Interleaved — Phase 2

Physics interleaved practice

printable — key stays hidden on paper

Instructions. Work each problem independently. The problems deliberately jump between subtopics — before solving, identify which framework applies (Lagrangian, Hamiltonian, Poisson-bracket, rigid-body, etc.). Show all steps. Use $g$ for gravitational acceleration. Marks in [ ]. Total: 72 marks.

1. [6] A bead of mass $m$ slides on a rigid wire bent into the shape $y = ax^2$ in a vertical plane. State how many degrees of freedom the system has, classify the constraint (holonomic/non-holonomic, scleronomic/rheonomic), choose a suitable generalized coordinate, and write the kinetic energy $T$ in that coordinate.

2. [7] For a particle in a central potential $V(r)$ in a plane, using polar coordinates $(r,\theta)$ : (a) Write the Lagrangian. (b) Identify any cyclic coordinate and state the corresponding conserved quantity. (c) Write the generalized momenta $p_r$ and $p_\theta$ .

3. [8] Two equal masses $m$ are connected by three identical springs (constant $k$ ) in a line between two fixed walls: wall–spring– $m$ –spring– $m$ –spring–wall. Find the normal mode frequencies and describe the normal coordinates.

4. [6] A disk of mass $M$ , radius $R$ rolls without slipping down an incline of angle $\alpha$ . Using a Lagrange multiplier for the rolling constraint, find the equation of motion and the constraint (friction) force.

5. [6] Given the Hamiltonian $H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2$ , write Hamilton's equations and sketch (describe) the phase-space trajectories. What is conserved and what is the shape of the trajectories?

6. [6] Compute the Poisson brackets (a) $\{q, p\}$ , (b) $\{L_x, L_y\}$ where $L_i = \epsilon_{ijk} x_j p_k$ , and (c) state the connection between the Poisson bracket and the quantum commutator.

7. [7] A symmetric top (moments $I_1 = I_2 \ne I_3$ ) undergoes torque-free rotation. Starting from Euler's equations, show that $\omega_3$ is constant and that the transverse angular velocity precesses at a fixed rate. Give that rate.

8. [7] For a Lagrangian $L$ invariant under time translation, use Noether's theorem (or the total time derivative of $L$ ) to show that the energy function $h = \sum_i \dot q_i \frac{\partial L}{\partial \dot q_i} - L$ is conserved. State the symmetry ↔ conservation law correspondence.

9. [6] The inertia tensor of a body in some frame is $I = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & 1 & 3 \end{pmatrix} \; (\text{units of } MR^2).$ Find the principal moments of inertia and the principal axes.

10. [7] Use the Hamilton–Jacobi method for a free particle in one dimension ( $H = p^2/2m$ ). Write the H–J equation, separate the time dependence, solve for Hamilton's principal function $S(x,t)$ , and recover $x(t)$ .

11. [6] A generating function $F_1(q,Q) = \tfrac{1}{2}m\omega q^2 \cot Q$ defines a canonical transformation for the harmonic oscillator. Find $p$ and $P$ , and show the new Hamiltonian gives action-angle-like variables.

Answer keyMark scheme & solutions

1. Subtopics 2.1.1, 2.1.2, 2.1.3. Why: cues "wire shape" + "classify constraint" force the constraints/DOF/KE combination.

The bead is confined to the curve $y=ax^2$ : this is one equation relating $x,y$ , time-independent → holonomic and scleronomic. In 2D there were 2 coordinates; one constraint leaves 1 DOF. Choose $q=x$ . $y = ax^2,\quad \dot y = 2ax\dot x.$ $T = \tfrac12 m(\dot x^2 + \dot y^2) = \tfrac12 m\dot x^2\left(1 + 4a^2x^2\right).$

2. Subtopics 2.1.4, 2.1.8, 2.1.7. Why: central potential → recognize $\theta$ cyclic.

(a) $L = \tfrac12 m(\dot r^2 + r^2\dot\theta^2) - V(r)$ . (b) $\theta$ does not appear in $L$ → cyclic. Conserved: $p_\theta = mr^2\dot\theta =$ angular momentum $\ell$ . (c) $p_r = \dfrac{\partial L}{\partial \dot r} = m\dot r$ , $\quad p_\theta = \dfrac{\partial L}{\partial\dot\theta}= mr^2\dot\theta$ .

3. Subtopic 2.1.20. Why: two coupled masses with symmetric spring chain — classic normal-mode problem.

Displacements $x_1,x_2$ . Equations: $m\ddot x_1 = -kx_1 + k(x_2-x_1) = -2kx_1 + kx_2,$ $m\ddot x_2 = -k(x_2-x_1) - kx_2 = kx_1 - 2kx_2.$ Matrix $\frac{k}{m}\begin{pmatrix}2 & -1\\-1 & 2\end{pmatrix}$ , eigenvalues $\omega^2 = \frac{k}{m}(2\mp1)$ .

Symmetric mode $Q_+ = x_1+x_2$ : $\omega_1 = \sqrt{k/m}$ (masses move together).
Antisymmetric mode $Q_- = x_1-x_2$ : $\omega_2 = \sqrt{3k/m}$ (masses move oppositely). Normal coordinates $Q_\pm = \tfrac{1}{\sqrt2}(x_1\pm x_2)$ decouple the equations.

4. Subtopic 2.1.10. Why: "find the constraint force" is the trigger for Lagrange multipliers rather than just eliminating the constraint.

Coordinates $s$ (down-slope distance) and $\phi$ (rotation), constraint $f = s - R\phi = 0$ . $L = \tfrac12 M\dot s^2 + \tfrac12 I\dot\phi^2 + MgS\sin\alpha,\quad I=\tfrac12 MR^2.$ E–L with multiplier $\lambda$ : $M\ddot s - Mg\sin\alpha = \lambda,\qquad I\ddot\phi = -\lambda R.$ Constraint: $\ddot s = R\ddot\phi$ . Substitute: $M\ddot s - Mg\sin\alpha = \lambda,\quad \tfrac12 MR^2 \frac{\ddot s}{R} = -\lambda R \Rightarrow \tfrac12 M\ddot s = -\lambda.$ So $M\ddot s - Mg\sin\alpha = -\tfrac12 M\ddot s \Rightarrow \tfrac32 M\ddot s = Mg\sin\alpha$ , $\boxed{\ddot s = \tfrac23 g\sin\alpha},\qquad \lambda = -\tfrac12 M\ddot s = -\tfrac13 Mg\sin\alpha.$ The friction (constraint) force magnitude is $\tfrac13 Mg\sin\alpha$ .

5. Subtopics 2.1.12, 2.1.13. Why: given $H$ → Hamilton's equations + phase portrait.

$\dot q = \frac{\partial H}{\partial p} = \frac{p}{m},\qquad \dot p = -\frac{\partial H}{\partial q} = -m\omega^2 q.$ $H = E$ is conserved. Trajectories are level sets $\frac{p^2}{2m} + \tfrac12 m\omega^2 q^2 = E$ = ellipses in $(q,p)$ space, traversed clockwise. Rescaling $p/\sqrt m$ , $q\sqrt m\omega$ makes them circles.

6. Subtopic 2.1.15. Why: direct PB computation and quantum connection.

(a) $\{q,p\} = \dfrac{\partial q}{\partial q}\dfrac{\partial p}{\partial p} - \dfrac{\partial q}{\partial p}\dfrac{\partial p}{\partial q} = 1$ . (b) $\{L_x,L_y\} = L_z$ (angular-momentum algebra), and cyclically. (c) Correspondence: $\{A,B\} \longrightarrow \dfrac{1}{i\hbar}[\hat A,\hat B]$ . Thus $\{L_x,L_y\}=L_z$ becomes $[\hat L_x,\hat L_y]=i\hbar\hat L_z$ .

7. Subtopics 2.1.21, 2.1.23. Why: "torque-free symmetric top, Euler's equations" specifically selects Euler's equations.

Euler's equations, $N_i=0$ : $I_1\dot\omega_1 = (I_2-I_3)\omega_2\omega_3,\;\; I_2\dot\omega_2=(I_3-I_1)\omega_3\omega_1,\;\; I_3\dot\omega_3=(I_1-I_2)\omega_1\omega_2.$ With $I_1=I_2$ : third gives $\dot\omega_3 = 0 \Rightarrow \omega_3=$ const. Define $\Omega = \dfrac{(I_3-I_1)}{I_1}\omega_3$ (constant). First two become $\dot\omega_1 = -\Omega\omega_2,\quad \dot\omega_2 = +\Omega\omega_1.$ So $\omega_1+i\omega_2 = A e^{i\Omega t}$ : the transverse angular velocity precesses at rate $\boxed{\Omega = \frac{I_3-I_1}{I_1}\omega_3}$ (body-frame precession).

8. Subtopics 2.1.9, 2.1.11. Why: time-translation symmetry → energy conservation (Noether).

$\frac{dL}{dt} = \sum_i\left(\frac{\partial L}{\partial q_i}\dot q_i + \frac{\partial L}{\partial\dot q_i}\ddot q_i\right)+\frac{\partial L}{\partial t}.$ Using E–L, $\frac{\partial L}{\partial q_i}=\frac{d}{dt}\frac{\partial L}{\partial\dot q_i}$ : $\frac{dL}{dt} = \sum_i\frac{d}{dt}\!\left(\dot q_i\frac{\partial L}{\partial\dot q_i}\right)+\frac{\partial L}{\partial t}.$ Thus $\frac{d}{dt}\Big(\sum_i \dot q_i\frac{\partial L}{\partial\dot q_i}-L\Big) = -\frac{\partial L}{\partial t}$ . If $L$ has no explicit $t$ (time-translation symmetry), $h=\sum\dot q_ip_i-L$ is conserved — this is the energy. Symmetry (time translation) ↔ energy conservation.

9. Subtopic 2.1.22. Why: given a matrix → diagonalize for principal axes.

The $x$ -axis is already principal: $I=4$ . Remaining $2\times2$ block $\begin{pmatrix}3&1\\1&3\end{pmatrix}$ has eigenvalues $3\pm1 = 4, 2$ , eigenvectors $(0,1,1)/\sqrt2$ and $(0,1,-1)/\sqrt2$ . Principal moments: $\{4, 4, 2\}\,MR^2$ . Principal axes: $\hat x$ ; $(0,1,1)/\sqrt2$ ; $(0,1,-1)/\sqrt2$ .

10. Subtopic 2.1.17. Why: "Hamilton–Jacobi" named explicitly.

H–J equation: $\dfrac{\partial S}{\partial t} + \dfrac{1}{2m}\left(\dfrac{\partial S}{\partial x}\right)^2 = 0.$ Separate $S = W(x) - Et$ : $\frac{1}{2m}(W')^2 = E \Rightarrow W' = \sqrt{2mE}=p$ , so $W = px$ . $S = px - \frac{p^2}{2m}t.$ New constant $\beta = \frac{\partial S}{\partial p} = x - \frac{p}{m}t \Rightarrow x = \beta + \frac{p}{m}t$ — uniform motion. ✓

11. Subtopics 2.1.16, 2.1.18. Why: given a generating function → canonical transformation, leading to action-angle.

$F_1(q,Q)=\tfrac12 m\omega q^2\cot Q$ . $p = \frac{\partial F_1}{\partial q} = m\omega q\cot Q,\qquad P = -\frac{\partial F_1}{\partial Q} = \frac{m\omega q^2}{2\sin^2 Q}.$ Invert: $q = \sqrt{\dfrac{2P}{m\omega}}\sin Q$ , $p=\sqrt{2Pm\omega}\cos Q$ . Then $H = \frac{p^2}{2m}+\tfrac12 m\omega^2 q^2 = \omega P$ . Since $H$ is independent of $Q$ , $Q$ is cyclic and $P=E/\omega$ is constant (the action variable); $\dot Q = \omega$ , $