Analytical Mechanics

Q1 (10 marks)

Setup. Virtual displacements consistent with constraints; write $\delta\mathbf r_i = \sum_j \frac{\partial \mathbf r_i}{\partial q_j}\delta q_j$ (holonomic ⇒ coordinates $q_j$ independent). (1)

Applied-force term. Generalized force $\sum_i \mathbf F_i\cdot\delta\mathbf r_i = \sum_j Q_j \delta q_j,\quad Q_j=\sum_i\mathbf F_i\cdot\frac{\partial\mathbf r_i}{\partial q_j}.$ (1) Conservative: $Q_j = -\partial V/\partial q_j$. (1)

Inertial term. $\sum_i \dot{\mathbf p}_i\cdot\delta\mathbf r_i = \sum_j\left[\frac{d}{dt}\frac{\partial T}{\partial\dot q_j}-\frac{\partial T}{\partial q_j}\right]\delta q_j.$ Key identities used:

• $\frac{\partial \mathbf r_i}{\partial q_j} = \frac{\partial \dot{\mathbf r}_i}{\partial \dot q_j}$ (cancellation of dots). (2)
• $\frac{d}{dt}\frac{\partial \mathbf r_i}{\partial q_j} = \frac{\partial \dot{\mathbf r}_i}{\partial q_j}$. (1) Product-rule manipulation gives the $T$-form above. (1)

Combine. D'Alembert ⇒ $\sum_j\left[\frac{d}{dt}\frac{\partial T}{\partial\dot q_j}-\frac{\partial T}{\partial q_j}+\frac{\partial V}{\partial q_j}\right]\delta q_j=0.$ Independence of $\delta q_j$ (holonomic) ⇒ each bracket $=0$. (1) With $V$ velocity-independent, $\partial V/\partial\dot q_j=0$, so define $L=T-V$: $\frac{d}{dt}\frac{\partial L}{\partial\dot q_j}-\frac{\partial L}{\partial q_j}=0. \quad\blacksquare$ (1)

Assumptions flagged: holonomic ⇒ independent $q_j$; scleronomic used in KE identities; conservative & velocity-independent $V$.

Q2 (12 marks)

(a) Constraint (radius $a$ fixed, rotation $\phi=\omega t$ imposed) is holonomic and rheonomic (explicit time dependence). (2)

(b) Position: $x=a\sin\theta\cos\omega t$, etc. Kinetic energy $T=\tfrac12 m a^2\dot\theta^2+\tfrac12 m a^2\omega^2\sin^2\theta,$ $V=-mga\cos\theta.$ $L=\tfrac12 m a^2\dot\theta^2+\tfrac12 m a^2\omega^2\sin^2\theta+mga\cos\theta.$ (4)

(c) EOM: $m a^2\ddot\theta = m a^2\omega^2\sin\theta\cos\theta - mga\sin\theta$ $\boxed{\ddot\theta=\omega^2\sin\theta\cos\theta-\frac{g}{a}\sin\theta.}$ (3)

(d) Equilibria ($\ddot\theta=0$): $\sin\theta_0=0$ ⇒ $\theta_0=0,\pi$; or $\cos\theta_0=g/(a\omega^2)$. (2) This third solution exists only when $g/(a\omega^2)\le1$, i.e. $\boxed{\omega_c=\sqrt{g/a}.}$ (1)

Q3 (10 marks)

(a) $L=\tfrac12 m(\dot r^2+r^2\dot\phi^2)-V(r)$. Momenta $p_r=m\dot r$, $p_\phi=mr^2\dot\phi$. (2) $H=p_r\dot r+p_\phi\dot\phi-L=\frac{p_r^2}{2m}+\frac{p_\phi^2}{2mr^2}+V(r).$ (2)

(b) $\dot r=\frac{p_r}{m},\quad \dot\phi=\frac{p_\phi}{mr^2},$ $\dot p_r=\frac{p_\phi^2}{mr^3}-V'(r),\quad \dot p_\phi=0.$ (3)

(c) $\dot p_\phi=\{p_\phi,H\}=-\partial H/\partial\phi=0$ since $H$ independent of $\phi$ (cyclic). So $p_\phi$ conserved. (2) By Noether's theorem, this follows from rotational invariance (symmetry under $\phi\to\phi+\epsilon$). (1)

Q4 (10 marks)

(a) $L=\tfrac12 m(\dot x_1^2+\dot x_2^2)-\tfrac12 k\left[x_1^2+(x_2-x_1)^2+x_2^2\right].$ (3)

(b) EOM: $m\ddot x_1=-2kx_1+kx_2,\quad m\ddot x_2=kx_1-2kx_2.$ Assume $x\sim e^{i\omega t}$: eigenvalues of $\frac{k}{m}\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$ ⇒ $\omega^2=\frac{k}{m}(2\mp1)$: $\boxed{\omega_1=\sqrt{k/m},\qquad \omega_2=\sqrt{3k/m}.}$ (4)

(c) Normal coordinates $Q_1=\tfrac1{\sqrt2}(x_1+x_2)$ (symmetric, in-phase, lower frequency $\omega_1$: masses move together, middle spring unstretched); $Q_2=\tfrac1{\sqrt2}(x_1-x_2)$ (antisymmetric, out-of-phase, higher frequency $\omega_2$: middle spring strongly compressed/stretched). (3)

Q5 (10 marks)

(a) Liouville: the phase-space distribution density is constant along trajectories; equivalently phase volume is conserved. Reason: flow velocity $(\dot q,\dot p)$ has zero divergence: $\frac{\partial\dot q}{\partial q}+\frac{\partial\dot p}{\partial p}=\frac{\partial^2 H}{\partial q\partial p}-\frac{\partial^2 H}{\partial p\partial q}=0.$ (4)

(b) Model snippet:

import numpy as np
from scipy.spatial import ConvexHull
th = np.random.uniform(1.0,1.1,2000)   # cloud of initial conditions
p  = np.random.uniform(0.0,0.1,2000)
dt=0.01
for _ in range(1000):
    p  = p - dt*np.sin(th)   # symplectic (semi-implicit) Euler
    th = th + dt*p           # use updated p -> symplectic
pts=np.column_stack([th,p])
print("area:", ConvexHull(pts).volume)

Comment: symplectic Euler conserves phase-space area (up to hull approximation) even though it distorts the cloud's shape — consistent with Liouville, unlike naive explicit Euler which grows the area. (6)

Q6 (8 marks)

Setup. Gravity torque about pivot $\tau=Mg\ell\sin\theta$, horizontal, perpendicular to the axis' vertical plane. (2) In fast-spin approximation, angular momentum ≈ spin part $L_s=I_3\dot\psi$ along symmetry axis; nutation neglected. (2) Precession: $\boldsymbol\tau=\dot{\mathbf L}=\boldsymbol\Omega\times\mathbf L$, magnitude $Mg\ell\sin\theta=\Omega\,L_s\sin\theta.$ (2) $\boxed{\Omega=\frac{Mg\ell}{I_3\dot\psi}.}$ Independent of $\theta$; valid when $\Omega\ll\dot\psi$ (large spin). (2)

[
  {"claim":"Bead critical angular velocity omega_c^2 = g/a from cos(theta)=g/(a w^2) boundary at theta=0",
   "code":"g,a,w=symbols('g a w',positive=True); crit=Eq(g/(a*w**2),1); sol=solve(crit,w); result=(sqrt(g/a) in sol)"},
  {"claim":"Coupled oscillator eigenvalues give omega1^2=k/m, omega2^2=3k/m",
   "code":"k,m=symbols('k m',positive=True); M=(k/m)*Matrix([[2,-1],[-1,2]]); evs=set(M.eigenvals().keys()); result=(evs=={k/m,3*k/m})"},
  {"claim":"Hamiltonian for central force: H = p_r^2/2m + p_phi^2/(2 m r^2)+V",
   "code":"m,r=symbols('m r',positive=True); pr,pphi=symbols('p_r p_phi'); rd,phid=symbols('rd phid'); V=symbols('V'); L=Rational(1,2)*m*(rd**2+r**2*phid**2)-V; pr_e=diff(L,rd); pphi_e=diff(L,phid); H=(pr_e*rd+pphi_e*phid-L); Hsub=H.subs({rd:pr/m,phid:pphi/(m*r**2)}); target=pr**2/(2*m)+pphi**2/(2*m*r**2)+V; result=simplify(Hsub-target)==0"},
  {"claim":"Gyroscope steady precession Omega = M g l/(I3 psidot)",
   "code":"Mg,l,I3,psid,th=symbols('Mg l I3 psid theta',positive=True); Om=symbols('Om'); eq=Eq(Mg*l*sin(th),Om*I3*psid*sin(th)); sol=solve(eq,Om)[0]; result=simplify(sol-Mg*l/(I3*psid))==0"}
]