Problem bank · Solutions

$\mathbf v=(\dot q,2cq\dot q+U)$ and $\delta\mathbf r=(1,2cq)\delta q$. Hence $T=m[\dot q^2+(2cq\dot q+U)^2]/2$, $V=mg(cq^2+Ut)$, and $Q_q^{(g)}=-2mgcq$ if gravity is kept outside $L$. With gravity in $V$, use $Q_q=0$. Then $L_{\dot q}=m(1+4c^2q^2)\dot q+2mcqU$. Its time derivative is $m(1+4c^2q^2)\ddot q+8mc^2q\dot q^2+2mcU\dot q$, while $L_q=4mc^2q\dot q^2+2mcU\dot q-2mgcq$. Thus $(1+4c^2q^2)\ddot q+4c^2q\dot q^2+2gcq=0$. Constant vertical support speed does not enter the acceleration equation.

$T=m\dot s^2/2$, $V=mga[1-\cos(s/a)]$, so $p_s=m\dot s$ and $H=p_s^2/(2m)+mga[1-\cos(s/a)]$. The equation is $\ddot s+g\sin(s/a)=0$. Since $p_\theta=ma^2\dot\theta=ap_s$, the canonical momenta have different dimensions but describe the same motion.

Choose $q$ downward for $m_1$. Then $T=[m_1+m_2+I/R^2]\dot q^2/2$ because $\omega=\dot q/R$, and $V=-(m_1-m_2)gq$. Thus $\ddot q=(m_1-m_2)g/[m_1+m_2+I/R^2]$. Pulley inertia reduces the acceleration; unequal tensions supply its torque.

$L=ma^2\dot\theta^2/2-mga\theta^2/2$ in the small-angle approximation. With $Q_\theta=\tau_0\cos\Omega t-b\dot\theta$, the equation is $ma^2\ddot\theta+b\dot\theta+mga\theta=\tau_0\cos\Omega t$. Resolving cosine and sine terms gives angular amplitude $\Theta=\tau_0/\sqrt{(mga-ma^2\Omega^2)^2+b^2\Omega^2}$. The approximation requires the resulting amplitude to remain small.

$T=m\dot X^2+m\dot s^2/4$, $V=ks^2/2$. Thus $P_X=2m\dot X$ is conserved and $X=X_0+V_0t$. The relative equation is $(m/2)\ddot s+ks=0$, with frequency $\sqrt{2k/m}$. The original positions are $x=X+s/2$, $y=X-s/2$.

$p=a\dot q$ and $H=p^2/(2a)+V$. Thus $\dot q=p/a$ and $\dot p=p^2a^{\prime}/(2a^2)-V^{\prime}$. But $\dot p=a\ddot q+a^{\prime}\dot q^2$, and $p^2/a^2=\dot q^2$. Therefore $a\ddot q+\tfrac12a^{\prime}\dot q^2+V^{\prime}=0$, agreeing with the Lagrangian calculation.

$L=m(\dot q+\dot X)^2/2$, $p=m(\dot q+\dot X)=m\dot x$. Thus $H=p^2/(2m)-p\dot X$. Hamilton’s equations give $\dot q=p/m-\dot X$ and $\dot p=0$. Laboratory momentum and energy remain constant, but $\dot H=-p\ddot X$ when the coordinate origin accelerates.

For $E<2mga$, two finite angle turning points give oscillation loops around the lower equilibrium. At equality the separatrix approaches the upper saddle asymptotically. Above it, momentum does not reverse and the rod pendulum rotates. A string can go slack unless tension remains nonnegative; the rod phase portrait alone does not enforce that inequality.

The circular condition is $\ell^2/\mu=anr_c^{n+2}$. Since $rV^{\prime\prime}/V^{\prime}=n-1$, $\omega_r^2/\Omega^2=n+2$. It is linearly stable when $n+2>0$, unstable when $n+2<0$, and marginal at equality. The given condition $an>0$ ensures attraction.

The quadratic curvature vanishes, so a linear frequency test says nothing. The leading quartic term is a local minimum. For sufficiently small excess energy, $\mu\dot\eta^2/2+A\eta^4$ confines the radial motion near zero. It is locally stable with amplitude-dependent, non-harmonic oscillations, assuming the higher terms do not remove the local minimum.

$E=3/4-1=-1/4$, $\ell=\sqrt{3/2}$, and $e^2=1+2(-1/4)(3/2)=1/4$. Thus $e=1/2$, $a=2$, and $r_a=3$. The initial point is periapsis. The period is $\tau=2\pi\sqrt{\mu a^3/k}=4\pi\sqrt2$ in the associated time units.

$k/\mu=2Gm$, so $\Omega^2=2Gm/d^3$ and $\tau=2\pi\sqrt{d^3/(2Gm)}$. Each body has speed $\Omega d/2=\sqrt{Gm/(2d)}$. Their relative speed is twice this value.

$p=r_c=a(1-e^2)=0.36a$, so $a=25r_c/9$. The period ratio is $(a/r_c)^{3/2}=(25/9)^{3/2}=125/27\simeq4.63$. Energy is less negative than the circular minimum at that fixed angular momentum.

$\mathbf L=(0,0,1)$ and $\mathbf p\times\mathbf L=(1,-1/2,0)$. Subtracting $\hat{\mathbf r}$ gives $\mathbf A=(0,-1/2,0)$, so $e=1/2$ and periapsis points along $-y$. Energy is $(1/4+1)/2-1=-3/8$, giving $a=4/3$. The positive radial speed confirms the particle is moving outward from that periapsis.

The relation $\tan(\chi/2)=k/(\mu b_{\rm imp}v_\infty^2)$ gives $b_{\rm imp,new}=b_{\rm imp,old}/4$ when speed doubles, for the same masses and force constant.

Angular momentum remains constant because torque is still zero; planarity and area rate survive. Total energy in the full perturbed potential remains constant because it is time-independent. The inverse-square Runge–Lenz vector generally changes, with $\dot{\mathbf A}=\delta F\hat{\mathbf r}\times\mathbf L$. Thus a fixed orbital plane does not guarantee a fixed apsidal direction.