Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 7D (Classical Dynamics)

The principle of stationary action says that the physical path makes the action stationary under smooth variations $q+\eta h$ with fixed endpoints $h(t_1)=h(t_2)=0$. It need not minimize the action. Differentiating and integrating by parts givesThe fundamental lemma of the calculus of variations then gives the Euler-Lagrange equation

Use $r\in (0,l)$, the bead's distance from the pivot, and $\theta$, the rod angle from downward vertical. The kinetic energy is $m({\dot{r}}^2+r^2{\dot{\theta }}^2)/2$. If the identical springs have common natural length $r_{\ast }$, their potential is $k[(r-r_{\ast }{)}^2+(l-r-r_{\ast }{)}^2]/2$. Its $r$-dependent part is $k(r-l/2{)}^2$; the unspecified common natural length changes only an additive constant. Gravity contributes $-mgr\cos \theta$. Consequently the Lagrangian may be taken asThis uses the illustrated symmetric two-spring arrangement without assuming an unstated natural length.

Apply the Euler-Lagrange equation separately to $r$ and $\theta$:The first equation balances radial inertia, centrifugal motion, gravity and the net spring force. The second is the angular-momentum equation $d(m{r}^2\dot{\theta })/dt=-mgr\sin \theta$. For $r>0$ it is equivalently $\ddot{\theta }+2\dot{r}\dot{\theta }/r+g\sin \theta /r=0$.