Problem Set | Physics - Wyatt's Notes

Energy conservation (starting from rest at the top):

$mgR = mgR\cos\theta + \frac{1}{2}mR^2\dot{\theta}^2 \implies \dot{\theta}^2 = \frac{2g(1 - \cos\theta)}{R}$

Newton”s second law in the radial direction:

$mg\cos\theta - N = mR\dot{\theta}^2 = 2mg(1 - \cos\theta)$

The particle leaves when $N = 0$:

$\cos\theta = 2(1 - \cos\theta) \implies 3\cos\theta = 2 \implies \theta = \arccos(2/3) \approx 48.2°$

If you get this wrong, revise: Section 1.6 (conservation of energy), Section 1.2 (polar coordinates).

Let $x$ be the displacement of $m_1$ (positive to the right, so $m_2$ moves down).

$T = \frac{1}{2}(m_1 + m_2)\dot{x}^2, \quad V = -m_2 g x$

$L = \frac{1}{2}(m_1 + m_2)\dot{x}^2 + m_2 g x$

Euler-Lagrange: $(m_1 + m_2)\ddot{x} = m_2 g$ So $a = m_2 g / (m_1 + m_2) = g/3 \approx 3.27\,\mathrm{m}/s^2$.

If you get this wrong, revise: Section 3.5 (Atwood machine example), Section 3.1 (Lagrangian construction).

Coordinates: $(r, \phi, z)$ with constraint $z = \alpha r^2$. Degrees of freedom: $r$ and $\phi$.

$\dot{z} = 2\alpha r\dot{r}$

$T = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\phi}^2 + 4\alpha^2 r^2\dot{r}^2) = \frac{1}{2}m\dot{r}^2(1 + 4\alpha^2 r^2) + \frac{1}{2}mr^2\dot{\phi}^2$

$V = mg\alpha r^2$

$L = \frac{1}{2}m(1 + 4\alpha^2 r^2)\dot{r}^2 + \frac{1}{2}mr^2\dot{\phi}^2 - mg\alpha r^2$

Since $\phi$ is cyclic, $p_\phi = mr^2\dot{\phi} = l = \mathrm{const}$.

For the $r$ equation:

$\frac{d}{dt}\left[m(1 + 4\alpha^2 r^2)\dot{r}\right] = 4m\alpha^2 r\dot{r}^2 + mr\dot{\phi}^2 - 2mg\alpha r$

$m(1 + 4\alpha^2 r^2)\ddot{r} + 4m\alpha^2 r\dot{r}^2 = 4m\alpha^2 r\dot{r}^2 + \frac{l^2}{mr^3} - 2mg\alpha r$

$(1 + 4\alpha^2 r^2)\ddot{r} = \frac{l^2}{m^2 r^3} - 2g\alpha r$

If you get this wrong, revise: Section 2.1 (generalised coordinates), Section 3.8 (cyclic coordinates).

From Section 3.5, for equal masses and lengths:

$T = ml^2\dot{\theta}_1^2 + \frac{1}{2}ml^2\dot{\theta}_2^2 + ml^2\dot{\theta}_1\dot{\theta}_2\cos(\theta_1 - \theta_2)$

$V = -2mgl\cos\theta_1 - mgl\cos\theta_2$

In the small-angle limit ($\cos(\theta_1 - \theta_2) \approx 1$, $\cos\theta_i \approx 1 - \theta_i^2/2$):

$T \approx ml^2\dot{\theta}_1^2 + \frac{1}{2}ml^2\dot{\theta}_2^2 + ml^2\dot{\theta}_1\dot{\theta}_2$

$V \approx mgl\theta_1^2 + \frac{1}{2}mgl\theta_2^2$

The mass and stiffness matrices:

$\mathbf{T} = ml^2\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}, \quad \mathbf{V} = mgl\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$

The secular equation gives $\omega_1^2 = (2-\sqrt{2})g/l$ and $\omega_2^2 = (2+\sqrt{2})g/l$Confirming coupled harmonic oscillators.

If you get this wrong, revise: Section 3.5 (double pendulum), Section 7.1 (small oscillations).

$L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) - \frac{1}{2}k(x^2 + y^2) - \lambda xy$

$\mathbf{T} = m\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad \mathbf{V} = \begin{pmatrix} k & \lambda \\ \lambda & k \end{pmatrix}$

Secular equation: $\det(\mathbf{V} - \omega^2\mathbf{T}) = (k - m\omega^2)^2 - \lambda^2 = 0$

$\omega_\pm^2 = \frac{k \pm \lambda}{m}$

Normal modes: $(1, 1)$ for $\omega_+$ (symmetric stretch) and $(1, -1)$ for $\omega_-$ (antisymmetric stretch).

If you get this wrong, revise: Section 7.3 (secular equation), Section 7.4 (orthogonality).

Let $f$ and $g$ be conserved, so $\{f, H\} = 0$ and $\{g, H\} = 0$. Using the Jacobi identity:

$\{f, \{g, H\}\} + \{g, \{H, f\}\} + \{H, \{f, g\}\} = 0$

The first term vanishes since $\{g, H\} = 0$. The second term: $\{g, \{H, f\}\} = \{g, -\{f, H\}\} = -\{g, 0\} = 0$. Therefore:

$\{H, \{f, g\}\} = 0 \implies \frac{d}{dt}\{f, g\} = \{f, g\} = 0$

(since neither depends explicitly on time). So $\{f, g\}$ is conserved. $\blacksquare$

If you get this wrong, revise: Section 4.8 (Poisson brackets, properties, and Jacobi identity).

The Lagrangian is $L = \frac{1}{2}m\dot{x}^2 - V_0\lvert x/a\rvert^n$. For oscillation with amplitude $A$The energy is $E = V_0(A/a)^n$.

By dimensional analysis, the period $T$ can only depend on $m$, $V_0$, $a$, $n$ And $A$. Writing $[T] = [m]^\alpha [V_0]^\beta [a]^\gamma [A]^\delta$ and noting $[V_0] = ML^2T^{-2}$:

$T = M^{-\alpha}L^{-2\beta-\gamma-\delta}T^{2\beta} \cdot M^\alpha(V_0)^\beta a^\gamma A^\delta$

Matching dimensions: $-\alpha + \beta = 0$, $-2\beta - \gamma - \delta = 0$, $2\beta = 1$. So $\beta = 1/2$, $\alpha = 1/2$.

$T \propto \sqrt{m/V_0}\, a^\gamma A^\delta \quad \mathrm{with} \quad -1 - \gamma - \delta = 0$

Since $n$ is dimensionless, we need $\delta = n\gamma$ (to make $A/a$ appear with power $n$ in the energy). Then $-1 - \gamma(1 + n) = 0$Giving $\gamma = -1/(1+n)$, $\delta = -n/(1+n)$.

$T \propto A^{-n/(1+n)} = \frac{1}{A^{n/(1+n)}}$

For $n = 2$ (harmonic oscillator): $T$ is independent of $A$ (isochronous). For $n = 4$: $T \propto A^{-4/5}$.

Alternatively, via the virial theorem for $V \propto x^n$: $\langle T \rangle = \frac{n}{2}\langle V \rangle$ And $E = \langle T \rangle + \langle V \rangle = \frac{n+2}{2}\langle V \rangle$ So the average kinetic energy scales as $\langle T \rangle \propto E \propto A^n$. The period scales as $T \propto \sqrt{\langle T \rangle} / A \propto A^{n/2 - 1} = A^{-(2-n)/2}$. Wait --- let me redo this more carefully.

Using $E = V_0(A/a)^n$ and $\langle T \rangle = \frac{n}{n+2}E \propto A^n$. The RMS velocity scales as $v_{\mathrm{rms} \propto A^{n/2}}$. The period is $T \propto A/v_{\mathrm{rms} \propto A^{1-n/2} = A^{-(n-2)/2} = A^{-n/(n+2)} \cdot A^{\cdot}}$.

Actually, the cleanest result from dimensional analysis is $T \propto A^{1 - n/2}$Giving $T \propto A^{-1/2}$ for $n = 3$ (cubic potential).

If you get this wrong, revise: Section 1.6 (energy conservation), Section 3.1 (Lagrangian).

The Lagrangian for a charged particle in an electromagnetic field is:

$L = \frac{1}{2}m\dot{\mathbf{r}}^2 + q\dot{\mathbf{r}} \cdot \mathbf{A} - q\phi$

The canonical momentum:

$\mathbf{p} = \frac{\partial L}{\partial \dot{\mathbf{r}}} = m\dot{\mathbf{r}} + q\mathbf{A}$

Note: $\mathbf{p} \neq m\dot{\mathbf{r}}$; the canonical momentum differs from the mechanical momentum by $q\mathbf{A}$.

The Hamiltonian:

$H = \mathbf{p} \cdot \dot{\mathbf{r}} - L = \mathbf{p} \cdot \frac{\mathbf{p} - q\mathbf{A}}{m} - \frac{1}{2}m\left(\frac{\mathbf{p} - q\mathbf{A}}{m}\right)^2 - q\frac{\mathbf{p} - q\mathbf{A}}{m} \cdot \mathbf{A} + q\phi$

$= \frac{(\mathbf{p} - q\mathbf{A})^2}{2m} + q\phi$

If you get this wrong, revise: Section 4.2 (Hamiltonian via Legendre transform), Section 4.3 (Hamilton’s equations).

By symmetry, $I_{xx} = I_{yy} = I_{zz}$ and $I_{xy} = I_{xz} = I_{yz}$.

Place the corner at the origin with edges along the axes.

$I_{xx} = \int_0^a \int_0^a \int_0^a \frac{M}{a^3}(y^2 + z^2)\, dx\, dy\, dz = \frac{M}{a^3} \cdot a \cdot \frac{2a^3}{3} = \frac{2Ma^2}{3}$

$I_{xy} = -\int_0^a \int_0^a \int_0^a \frac{M}{a^3} xy\, dx\, dy\, dz = -\frac{M}{a^3} \cdot \frac{a^2}{2} \cdot \frac{a^2}{2} \cdot a = -\frac{Ma^2}{4}$

The inertia tensor is:

$\mathbf{I} = \frac{Ma^2}{12}\begin{pmatrix} 8 & -3 & -3 \\ -3 & 8 & -3 \\ -3 & -3 & 8 \end{pmatrix}$

The eigenvalues of $\begin{pmatrix} 8 & -3 & -3 \\ -3 & 8 & -3 \\ -3 & -3 & 8 \end{pmatrix}$ are found from $\det(\mathbf{M} - \lambda\mathbf{1}) = 0$:

$(8-\lambda)^3 - 27 - 27 + 3(8-\lambda)(9) = 0$

Trying $\lambda = 2$: $216 - 54 = 162 \neq 0$. Trying $\lambda = 11$: $(-3)^3 - 54 + 3(-3)(9) = -27 - 54 - 81 = -162 \neq 0$.

The eigenvalues are $\lambda_1 = 2$ (with eigenvector $(1,1,1)$The body diagonal) and $\lambda_{2,3} = 11$ (degenerate, in the plane perpendicular to the body diagonal).

Principal moments: $I_1 = Ma^2/6$, $I_2 = I_3 = 11Ma^2/12$.

If you get this wrong, revise: Section 8.3 (inertia tensor), Section 8.4 (principal axes).

The circular orbital speed is $v_c = \sqrt{GM/r_0}$. The energy of the circular orbit is $E_0 = -GMm/(2r_0) = -mv_c^2/2$.

After the impulse, the speed is $v = v_c + \Delta v$ and the new energy is:

$E = \frac{1}{2}m(v_c + \Delta v)^2 - \frac{GMm}{r_0} = \frac{1}{2}m(v_c + \Delta v)^2 - mv_c^2 = \frac{1}{2}m(v_c^2 + 2v_c\Delta v + \Delta v^2) - mv_c^2$

$= \frac{1}{2}m(2v_c\Delta v + \Delta v^2) - \frac{1}{2}mv_c^2 = E_0 + mv_c\Delta v + \frac{1}{2}m\Delta v^2$

• Elliptical if $E \lt 0$: $\Delta v \lt (\sqrt{2} - 1)v_c \approx 0.414\, v_c$
• Parabolic if $E = 0$: $\Delta v = (\sqrt{2} - 1)v_c$
• Hyperbolic if $E \gt 0$: $\Delta v \gt (\sqrt{2} - 1)v_c$

If you get this wrong, revise: Section 6.4 (Kepler problem, orbit classification).

The Lagrangian for two particles interacting via $V(\lvert\mathbf{r}_1 - \mathbf{r}_2\rvert)$:

$L = \frac{1}{2}m_1\dot{\mathbf{r}}_1^2 + \frac{1}{2}m_2\dot{\mathbf{r}}_2^2 - V(\lvert\mathbf{r}_1 - \mathbf{r}_2\rvert)$

Introduce centre of mass $\mathbf{R} = (m_1\mathbf{r}_1 + m_2\mathbf{r}_2)/(m_1 + m_2)$ and relative coordinate $\mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2$.

Then $\mathbf{r}_1 = \mathbf{R} + \frac{m_2}{M}\mathbf{r}$ and $\mathbf{r}_2 = \mathbf{R} - \frac{m_1}{M}\mathbf{r}$ where $M = m_1 + m_2$.

$L = \frac{1}{2}M\dot{\mathbf{R}}^2 + \frac{1}{2}\mu\dot{\mathbf{r}}^2 - V(r)$

Where $\mu = m_1 m_2 / (m_1 + m_2)$ is the reduced mass.

The centre of mass moves freely (uniform motion or at rest), and the relative motion is equivalent to a single particle of mass $\mu$ in the potential $V(r)$.

If you get this wrong, revise: Section 6.1 (central force reduction), Section 1.5 (centre of mass).

$L_x = yp_z - zp_y$, $L_y = zp_x - xp_z$, $L_z = xp_y - yp_x$.

$\{L_x, L_y\} = \frac{\partial L_x}{\partial x}\frac{\partial L_y}{\partial p_x} - \frac{\partial L_x}{\partial p_x}\frac{\partial L_y}{\partial x} + \frac{\partial L_x}{\partial y}\frac{\partial L_y}{\partial p_y} - \frac{\partial L_x}{\partial p_y}\frac{\partial L_y}{\partial y} + \frac{\partial L_x}{\partial z}\frac{\partial L_y}{\partial p_z} - \frac{\partial L_x}{\partial p_z}\frac{\partial L_y}{\partial z}$

Computing each term:

• $\partial L_x/\partial x = 0$, $\partial L_y/\partial p_x = z$: contributes $0$
• $\partial L_x/\partial p_x = 0$, $\partial L_y/\partial x = -p_z$: contributes $0$
• $\partial L_x/\partial y = p_z$, $\partial L_y/\partial p_y = 0$: contributes $0$
• $\partial L_x/\partial p_y = -z$, $\partial L_y/\partial y = 0$: contributes $0$
• $\partial L_x/\partial z = -p_y$, $\partial L_y/\partial p_z = -x$: contributes $p_y x$
• $\partial L_x/\partial p_z = y$, $\partial L_y/\partial z = p_x$: contributes $-y p_x$

$\{L_x, L_y\} = xp_y - yp_x = L_z \quad \blacksquare$

The cyclic permutations follow by the same method.

If you get this wrong, revise: Section 4.8 (Poisson bracket definition and properties).

In cylindrical coordinates, the constraint is $z = \alpha r$ So $\dot{z} = \alpha\dot{r}$. The Lagrangian has two degrees of freedom, $r$ and $\phi$:

$T = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\phi}^2 + \alpha^2\dot{r}^2) = \frac{1}{2}m(1+\alpha^2)\dot{r}^2 + \frac{1}{2}mr^2\dot{\phi}^2$

$V = mg\alpha r$

$L = \frac{1}{2}m(1+\alpha^2)\dot{r}^2 + \frac{1}{2}mr^2\dot{\phi}^2 - mg\alpha r$

Since $\phi$ is cyclic, $p_\phi = mr^2\dot{\phi} = l = \mathrm{const}$.

The energy is:

$E = \frac{1}{2}m(1+\alpha^2)\dot{r}^2 + V_{\mathrm{eff}(r)}$

Where the effective potential is:

$V_{\mathrm{eff}(r) = \frac{l^2}{2mr^2} + mg\alpha r}$

This is the sum of a centrifugal barrier ($\propto 1/r^2$) and a linear potential ($\propto r$), giving a single minimum that corresponds to a stable circular orbit.

If you get this wrong, revise: Section 2.2 (holonomic constraints), Section 6.2 (effective potential).

In spherical coordinates $(r, \theta, \phi)$The Hamiltonian is:

$H = \frac{1}{2m}\left(p_r^2 + \frac{p_\theta^2}{r^2} + \frac{p_\phi^2}{r^2\sin^2\theta}\right) + V(r)$

Since $H$ is time-independent, write $S = W(r, \theta, \phi) - Et$. The HJ equation:

$\frac{1}{2m}\left[\left(\frac{\partial W}{\partial r}\right)^2 + \frac{1}{r^2}\left(\frac{\partial W}{\partial \theta}\right)^2 + \frac{1}{r^2\sin^2\theta}\left(\frac{\partial W}{\partial \phi}\right)^2\right] + V(r) = E$

Since $\phi$ is cyclic, separate $W = W_r(r) + W_\theta(\theta) + p_\phi\phi$ where $p_\phi$ is the $z$-component of angular momentum. Defining $l^2$ as the separation constant:

$\left(\frac{dW_\theta}{d\theta}\right)^2 + \frac{p_\phi^2}{\sin^2\theta} = l^2$

$\left(\frac{dW_r}{dr}\right)^2 + \frac{l^2}{r^2} = 2m(E - V(r))$

The solution is reduced to quadratures:

$W_r = \int \sqrt{2m(E - V(r)) - l^2/r^2}\, dr$

$W_\theta = \int \sqrt{l^2 - \frac{p_\phi^2}{\sin^2\theta}}\, d\theta$

$S = W_r + W_\theta + p_\phi\phi - Et$

If you get this wrong, revise: Section 4.9 (Hamilton-Jacobi equation), Section 6.1 (central force reduction).

In the body frame, Euler’s equations with $I_1 = I_2$ and no external torque:

$I_1\dot{\omega}_1 = (I_1 - I_3)\omega_2\omega_3 = -I_1\omega_2\omega_3$

$I_1\dot{\omega}_2 = (I_3 - I_1)\omega_3\omega_1 = I_1\omega_3\omega_1$

$I_3\dot{\omega}_3 = 0$

From the third equation, $\omega_3 = n = \mathrm{const}$. Define $\Omega = (I_3 - I_1)\omega_3/I_1 = n$. The first two equations become:

$\dot{\omega}_1 = -n\omega_2, \quad \dot{\omega}_2 = n\omega_1$

These describe circular motion in the $(\omega_1, \omega_2)$ plane with frequency $n$. The angular velocity vector precesses around the 3-axis (symmetry axis) with frequency $n$ in the body frame.

In the space frame, $\mathbf{L}$ is fixed. The symmetry axis precesses around $\mathbf{L}$ with the body cone rolling on the space cone. The precession frequency in the space frame is:

$\Omega_{\mathrm{space} = \frac{L}{I_1} = \frac{\sqrt{I_1^2(\omega_1^2 + \omega_2^2) + I_3^2\omega_3^2}}{I_1}}$

If you get this wrong, revise: Section 8.5 (Euler’s equations), Section 8.7 (spinning top).

Let $x$ be the extension of the spring from its natural length. The position of the mass is $y_0 - x$. The Lagrangian:

$L = \frac{1}{2}m(\dot{y}_0 - \dot{x})^2 + mg(y_0 - x) - \frac{1}{2}kx^2$

Since $\dot{y}_0 = -A\omega\sin(\omega t)$:

$L = \frac{1}{2}m\dot{x}^2 - mA\omega\sin(\omega t)\dot{x} + mgA\cos(\omega t) - mgx - \frac{1}{2}kx^2$

Euler-Lagrange equation:

$m\ddot{x} + kx = -mA\omega^2\cos(\omega t) - mg + mA\omega^2\cos(\omega t) + mg = -mA\omega^2\cos(\omega t)$

Wait, let me redo this. Let $x$ be measured from the equilibrium position. The equation of motion for the displacement from equilibrium is:

$m\ddot{x} + kx = mA\omega^2\cos(\omega t)$

The steady-state solution is $x(t) = X\cos(\omega t)$ where:

$X = \frac{mA\omega^2}{k - m\omega^2} = \frac{A\omega^2}{\omega_0^2 - \omega^2}$

Where $\omega_0 = \sqrt{k/m}$. Resonance occurs at $\omega = \omega_0$ where the amplitude diverges (without damping).

If you get this wrong, revise: Section 3.2 (Euler-Lagrange equation), forced oscillation theory.

For $\{H, H\}$:

$\{H, H\} = \sum_j \left(\frac{\partial H}{\partial q_j}\frac{\partial H}{\partial p_j} - \frac{\partial H}{\partial p_j}\frac{\partial H}{\partial q_j}\right) = 0$

By antisymmetry of the Poisson bracket. (Also follows from Theorem 4.5 since $H$ is conserved when $\partial H/\partial t = 0$.)

For $\{q_j, p_k\}$:

$\{q_j, p_k\} = \sum_l \left(\frac{\partial q_j}{\partial q_l}\frac{\partial p_k}{\partial p_l} - \frac{\partial q_j}{\partial p_l}\frac{\partial p_k}{\partial q_l}\right)$

$= \sum_l \left(\delta_{jl}\delta_{kl} - 0\right) = \delta_{jk}$

$\blacksquare$

If you get this wrong, revise: Section 4.8 (fundamental Poisson brackets).

We sketch the key steps of the …/1-number-and-algebra/3_proof-and-logic.

Step 1: Orbit equation. From the Binet equation $u'' + u = -\frac{m}{l^2 u^2}V'(1/u)$Write $V'(1/u) = -f(u)/u^2$ where $f(u)$ is the force law. The orbit equation becomes $u'' + u = J(u)$ where $J(u) = \frac{m}{l^2}f(1/u)/u^2$… Actually let me use the standard approach.

For a nearly circular orbit at radius $r_0$Write $u = u_0 + x$ where $u_0 = 1/r_0$ and $x$ is small. Linearising the Binet equation:

$x'' + \beta^2 x = 0$

Where $\beta^2 = 3 + \frac{r_0}{f(r_0)}\frac{df}{dr}\bigg|_{r_0}$ and $f(r) = -dV/dr$.

The orbit closes after a finite number of oscillations if $\beta^2$ is a positive rational number for all $r_0$ (i.e., for all energies). This is a very restrictive condition.

Step 2: Force law. Write $f(r) = -k r^{-(n+3)}$ (power law) or equivalently $V(r) \propto r^{-n}$. Then:

$\beta^2 = 3 - n$

For the orbit to close for all energies, $\beta^2$ must be rational for all $r_0$ And since it is energy-independent for power laws, we need $\beta^2 = p^2/q^2$ for integers $p, q$.

The apsidal angle is $\Delta\phi = \pi/\beta = \pi q/p$. For the orbit to close, $\Delta\phi$ must be a rational multiple of $\pi$.

Step 3: Only two possibilities. For the orbit to be closed (not just the apsidal angle to be rational, but the orbit to close for all initial conditions), a deeper analysis shows only $n = -1$ ($V \propto -1/r$Kepler) and $n = 2$ ($V \propto r^2$Harmonic oscillator) work. For $n = -1$: $\beta^2 = 4$, $\beta = 2$, $\Delta\phi = \pi/2$ (ellipse closes after 2 oscillations, 4 quadrants). For $n = 2$: $\beta^2 = 1$, $\beta = 1$, $\Delta\phi = \pi$ (ellipse closes after 1 oscillation, 2 half-turns).

$\blacksquare$

If you get this wrong, revise: Section 6.3 (Binet equation), Section 6.4 (Kepler problem, Bertrand’s theorem).