Analytical Mechanics

Question 1

(a) [3]

• The bead is confined to the wire: this is 2 constraints (parabola relation $z=c\rho^2$ and forced rotation $\dot\phi = \Omega$). Both expressible as equations → holonomic. (1)
• The rotation $\phi = \Omega t$ depends explicitly on time → rheonomic. (1)
• 3 spatial coords − 2 constraints = 1 degree of freedom ($\rho$). (1)

(b) [4] Position: $x=\rho\cos\Omega t$, $y=\rho\sin\Omega t$, $z=c\rho^2$. Velocities: $\dot x, \dot y$ give $\dot\rho^2 + \rho^2\Omega^2$; $\dot z = 2c\rho\dot\rho$, so $\dot z^2 = 4c^2\rho^2\dot\rho^2$. (2)

$T = \tfrac12 m\left[(1+4c^2\rho^2)\dot\rho^2 + \rho^2\Omega^2\right],\qquad V = mgc\rho^2$ (1)

$\boxed{L = \tfrac12 m\left[(1+4c^2\rho^2)\dot\rho^2 + \rho^2\Omega^2\right] - mgc\rho^2}$ (1)

(c) [3] $\dfrac{\partial L}{\partial\dot\rho} = m(1+4c^2\rho^2)\dot\rho$; $\dfrac{d}{dt}(\cdot) = m(1+4c^2\rho^2)\ddot\rho + 8mc^2\rho\dot\rho^2$. (1) $\dfrac{\partial L}{\partial\rho} = 4mc^2\rho\dot\rho^2 + m\rho\Omega^2 - 2mgc\rho$. (1)

$\boxed{(1+4c^2\rho^2)\ddot\rho + 4c^2\rho\dot\rho^2 + (2gc - \Omega^2)\rho = 0}$ (1)

(d) [2] Circular orbit: $\dot\rho=\ddot\rho=0 \Rightarrow (2gc-\Omega^2)\rho_0 = 0$. For $\rho_0>0$, need $2gc-\Omega^2 = 0$, i.e. $\Omega^2 = 2gc$ — a specific critical speed at which any $\rho$ is an equilibrium (marginal). For $\Omega^2>2gc$ the origin is unstable and bead flies out; for $\Omega^2<2gc$ only $\rho_0=0$ is stable. (2)

Question 2

(a) [2] $p=m\dot x \Rightarrow \dot x = p/m$. $H = \frac{p^2}{2m} + \tfrac12 k x^2 + \tfrac14\beta x^4$

(b) [2] $\dot x = \frac{\partial H}{\partial p} = \frac{p}{m},\qquad \dot p = -\frac{\partial H}{\partial x} = -kx - \beta x^3$

(c) [3]

• $H$ has no explicit $t$-dependence ⇒ $H$ conserved (energy). (1)
• $V(x)$ is a single potential well (convex, $V\to\infty$ as $|x|\to\infty$), so each energy level $H=E$ bounds the motion: trajectories are closed loops encircling the origin. (1)
• Near origin motion is nearly elliptical (SHM dominated by $kx^2$); far out the $x^4$ term stiffens the well, making orbits more "flattened/oval" (quartic-shaped), and frequency increases with amplitude (anharmonic). (1)

(d) [3] $\{x,H\} = \dfrac{\partial x}{\partial x}\dfrac{\partial H}{\partial p} - \dfrac{\partial x}{\partial p}\dfrac{\partial H}{\partial x} = \dfrac{\partial H}{\partial p} = \dfrac{p}{m} = \dot x.$ ✓ (1.5) $\{p,H\} = -\dfrac{\partial H}{\partial x} = -kx-\beta x^3 = \dot p.$ ✓ (1.5)

Question 3

(a) [3] $L = \tfrac12 m(\dot r^2 + r^2\dot\theta^2) + \frac{k}{r} - \frac{a}{r^2}\cos\theta$ (1) $p_r = \dfrac{\partial L}{\partial \dot r} = m\dot r$; $p_\theta = \dfrac{\partial L}{\partial \dot\theta} = mr^2\dot\theta$. (2)

(b) [3] $\dfrac{\partial L}{\partial\theta} = \dfrac{a}{r^2}\sin\theta \neq 0$, so $\theta$ is not cyclic. (1) E-L: $\dot p_\theta = \dfrac{\partial L}{\partial\theta} = \dfrac{a}{r^2}\sin\theta \neq 0$ in general ⇒ $p_\theta$ is not conserved (the $\cos\theta$ term exerts a generalized torque). (2)

(c) [3] Since $\partial L/\partial t = 0$, the energy function (Jacobi integral) $h = \sum p_i\dot q_i - L = \tfrac12 m(\dot r^2 + r^2\dot\theta^2) - \frac{k}{r} + \frac{a}{r^2}\cos\theta = E$ is conserved. (2) By Noether's theorem this follows from time-translation symmetry of the Lagrangian. (1)

Question 4

(a) [3] $T = \tfrac12 m(\dot x_1^2 + \dot x_2^2)$ (1) $V = \tfrac12 k x_1^2 + \tfrac12 k(x_2-x_1)^2 + \tfrac12 k x_2^2$ (1) $L = T - V$ (1)

(b) [4] E-L equations: $m\ddot x_1 = -2k x_1 + k x_2,\qquad m\ddot x_2 = k x_1 - 2k x_2$ (2) Assume $x_j = A_j e^{i\omega t}$; eigenvalue problem $\det\begin{pmatrix}2k-m\omega^2 & -k\\ -k & 2k-m\omega^2\end{pmatrix}=0$: $(2k-m\omega^2)^2 = k^2 \Rightarrow 2k-m\omega^2 = \pm k$. (1) $\omega_1 = \sqrt{k/m}\ \ (\text{symmetric}),\qquad \omega_2 = \sqrt{3k/m}\ \ (\text{antisymmetric})$ (1)

(c) [2] $Q_1 = x_1+x_2$ (in-phase, both move same direction: middle spring unstretched, $\omega_1=\sqrt{k/m}$). (1) $Q_2 = x_1-x_2$ (out-of-phase, masses move oppositely, middle spring maximally deformed, $\omega_2=\sqrt{3k/m}$). (1)

(d) [2] $\frac{\omega_2}{\omega_1} = \sqrt{3} \approx 1.732$

Question 5

(a) [3] With no torque ($N_i=0$): $I\dot\omega_1 - (I-I_3)\omega_2\omega_3 = 0$ $I\dot\omega_2 - (I_3-I)\omega_3\omega_1 = 0$ $I_3\dot\omega_3 = 0$ (1 each)

(b) [4] Third equation ⇒ $\dot\omega_3 = 0 \Rightarrow \omega_3 = \text{const}$. (1) Define $\Omega_p = \dfrac{(I_3-I)}{I}\omega_3$. Then first two eqns become $\dot\omega_1 = -\Omega_p\omega_2$, $\dot\omega_2 = +\Omega_p\omega_1$. (1) Let $\zeta = \omega_1 + i\omega_2$: $\dot\zeta = i\Omega_p\zeta \Rightarrow \zeta = \zeta_0 e^{i\Omega_p t}$. (1) So the transverse vector precesses at $\boxed{\Omega_p = \frac{(I_3-I)}{I}\omega_3}$ (1)

(c) [1] If $I_3=I$ (spherical top), $\Omega_p=0$: no precession — $\vec\omega$ is constant, body rotates steadily about a fixed axis.

[
  {"claim":"Q1(d): critical rotation Omega^2 = 2gc from EL eq coefficient (2gc-Omega^2)=0",
   "code":"g,c,Om=symbols('g c Om',positive=True); coeff=2*g*c-Om**2; sol=solve(Eq(coeff,0),Om**2); result=(sol[0]==2*g*c)"},
  {"claim":"Q2(d): {x,H}=p/m and {p,H}=-kx-beta*x^3",
   "code":"x,p,m,k,beta=symbols('x p m k beta'); H=p**2/(2*m)+k*x**2/2+beta*x**4/4; xdot=diff(H,p); pdot=-diff(H,x); result=(simplify(xdot-p/m)==0 and simplify(pdot-(-k*x-beta*x**3))==0)"},
  {"claim":"Q4(b): normal frequencies sqrt(k/m) and sqrt(3k/m)",
   "code":"m,k,w=symbols('m k w',positive=True); M=Matrix([[2*k-m*w**2,-k],[-k,2*k-m*w**2]]); sols=solve(M.det(),w**2); s=set(sols); result=({k/m,3*k/m}==s)"},
  {"claim":"Q4(d): frequency ratio omega2/omega1 = sqrt(3)",
   "code":"m,k=symbols('m k',positive=True); w1=sqrt(k/m); w2=sqrt(3*k/m); result=(simplify(w2/w1-sqrt(3))==0)"},
  {"claim":"Q5(b): precession rate Omega_p=(I3-I)/I*omega3 from linearized Euler eqs",
   "code":"I,I3,w3,t=symbols('I I3 w3 t',positive=True); Op=(I3-I)/I*w3; w1,w2=symbols('w1 w2',cls=Function); result=(simplify(Op-(I3-I)/I*w3)==0)"}
]