Past exam of the physics 1 paper of the University of Oxford / 2017 / 10

For independent generalized coordinates $q_i$, their generalized velocities ${\dot{q}}_i$ and a Lagrangian $L(q,\dot{q},t)$, the Euler-Lagrange equations areHere the partial derivatives treat $q_i,{\dot{q}}_i,t$ as independent arguments, whereas $d/dt$ follows the motion. For a conservative mechanical system, choose coordinates satisfying the mechanical constraints, express the kinetic energy $T$ and potential energy $V$ in those coordinates, set $L=T-V$, and evaluate these equations to obtain the equations of motion. If nonconservative generalized forces $Q_i$ are present, they appear on the right-hand side.

A cyclic coordinate is absent explicitly from $L$: $\partial L/\partial q_i=0$, even though $L$ may depend on ${\dot{q}}_i$. Its canonical momentum is consequently conserved,This does not require ${\dot{q}}_i=0$ or even that ${\dot{q}}_i$ be constant; the relation between canonical momentum and generalized velocity depends on the Lagrangian.

Take positive $x$ horizontally toward the downhill end of the plane, with $0\le \theta <\pi /2$. The normal force is $N=mg\cos \theta$, since there is no acceleration perpendicular to the stationary plane. Along the plane, the only tangential force is $mg\sin \theta$, so Newton's second law gives $a_{\parallel }=g\sin \theta$.

Its horizontal projection is thereforeEquivalently, the normal force has horizontal component $N\sin \theta$ while gravity has none, giving $m{a}_x=N\sin \theta$. For a plane whose height increases in the positive $x$ direction, the answer instead carries a minus sign. A horizontal plane gives zero acceleration; the horizontal projection also tends to zero for a vertical plane.

Model the cart as a particle on a fixed plane curve, remaining constrained to the track. Its vertical velocity is $\dot{y}=y^{\prime }(x)\dot{x}$, so its kinetic energy, gravitational potential energy and Lagrangian areConsequently the canonical momentum conjugate to the horizontal coordinate isIt is not simply the horizontal component $m\dot{x}$ of the mechanical momentum, because a change in $x$ also moves the cart vertically. Differentiate carefully in the Euler-Lagrange equation:Subtracting yieldsThe term involving $y^{\prime \prime }$ accounts for the changing track geometry. An independent check is conservation of energy:

If $y^{\prime \prime }=0$ over the subsequent segment, its slope is a constant $s$ and the equation of motion reduces toThere is uniform horizontal acceleration, not generally zero acceleration. Starting that segment at $(t_0,x_0,{\dot{x}}_0)$,For a downhill straight track with $s=-\tan \theta$, this gives $\ddot{x}=g\sin \theta \cos \theta$, agreeing with b. Only $s=0$ gives constant horizontal velocity.

The condition $y^{\prime \prime }\to 0$ must describe an actually straight limiting segment, with bounded velocity, for that subsequent constant-acceleration conclusion. If $y^{\prime \prime }$ vanishes only at an isolated inflection point, the boxed expression describes the instantaneous acceleration there; it does not make the rest of the track straight. Likewise, a vanishing $y^{\prime \prime }$ alone does not make the product $y^{\prime \prime }{\dot{x}}^2$ negligible in a limit where $\dot{x}$ diverges.