Past exam of the physics 1 paper of the University of Oxford / 2019 / 9

For a Lagrangian $L(q_i,{\dot{q}}_i,t)$ expressed in independent generalized coordinates, stationary action gives the Euler-Lagrange equationsIdeal constraints can either be incorporated into the independent coordinates or treated using Lagrange multipliers, as in solution.

A cyclic coordinate is absent from the Lagrangian itself: $\partial L/\partial q_i=0$. Its conjugate momentum $p_i=\partial L/\partial {\dot{q}}_i$ therefore obeys ${\dot{p}}_i=0$. In the rolling-wedge calculation, the cyclic horizontal translation gives conservation of the system's horizontal momentum.

A holonomic constraint can be expressed as $f(q_1,\dots ,q_n,t)=0$, rather than as a genuinely nonintegrable relation among velocities. A fixed rod length is an example. Rolling along a single straight incline also gives an integrable relation $s+R\theta =\text{constant}$ with the sign convention used below; arbitrary rolling in more dimensions need not be holonomic.

Take $s$ positive uphill, $X$ positive to the right, and the cylinder's rotation positive anticlockwise. Contact with the fixed-angle surface determines its normal position; no slipping gives $\dot{\theta }=-\dot{s}/R$. These two constraints leave $X,s$ as the two independent coordinates of the cylinder rolling on a freely translating wedge.

The cylinder centre is at $(X+s\cos \alpha -R\sin \alpha ,s\sin \alpha +R\cos \alpha )$. Since the angle is constant, its squared speed is ${\dot{X}}^2+2\dot{X}\dot{s}\cos \alpha +{\dot{s}}^2$. Including the block's kinetic energy and $I_c=M_c{R}^2/2$, and dropping a constant from the potential, givesThe cyclic coordinate is $X$. Its conserved conjugate momentum is $p_X=(M_b+M_c)\dot{X}+M_c\cos \alpha \dot{s}$, the total horizontal momentum. The two Euler-Lagrange equations areEliminating the accelerations, with $D=3(M_b+M_c)-2M_c{\cos }^2\alpha >0$, givesThus the cylinder rolls downhill and the wedge accelerates rightward. For $M_b\gg M_c$, $\ddot{X}\to 0$ and $\ddot{s}\to -(2/3)g\sin \alpha$, the ordinary fixed-incline result.

With the wedge fixed, let $r$ be the signed gap between the cylinder bottom and the surface, positive outwards. The centre height is $s\sin \alpha +(R+r)\cos \alpha$. Retaining all three coordinates before imposing constraints givesUse the holonomic constraints $f_1=r=0$ and $f_2=s+R\theta -s_0=0$. With the convention $d(\partial L/\partial \dot{q})/dt-\partial L/\partial q={\sum }_j{\lambda }_j\partial f_j/\partial q$, the Lagrange multiplier equations areThe constraint gradients identify ${\lambda }_1=N$, the normal reaction, and ${\lambda }_2=f$, the uphill static friction force. The latter supplies the rolling torque. Setting $\ddot{r}=0$ and $\ddot{\theta }=-\ddot{s}/R$ yieldsThe static friction condition $|f|\le {\mu }_{s}N$ therefore becomesfor $0\le \alpha <\pi /2$. Static friction is required to create rotation, but does not dissipate energy here because the contact point is instantaneously at rest on the fixed plane.