Past exam of the options I course of the University of Oxford / 2013 / H (Classical Mechanics)

Let $s$ be distance along the rod from its fixed end. The position of that material point is $sn(t)$, where $n=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,-\cos \theta )$ is a unit vector. Its velocity is $s\dot{n}$, and direct differentiation givesThe uniform linear mass density is $M/a$. Integrating the elemental kinetic energy therefore proves, rather than assumes, the rod formula:Equivalently, its moment of inertia about a perpendicular axis through the pivot is $M{a}^2/3$.

The centre of mass is at height $-a\cos \theta /2$. With gravitational potential energy zero at the pivot level,Hence the Lagrangian isAdding a constant to the potential would change the numerical energy reference but would not change the motion.

For an autonomous Lagrangian, the Euler-Lagrange equations implyIndeed, differentiating the expression cancels the acceleration terms, leaving ${\sum }_q\dot{q}(d{L}_{\dot{q}}/dt-L_q)-L_t$, which vanishes. Here $T$ is quadratic in the velocities, so $\sum \dot{q}{L}_{\dot{q}}=2T$. Thus the conserved energy isThe azimuth $\varphi$ is a cyclic coordinate. Its conjugate momentum is constant:Therefore $\boxed{{\sin }^2\theta \dot{\varphi }=h}$ is constant as well. Physically, $p_{\varphi }$ is the vertical component of angular momentum.

Away from the coordinate poles, use $\dot{\varphi }=h/{\sin }^2\theta$ in the conserved energy:Rearranging gives the first-order equationWhere the right side is positive, either sign of its square root is possible, according to the direction of motion. Its zeros are turning points. For $h\ne 0$, the centrifugal term excludes the poles; if $h=0$, the polar-coordinate formula is interpreted through the original rod equations.

The polar Euler-Lagrange equation isFor a constant angle with $0<{\theta }_0<\pi$, conservation of $h$ makes $\dot{\varphi }$ constant, say $\Omega$. The remaining equation isIf $g>0$, every nonvertical constant-angle motion is a conical pendulum motion:The rod sweeps out a cone about the downward vertical, in either azimuthal direction. Equivalently, $h^2=(3g/(2a)){\sin }^4{\theta }_0/\cos {\theta }_0$. The exactly downward and exactly upward vertical rods are also static solutions; azimuth has no physical meaning at either pole. No static tilted rod is possible under nonzero gravity.

If $g=0$, the condition is $\sin {\theta }_0\cos {\theta }_0{\Omega }^2=0$. Thus a rod may remain at rest in any orientation, or rotate uniformly in the horizontal plane with ${\theta }_0=\pi /2$ and arbitrary constant $\Omega$. The vertical cases are included among static orientations; changing only their azimuth does not move the rod.