Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / H (Classical Mechanics)

Measure the axial coordinate $z$ from the vertex. The slice radius is $z\tan \alpha$ and its mass is $dm=3m{z}^2{H}^{-3}dz$, since cross-sectional area is proportional to $z^2$. Its axial moment is $\frac{1}{2}(z\tan \alpha {)}^{2}dm$. ThereforeFor an axis through the vertex perpendicular to the cone axis, a thin disc contributes $[z^2+\frac{1}{4}{z}^2{\tan }^2\alpha ]dm$: the second term is its diametral moment. Integration gives the two equal transverse principal momentsRotational symmetry eliminates the mixed moments and makes these the principal axes. The centre of mass is at $3H/4$, so the parallel-axis theorem would give $A_G=3m{H}^2/80+3m{H}^2{\tan }^2\alpha /20$ and $C_G=C_O$.

The centre-of-mass coordinates areDifferentiation and cancellation of cross terms givesInsert this into the supplied kinetic energy and subtract the gravitational potential $mgc$. Thus

The coordinate $\psi$ is cyclic, so $p_{\psi }=C(\dot{\psi }+\dot{\varphi }\cos \theta )$ is conserved. Initially the spin is zero, hence $\boxed{\dot{\psi }+\dot{\varphi }\cos \theta =0}$ throughout. The coordinate $\varphi$ is also cyclic. After imposing this zero spin, its conserved momentum becomes $(A+m{h}^2)\dot{\varphi }{\sin }^2\theta$. Initially $\theta =\pi /2$ and $\dot{\varphi }=\omega$, so $\boxed{\dot{\varphi }{\sin }^2\theta =\omega }$.

The equation for $z$ isThe quantity in parentheses is the vertical velocity of the centre of mass. Initially that velocity and its height are both zero: the initial rotation is horizontal and the vertex starts at rest at the reference height. Twice integrating givesIndependently, the frictionless vertical wire exerts only horizontal force. Gravity is therefore the only vertical external force on the body, and the centre of mass undergoes vertical free fall. This gives the same equation without the angular equations.