Unit 3 · Hints & solutions

The velocity Hessian $L_{\dot q_i\dot q_j}$ must be nonsingular locally. Then $p_i=L_{\dot q_i}$ can be solved for all velocities. A singular matrix requires additional analysis beyond the regular construction.

$p=m\dot x$, so $\dot x=p/m$. Then $H=p(p/m)-[p^2/(2m)-mgx]=p^2/(2m)+mgx$.

$p_\theta=I\dot\theta$, so $H=p_\theta^2/I-[p_\theta^2/(2I)-U]=p_\theta^2/(2I)+U(\theta)$. The units of $p_\theta$ are angular momentum.

$H$ must be a function of $q,p,t$. An unresolved $\dot q$ makes partial derivatives ambiguous and means the Legendre transformation is incomplete. Prescribed support speeds are parameters, not unresolved dynamical velocities.

$p=a(q)\dot q$ and $\dot q=p/a(q)$. Thus $H=p^2/[2a(q)]+V(q)$. A position-dependent inertia remains in the denominator.

$p=m\dot q+B$, so $\dot q=(p-B)/m$. Substitution gives $H=(p-B)^2/(2m)+V(q)$. This is not $p^2/(2m)+V$ in the shifted canonical momentum.

$\dot q_1=(3p_1-p_2)/(5m)$, $\dot q_2=(-p_1+2p_2)/(5m)$. For homogeneous quadratic kinetic energy, $H=T=\tfrac12\sum_ip_i\dot q_i=(3p_1^2-2p_1p_2+2p_2^2)/(10m)$.

$L=m(\dot q+U)^2/2$, $p=m(\dot q+U)$, and $\dot q=p/m-U$. Thus $H=p(p/m-U)-p^2/(2m)=p^2/(2m)-Up$.

$H_r=-p_\theta^2/(mr^3)+V^{\prime}(r)$. Holding angular velocity fixed instead would give a different derivative and is not the canonical operation.

Changing the sign of $V$ works as shorthand only when $L=T_2-V$ and velocities are correctly reexpressed in momenta. Moving coordinates add linear and constant kinetic terms; the full transformation gives $H=T_2-T_0+V$, with shifted momentum inversion.

$H=p^2/(2m)+kx^2/2=36/6+2\times4/2=10\,\mathrm J$. Velocity is $p/m=2\,\mathrm{m\,s^{-1}}$.

At $r=0$, the equation loses all dependence on $\dot\theta$ and cannot be inverted. The angle does not specify a distinct position there. Cartesian coordinates have a nonsingular inertia matrix and describe the same physical problem regularly.

Four: $(q_1,q_2,p_1,p_2)$. A configuration-space point needs two numbers, while a state needs four initial values.

The same position can have different momenta and hence different futures. The same complete phase point in a smooth autonomous system has one locally unique solution. Crossings of distinct phase trajectories would violate that uniqueness.

$\dot x=p/m$ and $\dot p=-V^{\prime}(x)$. Hence $m\ddot x=\dot p=-V^{\prime}(x)$, for constant mass.

$x_{\max}=\sqrt{2E/k}=1\,\mathrm m$ and $p_{\max}=\sqrt{2mE}=4\,\mathrm{kg\,m\,s^{-1}}$. Thus $x^2/1^2+p^2/4^2=1$ in those units.

No. $p=0$ makes $\dot x=0$, but $\dot p=-kx<0$ if $x>0$. The phase point immediately moves downward. Only $(0,0)$ is an equilibrium.

$\dot H=\sum_i(H_{q_i}\dot q_i+H_{p_i}\dot p_i)=\sum_i(H_{q_i}H_{p_i}-H_{p_i}H_{q_i})=0$. The proof requires no explicit $t$ term and no remaining forces.

It is a cylinder: $\theta$ lies on a circle, while $p_\theta$ ranges over the real line. A rectangular phase plot has matching edges at angles differing by $2\pi$.

$mga=mga(1-\cos\theta)$ gives $\cos\theta=0$, so the oscillation turns at $\theta=\pm\pi/2$ around the bottom.

At $E=2mga$, write $\theta=\pi+\eta$. Then $|\dot\eta|\simeq\sqrt{g/a}|\eta|$, so elapsed time involves $\int d\eta/|\eta|$, which diverges logarithmically as $\eta\to0$. The upper fixed point is not reached in finite time from another state.

$p^2/(2m)-kx^2/2=0$ gives $p=\pm\sqrt{mk}\,x$. They are the approaching and departing separatrix directions of the saddle at the origin.

Generally no. A regular energy surface has dimension three inside four-dimensional phase space. Many trajectories can lie on it; further initial data or conserved quantities are needed.

$\dot p=0$ and $\dot x=p/m<0$. The state moves left along a horizontal constant-momentum line. It is not a closed phase orbit.

For a regular $L=T-V$ with a velocity-independent potential and stationary coordinate transformations, $T$ is quadratic homogeneous, so $\sum_ip_i\dot q_i=2T$ and $H=T+V$. This equality does not by itself imply conservation.

$\dot q_i=H_{p_i}$, $\dot p_i=-H_{q_i}+Q_i$. Hence $\dot H=H_t+\sum_i[H_{q_i}H_{p_i}+H_{p_i}(-H_{q_i}+Q_i)]=H_t+\sum_iQ_i\dot q_i$.

$H=p^2/(2m)$ and $Q=-c\dot x=-cp/m$. Therefore $\dot x=p/m$ and $\dot p=-cp/m$. Omitting $Q$ would incorrectly make momentum constant.

$\dot H=-3\times16/4=-12\,\mathrm W$. The loss depends on velocity at the instant; the spring displacement does not enter this drag power directly.

Hamilton’s equations cancel the phase-derivative terms, leaving $\dot H=H_t=\dot kx^2/2$. The changing spring stiffness represents external energy exchange.

For the rotating wire, $H=p^2/(2m)-m\Omega^2q^2/2$ is autonomous and conserved. Laboratory kinetic energy is $p^2/(2m)+m\Omega^2q^2/2$, which changes through actuator work.

$\dot p_i=F_0$, so $p_i=p_i(0)+F_0t$. It is conserved only when the remaining force vanishes.

With $\dot x=p/m$, $\dot p=-kx-cp/m$, the divergence is $\partial_x(p/m)+\partial_p(-kx-cp/m)=-c/m$. Negative divergence indicates area contraction in this model.

The integral of frictional work depends on the path and elapsed motion, not generally on the instantaneous physical $(q,p)$ alone. Generic dissipation is represented here by $Q_i$ in the forced equations. Special enlarged or time-dependent formulations require their own definitions.

$df/dt=\sum_if_{q_i}\dot q_i+f_t$, so $P_i=p_i+f_{q_i}$. With corresponding momenta, the Hamiltonian changes by $H^{\prime}=H-f_t$. It must also be expressed in the shifted variables.

$H^{\prime}=(P-at)^2/(2m)+V-ax$ gives $\dot P=-V^{\prime}+a$. Since $m\ddot x=\dot P-a$, the equation is still $m\ddot x=-V^{\prime}$.

Yes. For a damped spring, $H=p^2/(2m)+kx^2/2=T+V$ at every instant, yet $\dot H=-cp^2/m^2$. Equality and constancy are logically separate properties.

They are first-order equations requiring $2n$ initial numbers. Lagrange’s $n$ second-order equations require the same number, $q_i(0)$ and $\dot q_i(0)$. Momentum inversion relates the two sets.

With $M=m_1+m_2$ and $\Delta m=m_1-m_2$, $L=M\dot q^2/2+\Delta m gq$, $p=M\dot q$, $H=p^2/(2M)-\Delta m gq$. Thus $\ddot q=\dot p/M=\Delta m g/M$.

$H=p_r^2/(2m)+p_\theta^2/(2mr^2)+V$. Therefore $\dot r=p_r/m$, $\dot\theta=p_\theta/(mr^2)$, $\dot p_r=p_\theta^2/(mr^3)-V^{\prime}$, and $\dot p_\theta=0$.

$\phi$ is cyclic and $p_\phi=ma^2\sin^2\theta\dot\phi=J$ is conserved. The coordinate $\theta$ is generally not cyclic.

The angular contribution is $J^2/(2ma^2\sin^2\theta)$, which diverges positively as $\sin\theta\to0$ for $J\ne0$. Finite total energy cannot reach that region.

$mv_0^2/2\geq2mga$, so $v_0\geq2\sqrt{ga}$. Equality reaches the top only asymptotically. A speed strictly above the threshold produces passage in finite time in the ideal rod model.

At the top $mv_t^2/a=mg+\mathcal T$, so $v_t^2\geq ga$. Energy gives $v_0^2=v_t^2+4ga$, hence $v_0^2\geq5ga$. This exceeds the rod energy threshold because the string cannot push.

The naive substituted $L=m\dot r^2/2+\ell^2/(2mr^2)-V$ produces the wrong sign for the centrifugal term. The correct reduced expression subtracts $\ell\dot\theta$, giving $L_{\rm red}=m\dot r^2/2-\ell^2/(2mr^2)-V$ and $m\ddot r=\ell^2/(mr^3)-V^{\prime}$.

Newton’s equation for an individual mass or its radial component directly recovers tension. Lagrangian or Hamiltonian coordinates efficiently determine motion first; switching methods afterward uses the same physical model.

$\dot p=F$ gives $p=p_0+Ft$. Integrating $\dot q=p/M$ gives $q=q_0+p_0t/M+Ft^2/(2M)$.

The physical orbit shows the position vector in the orbital plane. The radial phase curve shows $(r,p_r)$ and records inward or outward radial motion at each radius. Angular motion continues even at $p_r=0$, so the two pictures are not the same curve.

$\dot p_\theta=-mga\sin\theta$, so $\ddot\theta=\dot p_\theta/(ma^2)=-(g/a)\sin\theta$. No small-angle approximation has entered.