origin at the centre of mass, so ${\int }_B\rho \mathbf{r}{d}^{3}r=0$, and write $r_i=\mathbf{r}\cdot {\mathbf{e}}_i$ in the specified orthonormal basis. The inertia tensor isHere ${\delta }_{ij}$ is the Kronecker delta. For a rigid body with angular velocity $\mathbf{\omega }$, this tensor gives $\mathbf{L}=\mathbf{I}\mathbf{\omega }$ and rotational kinetic energy $T=\mathbf{\omega }\cdot \mathbf{I}\mathbf{\omega }/2$.

The inertia tensor is a real symmetric matrix, so it admits an orthonormal basis of eigenvectors. The lines in these eigenvector directions through the centre of mass are its principal axes; the corresponding eigenvalues $I_1,I_2,I_3$ are its principal moments of inertia:If an eigenvalue repeats, the corresponding principal axes need not be unique.

Put $\mathbf{r}=(x,y,z)$ with ${\mathbf{e}}_3$ along $z$, and interpret the symmetry as axisymmetry of the whole mass density. A half-turn about ${\mathbf{e}}_3$ sends $(x,y,z)$ to $(-x,-y,z)$ and leaves the integration domain and mass density invariant. Therefore $\int \rho xz{d}^{3}r=\int \rho yz{d}^{3}r=0$. A quarter-turn sends $(x,y,z)$ to $(-y,x,z)$; it givesAll off-diagonal entries of the inertia tensor consequently vanish. Writing $s^2=x^2+y^2$, the principal moments of inertia satisfyThus every orthonormal basis with third vector along the symmetry axis is a principal body frame. This is axisymmetry of the inertia tensor. In general there is no requirement that $I_3=I_1$.

For a vector $\mathbf{A}={\sum }_i{A}_i{\mathbf{e}}_i$ in the principal body frame, the derivative of a body-fixed basis vector is $(d{\mathbf{e}}_i/dt{)}_{\hat{S}}=\mathbf{\omega }\times {\mathbf{e}}_i$. Differentiating the expansion gives the rotating-frame derivative formula, also called the Coriolis formula:Apply this to angular momentum $\mathbf{L}={\sum }_i{I}_i{\omega }_i{\mathbf{e}}_i$. The principal moments of inertia are constant in the principal body frame, so external torque balance becomesResolving the cross product gives Euler's equations:All components and dotted components in this display refer to the principal body frame.

For the symmetric top, set $I_1=I_2=J$. The third of Euler's equations reduces to $I_3{\dot{\omega }}_3={\tau }_3=0$. Thus the axial angular velocity component is a conserved quantity:The transverse angular velocity components are generally variable. A torque constant in the principal body frame generally rotates in the inertial frame; constancy in the two frames is different.

Let $J=I_1=I_2$, ${\omega }_3=c$ andHere $c$ and $\nu$ are constant. The first two of Euler's equations give ${\dot{\omega }}_1=-\nu {\omega }_2+{\tau }_0/J$ and ${\dot{\omega }}_2=\nu {\omega }_1+{\tau }_0/J$. Combining them producesMultiplication by the integrating factor $e^{-i\nu t}$ gives, for $\nu \ne 0$,For completeness, $\nu =0$ instead gives $z(t)=z(0)+{\tau }_0(1+i)t/J$. These formulas describe a symmetric top under constant transverse body torque.

The stated initial conditions require $b\ne 0$ and $I_3\ne J$. They give $c=b/(I_3-J)$, $\nu =b/J$ and $z(0)=i{\tau }_0/b$. Substitution and simplification yieldTaking the real part and imaginary part, the angular velocity in the principal body frame isIts value at $t=0$ is the required initial condition; differentiating the first two components verifies both forced Euler's equations.

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.