Lagrangian mechanics: coordinates stop mattering

Idea. Choose the coordinate in which the constraints hold by construction; then the tension in the string, the reaction at the axle and the force in the rod never appear.

1. With \(x\) the descent of \(m_2\), the speeds are \(\dot x\) for both masses and \(\dot x/a\) for the pulley's rotation, so \[ L = \tfrac12\Big(m_1 + m_2 + \frac{I}{a^2}\Big)\dot x^2 + (m_2 - m_1)gx . \] The Euler–Lagrange equation is \((m_1 + m_2 + I/a^2)\ddot x = (m_2 - m_1)g\), so \[ \ddot x = \frac{(m_2 - m_1)\,g}{m_1 + m_2 + I/a^2} = \frac{0.5 \times 9.81}{2.7} = \mathbf{1.817\ m\,s^{-2}} , \] against \(0.5 \times 9.81/2.5 = 1.962\ \mathrm{m\,s^{-2}}\) with a light pulley. The pulley's inertia simply adds \(I/a^2\) to the mass being accelerated.

2. Squaring the velocity in rung 3, \[ \lvert\dot{\vec x}\rvert^2 = l^2\dot\theta^2 - 2bl\gamma\,\dot\theta\sin\gamma t\cos\theta + b^2\gamma^2\sin^2\gamma t , \] and \(V = -mgl\cos\theta\). So \[ L = \tfrac12 ml^2\dot\theta^2 - mbl\gamma\,\dot\theta\sin\gamma t\cos\theta + \tfrac12 mb^2\gamma^2\sin^2\gamma t + mgl\cos\theta . \] Now \(p_\theta = ml^2\dot\theta - mbl\gamma\sin\gamma t\cos\theta\), with time derivative \(ml^2\ddot\theta - mbl\gamma^2\cos\gamma t\cos\theta + mbl\gamma\,\dot\theta\sin\gamma t\sin\theta\), and \(\partial L/\partial\theta = mbl\gamma\,\dot\theta\sin\gamma t\sin\theta - mgl\sin\theta\). The \(\dot\theta\sin\gamma t\sin\theta\) terms cancel, leaving \[ \ddot\theta = -\frac gl\sin\theta + \frac{b\gamma^2}{l}\cos\gamma t\,\cos\theta . \] Interpretation. In the pivot's frame, which accelerates at \(\ddot x_{\mathrm p} = -b\gamma^2\cos\gamma t\), the bob feels a horizontal fictitious force \(-m\ddot x_{\mathrm p} = mb\gamma^2\cos\gamma t\); its lever arm about the pivot is the vertical distance \(l\cos\theta\), and torque divided by \(ml^2\) is the extra term.

3. \(\tfrac12 mb^2\gamma^2\sin^2\gamma t = \frac{\mathrm d}{\mathrm dt}F(t)\) with \(F(t) = \int_0^t\tfrac12 mb^2\gamma^2\sin^2\gamma s\,\mathrm ds\): a total time derivative of a function of \(t\), which the deck's gauge-freedom theorem says changes no equation. Directly: it contributes nothing to \(\partial L/\partial\theta\) or \(\partial L/\partial\dot\theta\).

Check. Dimensions: \(b\gamma^2/l\) is \(\mathrm{s^{-2}}\), as is \(g/l\) ✓. \(b = 0\) gives the plain pendulum ✓. \(\gamma = 0\): the pivot is fixed at \(x = b\), and the extra term vanishes ✓. Part 1 with \(I = 0\) and \(m_1 = m_2\): no acceleration ✓. Note also that \(L\) in part 2 depends on \(t\), so \(h\) is not conserved: the shaker feeds energy in.

Idea. The constraint depends on time, so the energy function is conserved but the energy is not; the difference is the work of the constraint, which is ideal (no virtual work) but not workless.

1. \(L = \tfrac12 m(\dot r^2 + \Omega^2r^2)\), and \(\mathcal E_r = m\ddot r - m\Omega^2r = 0\). The solution with \(r(0) = r_0\), \(\dot r(0) = 0\) is \(r(t) = r_0\cosh\Omega t\) (unit 06: the characteristic roots are \(\pm\Omega\)).

2. \(h = \dot r\,m\dot r - L = \tfrac12 m\dot r^2 - \tfrac12 m\Omega^2r^2\). \(L\) has no explicit \(t\), so \(h\) is conserved: \(h = -\tfrac12 m\Omega^2r_0^2\), that is, \(\dot r^2 = \Omega^2(r^2 - r_0^2)\). The energy is \(E = \tfrac12 m(\dot r^2 + \Omega^2r^2)\), and using the conserved \(h\), \[ E = h + m\Omega^2r^2 = m\Omega^2r^2 - \tfrac12 m\Omega^2r_0^2 , \] which grows as the bead moves out; \(E - E(0) = m\Omega^2(r^2 - r_0^2)\). The wire's force. It is horizontal and perpendicular to the wire (smooth wire; there is nothing radial for it to do, as \(m\ddot r = m\Omega^2r\) is exactly the radial inertial requirement \(m(\ddot r - r\dot\varphi^2) = 0\)). Its magnitude is \(N = m(r\ddot\varphi + 2\dot r\dot\varphi) = 2m\Omega\dot r\), the Coriolis term. Its point of application moves azimuthally at speed \(r\Omega\), so its power is \(P = 2m\Omega^2r\dot r = \frac{\mathrm d}{\mathrm dt}(m\Omega^2r^2)\), which integrates to \(m\Omega^2(r^2 - r_0^2)\) ✓. The wire does no virtual work (virtual displacements are along the wire, the force is across it) but it does real work, because the wire itself moves.

3. The bead reaches \(r = 1.0\) m when \(\cosh\Omega t = 10\): \(t = \operatorname{arccosh}(10)/2 = 2.9932/2 = \mathbf{1.497\ s}\). Then \(\dot r = \Omega\sqrt{r^2 - r_0^2} = 2\sqrt{0.99} = \mathbf{1.990\ m\,s^{-1}}\). Work done: \(m\Omega^2(r^2 - r_0^2) = 0.050 \times 4 \times 0.99 = \mathbf{0.198\ J}\).

Check. Dimensions: \([h] = \mathrm{kg\,m^2\,s^{-2}} = \mathrm J\) ✓. Energy bookkeeping at the end: \(E = \tfrac12(0.050)(1.990^2 + 2^2\cdot1^2) = 0.199\ \mathrm J\) and \(E(0) = \tfrac12(0.050)(4)(0.01) = 0.001\ \mathrm J\), difference \(0.198\) J ✓. \(\Omega \to 0\): the bead stays put and no work is done ✓.

Another way

Part 1 without solving the differential equation at all. \(L\) has no explicit \(t\), so \(h\) is conserved before anything is integrated, and \(h = -\tfrac12 m\Omega^2r_0^2\) gives \(\dot r = \Omega\sqrt{r^2 - r_0^2}\) straight away — unit 00's \(F(x)\) case, already reduced to quadrature. Separating, \[ \Omega t = \int_{r_0}^{r}\frac{\mathrm ds}{\sqrt{s^2 - r_0^2}} = \operatorname{arccosh}\frac{r}{r_0} , \] so \(r = r_0\cosh\Omega t\), the same answer, and part 3's time is read off the same line.

Idea. The lemma says the Euler–Lagrange expressions pair with the columns of the Jacobian; doing it once by hand shows where the centripetal term comes from.

1. \(\mathcal E_x = m\ddot x + V_x\), \(\mathcal E_y = m\ddot y + V_y\). In polar coordinates, \(\partial\tilde L/\partial\dot r = m\dot r\), \(\partial\tilde L/\partial r = mr\dot\theta^2 - V_r\), \(\partial\tilde L/\partial\dot\theta = mr^2\dot\theta\), \(\partial\tilde L/\partial\theta = -V_\theta\), so \[ \tilde{\mathcal E}_r = m\ddot r - mr\dot\theta^2 + V_r,\qquad \tilde{\mathcal E}_\theta = \frac{\mathrm d}{\mathrm dt}(mr^2\dot\theta) + V_\theta = m(r^2\ddot\theta + 2r\dot r\dot\theta) + V_\theta . \]

2. Differentiating twice, \[ \begin{aligned}\ddot x &= \ddot r\cos\theta - 2\dot r\dot\theta\sin\theta - r\ddot\theta\sin\theta - r\dot\theta^2\cos\theta,\\ \ddot y &= \ddot r\sin\theta + 2\dot r\dot\theta\cos\theta + r\ddot\theta\cos\theta - r\dot\theta^2\sin\theta.\end{aligned} \] Then \(\ddot x\cos\theta + \ddot y\sin\theta = \ddot r - r\dot\theta^2\) and \(-\ddot x\,r\sin\theta + \ddot y\,r\cos\theta = 2r\dot r\dot\theta + r^2\ddot\theta\). By the chain rule, \(V_r = V_x\cos\theta + V_y\sin\theta\) and \(V_\theta = -V_xr\sin\theta + V_yr\cos\theta\). Adding, both identities hold. The matrix is \(J^{\top}\), with \[ J = \frac{\partial(x, y)}{\partial(r, \theta)} = \begin{pmatrix}\cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta\end{pmatrix}, \] so \((\tilde{\mathcal E}_r, \tilde{\mathcal E}_\theta)^{\top} = J^{\top}(\mathcal E_x, \mathcal E_y)^{\top}\) — the deck's lemma.

3. \(-V_\theta = \vec F\cdot\partial\vec x/\partial\theta = \vec F\cdot(r\hat\theta) = rF_\theta\): the torque about the origin, in \(\mathrm{N\,m}\). The generalised force pairs \(\vec F\) with the coordinate direction \(\partial\vec x/\partial\theta\), which has length \(r\), not with the unit vector.

4. \(\tilde{\mathcal E}_r = 0\) reads \(m\ddot r = F_r + mr\dot\theta^2\): the Euler–Lagrange form is preserved because \(\partial\tilde L/\partial r\) picks up the \(r\)-dependence of the kinetic energy, whereas "mass times second derivative of the coordinate" is not a coordinate-free operation — \(\ddot r\) is not the radial component of the acceleration.

Check. Dimensions: \(\tilde{\mathcal E}_r\) in newtons and \(\tilde{\mathcal E}_\theta\) in newton-metres, matching \(J^{\top}\), whose second row carries a factor \(r\) ✓. Special case \(V = 0\), circular motion at constant \(\dot\theta\): \(\mathcal E = 0\) needs \(\ddot r = r\dot\theta^2 \ne 0\), so a free particle cannot move in a circle ✓.

Idea. The lemma is the chain rule plus Clairaut; invertibility is used only to pass from \(\tilde{\mathcal E} = 0\) back to \(\mathcal E = 0\), and when it fails the lost components are constraint forces.

1. Along a path, \(\dot q_i = \sum_b(\partial q_i/\partial Q_b)\dot Q_b + \partial q_i/\partial t\). As a function of \((Q, \dot Q, t)\) this is affine in \(\dot Q\), so \(\partial\dot q_i/\partial\dot Q_a = \partial q_i/\partial Q_a\). Differentiating in \(Q_a\), \(\partial\dot q_i/\partial Q_a = \sum_b\frac{\partial^2q_i}{\partial Q_a\partial Q_b}\dot Q_b + \frac{\partial^2q_i}{\partial Q_a\partial t}\), and by Clairaut (\(\phi\in C^2\)) this equals \(\frac{\mathrm d}{\mathrm dt}\frac{\partial q_i}{\partial Q_a}\). The chain rule gives \[ \frac{\partial\tilde L}{\partial\dot Q_a} = \sum_i\frac{\partial L}{\partial\dot q_i}\frac{\partial q_i}{\partial Q_a},\qquad \frac{\partial\tilde L}{\partial Q_a} = \sum_i\Big(\frac{\partial L}{\partial q_i}\frac{\partial q_i}{\partial Q_a} + \frac{\partial L}{\partial\dot q_i}\frac{\mathrm d}{\mathrm dt}\frac{\partial q_i}{\partial Q_a}\Big). \] Differentiating the first in \(t\) by the product rule produces \(\sum_i\big(\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q_i}\big)\frac{\partial q_i}{\partial Q_a} + \sum_i\frac{\partial L}{\partial\dot q_i}\frac{\mathrm d}{\mathrm dt}\frac{\partial q_i}{\partial Q_a}\), and subtracting the second cancels the last sum, leaving \(\tilde{\mathcal E}_a = \sum_i\mathcal E_i\,\partial q_i/\partial Q_a\), i.e. \(\tilde{\mathcal E} = J^{\top}\mathcal E\). If \(J\) is invertible, so is \(J^{\top}\), and \(\tilde{\mathcal E} = 0 \iff \mathcal E = 0\): a path satisfies the equations in one system exactly when its image does in the other. ∎

2. \(\tilde L = \tfrac12 m\dot Q^2\), so \(\tilde{\mathcal E} = m\ddot Q\), zero for \(Q = vt\). Its image \((x, y) = (vt, 0)\) has \(\mathcal E = (m\ddot x,\ m\ddot y + mg) = (0, mg) \ne 0\). Here \(J = (1, 0)^{\top}\) and \(J^{\top}\mathcal E = \mathcal E_x = 0\): the lemma holds, but only the \(x\)-component is seen. Physically this is a particle sliding on a smooth horizontal table \(y = 0\): the unseen component \(\mathcal E_y = mg\) is the table's normal force (Newton says \(\mathcal E = \vec R\)). A non-invertible \(\phi\) is a parametrisation of a constraint, and what it loses is exactly the constraint force — which is how Lagrange's equations for constrained systems arise.

3. \(\dot x = \dot Q + at\), so \(\tilde L = \tfrac12 m(\dot Q + at)^2 - V(Q + \tfrac12 at^2)\). Then \(\partial\tilde L/\partial\dot Q = m(\dot Q + at)\), with time derivative \(m\ddot Q + ma\), and \(\partial\tilde L/\partial Q = -V'\). So \(m\ddot Q = -V' - ma\): the fictitious force \(-ma\) of an accelerating frame, obtained with no argument about frames. Expanding, \[ \tilde L = \tfrac12 m\dot Q^2 + mat\,\dot Q + \tfrac12 ma^2t^2 - V = \tfrac12 m\dot Q^2 - maQ - V + \frac{\mathrm d}{\mathrm dt}\Big(maQt + \tfrac16 ma^2t^3\Big), \] using \(mat\dot Q = \frac{\mathrm d}{\mathrm dt}(maQt) - maQ\). Dropping the total derivative, the accelerated observer's Lagrangian is that of a particle in an extra uniform gravitational field of strength \(a\), potential energy \(maQ\). An accelerating frame and a uniform gravitational field are indistinguishable by their equations of motion — the observation unit 30 builds the equivalence principle on.

Check. Dimensions in part 3: \(maQ\) is \(\mathrm{kg\,m\,s^{-2}\,m} = \mathrm J\) ✓, and \(\frac{\mathrm d}{\mathrm dt}(\tfrac16 ma^2t^3) = \tfrac12 ma^2t^2\) ✓. Limit \(a = 0\): \(Q = x\) and nothing changes ✓. Part 1 with \(\phi\) the identity: \(J = I\) ✓.

Idea. The angular chart builds the constraint in; the Cartesian description keeps it as an equation and pays with a multiplier, which turns out to be the tension. Pairing the Cartesian equation with the chart's coordinate directions discards the multiplier and returns the angular equations.

1. \(L = \tfrac12 ml^2(\dot\theta^2 + \sin^2\theta\,\dot\varphi^2) + mgl\cos\theta\). \[ ml^2\ddot\theta = ml^2\sin\theta\cos\theta\,\dot\varphi^2 - mgl\sin\theta,\qquad \frac{\mathrm d}{\mathrm dt}\big(ml^2\sin^2\theta\,\dot\varphi\big) = 0 . \] Conserved: \(p_\varphi = ml^2\sin^2\theta\,\dot\varphi\) (\(\varphi\) cyclic: angular momentum about the vertical), and \(h = T + V = \tfrac12 ml^2(\dot\theta^2 + \sin^2\theta\,\dot\varphi^2) - mgl\cos\theta\) (no explicit \(t\), scleronomic constraint).

2. \(L = \tfrac12 m\lvert\dot{\vec x}\rvert^2 - mgz\) and \(\nabla G = 2\vec x\), so \(m\ddot{\vec x} = -mg\hat z + 2\lambda\vec x\). Differentiating \(\lvert\vec x\rvert^2 = l^2\) twice: \(\vec x\cdot\ddot{\vec x} + \lvert\dot{\vec x}\rvert^2 = 0\). Dot the equation of motion with \(\vec x\): \(-m\lvert\dot{\vec x}\rvert^2 = -mgz + 2\lambda l^2\), so \[ \lambda = \frac{m\big(gz - \lvert\dot{\vec x}\rvert^2\big)}{2l^2} . \] The rod's force is \(2\lambda\vec x\), of magnitude \(2\lvert\lambda\rvert l\), pulling towards the pivot when \(\lambda \lt 0\) — which it is whenever the bob is below the pivot (\(z \lt 0\)).

3. Substitute \(\vec x(\theta, \varphi)\) into the Cartesian equation and dot it with \(\partial\vec x/\partial\theta = l(\cos\theta\cos\varphi, \cos\theta\sin\varphi, \sin\theta)\) and \(\partial\vec x/\partial\varphi = l(-\sin\theta\sin\varphi, \sin\theta\cos\varphi, 0)\). Both are orthogonal to \(\vec x\) (check: \(\vec x\cdot\partial\vec x/\partial\theta = l^2(\sin\theta\cos\theta - \cos\theta\sin\theta) = 0\), and \(\vec x\cdot\partial\vec x/\partial\varphi = 0\)), so the multiplier term drops out. Gravity gives \(-mg\hat z\cdot\partial\vec x/\partial\theta = -mgl\sin\theta\) and \(-mg\hat z\cdot\partial\vec x/\partial\varphi = 0\). On the left, by the two identities of the deck's d'Alembert slide, \(m\ddot{\vec x}\cdot\partial\vec x/\partial q = \frac{\mathrm d}{\mathrm dt}\partial T/\partial\dot q - \partial T/\partial q\) with \(T = \tfrac12 ml^2(\dot\theta^2 + \sin^2\theta\,\dot\varphi^2)\). For \(q = \theta\) this is \(ml^2\ddot\theta - ml^2\sin\theta\cos\theta\,\dot\varphi^2\); for \(q = \varphi\) it is \(\frac{\mathrm d}{\mathrm dt}(ml^2\sin^2\theta\,\dot\varphi)\). So the two paired equations are exactly part 1's. The third direction, \(\vec x\) itself, carries the remaining Cartesian equation, which only determines \(\lambda\). Conversely, part 1's equations plus the constraint give the motion, and \(\lambda\) is then computed from part 2's formula: the descriptions have the same solutions.

Where each fails. The angular chart fails at \(\theta = 0\) (and \(\pi\)): \(\varphi\) is undefined there and \(\partial\vec x/\partial\varphi = 0\). A plane pendulum swinging through the bottom has \(\varphi\) jumping by \(\pi\) and \(\dot\varphi\) infinite in this chart, though nothing physical happens. The Cartesian description has no such point: it uses the configuration space as a subset of \(\mathbb R^3\), which is the price of the multiplier.

4. Constant \(\theta = \alpha\) in part 1's \(\theta\) equation: \(0 = \sin\alpha\cos\alpha\,\dot\varphi^2 - (g/l)\sin\alpha\), so \[ \dot\varphi^2 = \frac{g}{l\cos\alpha} = \frac{9.81}{1.2 \times 0.86603} = 9.4397,\qquad \dot\varphi = \mathbf{3.072\ s^{-1}},\qquad \text{period } \mathbf{2.045\ s}. \] Tension from part 2: \(z = -l\cos\alpha\) and \(\lvert\dot{\vec x}\rvert^2 = l^2\sin^2\alpha\,\dot\varphi^2 = lg\sin^2\alpha/\cos\alpha\), so \(\lambda = -\frac{mg}{2l}\big(\cos\alpha + \sin^2\alpha/\cos\alpha\big) = -\frac{mg}{2l\cos\alpha}\), and the tension is \(2\lvert\lambda\rvert l = mg/\cos\alpha = 0.50 \times 9.81/0.86603 = \mathbf{5.664\ N}\). From the angular description it is not available directly — no multiplier — but a vertical force balance on the steadily circling bob gives \(T\cos\alpha = mg\), the same \(5.664\) N.

5. Form invariance: for a \(C^2\) point transformation with invertible Jacobian the Euler–Lagrange expressions satisfy \(\tilde{\mathcal E} = J^{\top}\mathcal E\), so their zero sets correspond. Here the relevant version is the non-invertible one of problem P4 part 2: the chart \((\theta, \varphi) \mapsto \vec x\) has \(3\times2\) Jacobian of rank 2 away from the poles, and the lemma says the angular equations are the Cartesian residuals paired with its columns — which is part 3.

Check. Dimensions: \([\lambda] = \mathrm{kg\,m\,s^{-2}}/\mathrm m = \mathrm{N\,m^{-1}}\), so \(2\lambda\vec x\) is in newtons ✓. At rest hanging straight down, \(\lambda = -mg/(2l)\) and the tension is \(mg\) ✓. \(\alpha \to 0\): \(\dot\varphi \to \sqrt{g/l}\), the small-oscillation frequency of a plane pendulum, as a conical pendulum of tiny radius is two perpendicular small oscillations a quarter period apart ✓.

Idea. A total derivative adds nothing to the equations but shifts the momenta. So a conservation law for a canonical momentum is a statement about a gauge-dependent quantity: the physics is in the equations, and in the gauge-invariant combinations.

1. With \(F = \sum_bf_{q_b}\dot q_b + f_t\): \(\partial F/\partial\dot q_a = f_{q_a}\), whose time derivative along any path is \(\sum_bf_{q_aq_b}\dot q_b + f_{q_at}\); and \(\partial F/\partial q_a = \sum_bf_{q_bq_a}\dot q_b + f_{tq_a}\). By Clairaut (\(f\in C^2\)) these are equal, so \(\mathcal E_a[F] = 0\) identically, for every path. Since \(\mathcal E\) is linear in \(L\), \(\mathcal E[L'] = \mathcal E[L]\). ∎

2. \(p'_a = \partial L'/\partial\dot q_a = p_a + \partial F/\partial\dot q_a = p_a + f_{q_a}\). And \(h' = \sum_a\dot q_ap'_a - L' = h + \sum_a\dot q_af_{q_a} - \big(\sum_af_{q_a}\dot q_a + f_t\big) = h - f_t\).

3. \(\frac{\mathrm d}{\mathrm dt}(mgtz) = mgz + mgt\dot z\), and \(L_1 + \frac{\mathrm d}{\mathrm dt}(mgtz) = \tfrac12 m\dot z^2 + mgt\dot z = L_2\) ✓. In \(L_2\), \(z\) does not appear: it is cyclic, so \(p_2 = m\dot z + mgt\) is conserved — which says \(\dot z = -gt + \text{const}\), true. In \(L_1\), \(z\) is not cyclic, \(p_1 = m\dot z\) is not conserved, but \(L_1\) has no explicit \(t\), so \(h_1 = \tfrac12 m\dot z^2 + mgz\) is conserved — energy conservation, true. In \(L_2\), \(h_2 = \dot z(m\dot z + mgt) - L_2 = \tfrac12 m\dot z^2\), not conserved, consistent with \(\mathrm dh_2/\mathrm dt = -\partial L_2/\partial t = -mg\dot z\). Both descriptions are right; they choose different functions to call "momentum" and "energy function", and the conservation laws each gets for free are different — though each is a consequence of the other's equations.

4. The canonical momentum \(p = \partial L/\partial\dot q\) is not gauge-invariant (nor is \(h\)). What is gauge-invariant is anything built from the motion itself: here the kinetic momentum \(m\dot z\) and the energy \(\tfrac12 m\dot z^2 + mgz\), which have the same values whichever Lagrangian was used to find the motion. Section 5 of the deck meets the important case: for a charge, \(p = m\dot{\vec x} + q\vec A\) changes under \(\vec A \to \vec A + \nabla\chi\), and \(m\dot{\vec x}\) does not.

Check. Dimensions: \(mgt\dot z\) is \(\mathrm{kg\,m\,s^{-2}\cdot s\cdot m\,s^{-1}} = \mathrm J\) ✓, so \(L_2\) is a Lagrangian. Both give \(\ddot z = -g\): for \(L_2\), \(\frac{\mathrm d}{\mathrm dt}(m\dot z + mgt) = 0\) ✓. Limit \(g \to 0\): \(L_1 = L_2\), \(f = 0\) ✓.

Idea. The two gauges describe the same force, so the same motion; but a gauge change alters \(L\) by a total derivative, which can make a different coordinate cyclic and changes what "the conserved momentum" is.

1. \(\nabla\times(-By, 0, 0) = (0, 0, -\partial_y(-By)) = (0, 0, B)\), and \(\nabla\times\tfrac12 B(-y, x, 0) = (0, 0, \tfrac12 B + \tfrac12 B) = (0, 0, B)\) ✓. The difference is \(\vec A_{\mathrm L} - \vec A_{\mathrm S} = (-\tfrac12 By, -\tfrac12 Bx, 0) = \nabla\chi\) with \(\chi = -\tfrac12 Bxy\). So \(L_{\mathrm L} = L_{\mathrm S} + q\,\mathrm d\chi/\mathrm dt\).

2. \(L = \tfrac12 m(\dot x^2 + \dot y^2 + \dot z^2) - qBy\dot x\); \(x\) and \(z\) are cyclic. \(p_x = m\dot x - qBy\) is conserved, and the \(y\) equation is \(m\ddot y = \partial L/\partial y = -qB\dot x\). Substituting \(\dot x = (p_x + qBy)/m\): \[ \ddot y = -\frac{qB}{m^2}(p_x + qBy) = -\omega_c^2\,(y - y_0),\qquad y_0 = -\frac{p_x}{qB}, \] with \(\omega_c = \lvert q\rvert B/m\): simple harmonic motion in \(y\) about \(y_0\). Then \(\dot x = (qB/m)(y - y_0)\), so \(x\) oscillates at the same frequency a quarter period out of phase: a circle, centred on the line \(y = y_0\). (Its \(x\)-centre is fixed by initial data; \(p_x\) knows only \(y_0\) — the other component of the centre is conserved too, but is not a canonical momentum in this gauge.) Along \(z\), \(m\dot z\) is constant.

3. In cylindrical coordinates \(\vec A_{\mathrm S} = \tfrac12 Br\,\hat\varphi\) and \(q\dot{\vec x}\cdot\vec A_{\mathrm S} = \tfrac12 qBr^2\dot\varphi\), so \(L = \tfrac12 m(\dot r^2 + r^2\dot\varphi^2 + \dot z^2) + \tfrac12 qBr^2\dot\varphi\). Now \(\varphi\) is cyclic, and the conserved quantity is \(p_\varphi = mr^2\dot\varphi + \tfrac12 qBr^2\) — the canonical angular momentum, not the mechanical \(mr^2\dot\varphi\), which is not conserved about the origin for a circle that does not surround it. In this gauge \(x\) is not cyclic; in the other, \(\varphi\) is not. The motions are identical (the Lorentz force \(q\dot{\vec x}\times\vec B\) is the same); the kinetic momentum \(m\dot{\vec x}\) is the same in both; the canonical momenta differ by \(q\nabla\chi\).

4. \(\omega_c = eB/m_e = 1.60217663\times10^{-19}/9.1093837\times10^{-31} = \mathbf{1.759\times10^{11}\ s^{-1}}\); period \(2\pi/\omega_c = \mathbf{3.572\times10^{-11}\ s}\); radius \(v/\omega_c = m_ev/(eB) = \mathbf{5.686\ \mu m}\).

Check. Dimensions: \(qB/m\) is \(\mathrm{C\,T\,kg^{-1}} = \mathrm{C\,(kg\,s^{-1}C^{-1})\,kg^{-1}} = \mathrm{s^{-1}}\) ✓; \(qBy\) is a momentum, as \(p_x\) requires ✓. Limit \(B \to 0\): \(\omega_c \to 0\), radius \(\to\infty\), straight-line motion ✓. Circular orbit centred on the origin in part 3: \(r = \rho\), \(\dot\varphi = -qB/m\) gives \(p_\varphi = -\tfrac12 qB\rho^2\), constant ✓.

Idea. The multiplier is the normal force, and a one-sided contact lasts exactly as long as it has the right sign.

1. \(\partial L/\partial\dot r = m\dot r\), \(\partial L/\partial r = mr\dot\theta^2 - mg\cos\theta\), so \(m\ddot r - mr\dot\theta^2 + mg\cos\theta = \lambda\cdot\partial G/\partial r = \lambda\). On the sphere, \(\lambda = mg\cos\theta - mR\dot\theta^2\). The constraint is scleronomic, so \(h = T + V\): \(\tfrac12 mR^2\dot\theta^2 + mgR\cos\theta = mgR\), giving \(R\dot\theta^2 = 2g(1 - \cos\theta)\). Hence \[ \lambda(\theta) = mg\,(3\cos\theta - 2) . \] The force \(\lambda\,\partial G/\partial r\) acts along \(\hat r\), outward: the sphere's push.

2. The sphere can supply only \(\lambda \ge 0\). \(\lambda\) falls from \(mg\) at the top to zero at \(\cos\theta = 2/3\): \(\theta = \mathbf{48.19^\circ}\). Beyond it the constrained solution would need \(\lambda \lt 0\), a pull, so the block leaves and flies freely. Its speed there is \(v^2 = R^2\dot\theta^2 = 2gR(1 - \tfrac23) = \tfrac23 gR\), so \(v = \sqrt{2 \times 9.81/3} = \mathbf{2.557\ m\,s^{-1}}\) for \(R = 1.0\) m, after a drop of \(R/3\).

Check. Dimensions: \(\lambda\) in newtons ✓. At the top, \(\lambda = mg\): the sphere holds the block's weight ✓. At departure the radial equation reads \(mv^2/R = mg\cos\theta\): gravity alone supplies the centripetal force, the classic condition ✓. The angle is independent of \(g\), \(R\) and \(m\), as dimensional analysis (unit 00) demands: no dimensionless combination of them exists.

Idea. Lagrange–d'Alembert restricts the virtual displacements; the vakonomic route restricts the varied curves. For a non-holonomic constraint these are different conditions, and only the first is mechanics.

1. \(m\ddot x = \lambda\sin\varphi\), \(m\ddot y = -\lambda\cos\varphi\), \(I\ddot\varphi = 0\), plus the constraint. So \(\dot\varphi = \omega\), constant. The constraint gives \(\dot x = v\cos\varphi\), \(\dot y = v\sin\varphi\), and \(\dot v = \ddot x\cos\varphi + \ddot y\sin\varphi + \dot\varphi(-\dot x\sin\varphi + \dot y\cos\varphi) = \tfrac\lambda m(\sin\varphi\cos\varphi - \cos\varphi\sin\varphi) + 0 = 0\). Constant speed and constant turning rate: a circle of radius \(v/\omega\) (a straight line if \(\omega = 0\)). Projecting sideways, \(m(\ddot x\sin\varphi - \ddot y\cos\varphi) = \lambda\), and the differentiated constraint gives \(\ddot x\sin\varphi - \ddot y\cos\varphi = -\omega v\), so \(\lambda = -mv\omega\): the magnitude \(mv\omega = mv^2/(v/\omega)\) is the centripetal force, supplied by the blade's sideways grip.

2. The Euler–Lagrange equations of the augmented integrand: \[ \frac{\mathrm d}{\mathrm dt}(m\dot x + \mu\sin\varphi) = 0,\quad \frac{\mathrm d}{\mathrm dt}(m\dot y - \mu\cos\varphi) = 0,\quad I\ddot\varphi = \mu(\dot x\cos\varphi + \dot y\sin\varphi) = \mu v, \] and the \(\mu\)-equation is the constraint. The first two give \(m\ddot x = -\dot\mu\sin\varphi - \mu\dot\varphi\cos\varphi\), \(m\ddot y = \dot\mu\cos\varphi - \mu\dot\varphi\sin\varphi\). Forward projection: \(m(\ddot x\cos\varphi + \ddot y\sin\varphi) = -\mu\dot\varphi\), and since the \(\dot\varphi(-\dot x\sin\varphi + \dot y\cos\varphi)\) term of \(\dot v\) vanishes by the constraint, \(m\dot v = -\mu\dot\varphi\). Sideways projection: \(m(\ddot x\sin\varphi - \ddot y\cos\varphi) = -\dot\mu\); with the differentiated constraint, \(-m\dot\varphi v = -\dot\mu\), so \(\dot\mu = m\dot\varphi v\).

3. \(\frac{\mathrm d}{\mathrm dt}(m^2v^2 + \mu^2) = 2mv(-\mu\dot\varphi) + 2\mu(m\dot\varphi v) = 0\). So \(mv = \rho\cos\alpha\), \(\mu = \rho\sin\alpha\) with \(\rho\) constant, and either equation gives \(\dot\alpha = \dot\varphi\): \(\alpha = \varphi - c\). Then \[ I\ddot\varphi = \mu v = \frac{\rho^2}{m}\sin(\varphi - c)\cos(\varphi - c) = \frac{\rho^2}{2m}\sin\big(2(\varphi - c)\big), \] a pendulum equation for the heading (in the variable \(2(\varphi - c)\)), with the speed \(v = (\rho/m)\cos(\varphi - c)\) rising and falling as the heading swings. A Lagrange–d'Alembert circle has \(\dot v = 0\) and \(\dot\varphi = \omega \ne 0\); the vakonomic \(\dot v = -\mu\omega/m\) then forces \(\mu \equiv 0\), and \(\dot\mu = m\omega v\) forces \(\omega v = 0\). So the only common motions are straight lines (\(\omega = 0\)) and turning on the spot (\(v = 0\)).

4. A real skate given a push while turning traces a circle at constant speed; that is Lagrange–d'Alembert, and it is what experiment finds (Lewis and Murray, 1995, cited in the deck). The vakonomic system predicts a heading that oscillates like a pendulum and a speed that varies, with no force along the motion to cause it. And it has an extra initial datum: the physical state \((x, y, \varphi, v, \dot\varphi)\) at \(t = 0\) does not determine \(\mu(0)\), so it does not determine the motion — a failure of the determinism unit 01 proved Newtonian mechanics has. The vakonomic equations describe an optimal-control problem (the cheapest way to steer a skate between two configurations), not a skate.

Check. Energy \(E = \tfrac12 mv^2 + \tfrac12 I\dot\varphi^2\) in the vakonomic system: \(\dot E = mv\dot v + I\dot\varphi\ddot\varphi = -v\mu\dot\varphi + \dot\varphi\mu v = 0\), so both systems conserve energy, and energy cannot tell them apart; the circle can. Dimensions: \([\mu] = [mv] = \mathrm{kg\,m\,s^{-1}}\), and \([\rho^2/2m] = (\mathrm{kg\,m\,s^{-1}})^2/\mathrm{kg} = \mathrm{kg\,m^2\,s^{-2}}\), the units of \(I\ddot\varphi\) ✓. Limit \(I \to \infty\): \(\dot\varphi\) constant in both, and the vakonomic \(\dot v = -\mu\omega/m\), \(\dot\mu = m\omega v\) rotate \((mv, \mu)\) at rate \(\omega\), so \(v\) still oscillates unless \(\omega = 0\) ✓ — the discrepancy is not an artefact of the heading's dynamics.

Idea. The frequency at a stable equilibrium is the square root of the curvature of \(V_{\mathrm{eff}}\) there; at the bifurcation that curvature passes through zero, so the frequency does too. The normal form is the leading term of a series whose next terms are computable.

1. \(f'(\theta) = \Omega^2\cos2\theta - \omega_0^2\cos\theta\). At the bottom \(f'(0) = \Omega^2 - \omega_0^2 \lt 0\) for \(\Omega \lt \omega_0\), frequency \(\sqrt{\omega_0^2 - \Omega^2}\). At \(\theta_*\), \(f'(\theta_*) = \Omega^2(2\cos^2\theta_* - 1) - \Omega^2\cos^2\theta_* = -\Omega^2\sin^2\theta_*\), frequency \(\Omega\sin\theta_* = \sqrt{\Omega^4 - \omega_0^4}/\Omega\). Both vanish as \(\Omega \to \omega_0\) — critical slowing down.

2. With \(\omega_0^2 = 28.0286\ \mathrm{s^{-2}}\):

At \(\Omega = 5.5\) the bottom has \(f'(0) = \mu = 2.2214\ \mathrm{s^{-2}}\): a displacement grows like \(e^{\sqrt\mu\,t}\), rate \(1.4904\ \mathrm{s^{-1}}\), \(e\)-folding time \(\mathbf{0.671\ s}\). Just above \(\Omega_c = 5.294\), at \(5.3\), the tilted wells are only \(3.8^\circ\) from the bottom and a rocking period is nearly 18 s.

3. \(\sin\theta = \theta - \tfrac16\theta^3 + \tfrac1{120}\theta^5\) and \(\Omega^2\cos\theta - \omega_0^2 = \mu - \tfrac12\Omega^2\theta^2 + \tfrac1{24}\Omega^2\theta^4\). The product to \(\theta^5\) is \[ f = \mu\theta - \Big(\frac{\Omega^2}{2} + \frac\mu6\Big)\theta^3 + \Big(\frac{\Omega^2}{8} + \frac{\mu}{120}\Big)\theta^5 , \] the quintic collecting \(\tfrac1{24}\Omega^2 + \tfrac1{12}\Omega^2 = \tfrac18\Omega^2\) and \(\tfrac1{120}\mu\). At \(\Omega = 5.5\): \(\mu = 2.2214\), cubic coefficient \(15.4952\), quintic \(3.7998\).

• Leading order, cubic coefficient at \(\mu = 0\) (\(\tfrac12\omega_0^2\)): \(\theta_*^2 = 2\mu/\omega_0^2\), \(\theta_* = \mathbf{22.81^\circ}\).
• Full cubic: \(\theta_*^2 = \mu/15.4952\), \(\theta_* = \mathbf{21.69^\circ}\).
• With the quintic: the smaller root of \(\mu - 15.4952\,x + 3.7998\,x^2 = 0\), \(x = \theta_*^2\), gives \(\theta_* = \mathbf{22.10^\circ}\).

Exact: \(\arccos(28.0286/30.25) = \mathbf{22.09^\circ}\). The first two estimates straddle it, the first \(0.72^\circ\) above and the second \(0.40^\circ\) below — an \(O(\mu)\) relative error, as the normal form promises — and the quintic brings it within \(0.01^\circ\).

Check. Dimensions: \(\Omega\sin\theta_*\) is \(\mathrm{s^{-1}}\) ✓. The two formulas of part 1 at \(\Omega = \omega_0\) both give 0 ✓. Large \(\Omega\): \(\Omega\sin\theta_* \to \Omega\), the \(g = 0\) oscillation about the equator found in the deck's worked Lagrangian ✓.

Idea. After scaling out units, the whole problem depends on one number \(\beta\), and the determinant factorises as a difference of squares.

1. With the scaled matrices, \(\det\begin{pmatrix}1 - \lambda & -\beta\lambda\\ -\beta\lambda & \beta(1 - \lambda)\end{pmatrix} = \beta(1 - \lambda)^2 - \beta^2\lambda^2 = 0\), so \(1 - \lambda = \pm\sqrt\beta\,\lambda\) and \(\lambda_\mp = 1/(1 \pm \sqrt\beta)\). Write \(s = \sqrt\beta\). The first row gives \((1 - \lambda)v_1 = \beta\lambda v_2\), so \(v_2/v_1 = (1 - \lambda)/(\beta\lambda) = \pm s/\beta = \pm1/s\): the slow mode \(\lambda = 1/(1 + s)\) is \((1, 1/s)\), in phase; the fast mode \(\lambda = 1/(1 - s)\) is \((1, -1/s)\), in antiphase.

2. \(\omega_{\mathrm{fast}}/\omega_{\mathrm{slow}} = \sqrt{(1 + s)/(1 - s)} = 2\) gives \(1 + s = 4 - 4s\), \(s = 3/5\), \(\beta = 9/25 = 0.36\), and \(m_2/m_1 = \beta/(1 - \beta) = 9/16 = \mathbf{0.5625}\). Then \(\lambda = 1/1.6 = 0.625\) and \(1/0.4 = 2.5\): \(\omega_{\mathrm{slow}} = \sqrt{0.625 \times 9.81} = 2.4761\ \mathrm{s^{-1}}\), period \(\mathbf{2.537\ s}\); \(\omega_{\mathrm{fast}} = 4.9523\ \mathrm{s^{-1}}\), period \(\mathbf{1.269\ s}\), exactly half ✓. The modes are \((1, \tfrac53)\) and \((1, -\tfrac53)\).

3. With the scaled \(M\), \(u^{\top}Mv\) for \(u = (1, 1/s)\), \(v = (1, -1/s)\) is \(1 + \beta(-1/s) + \beta(1/s) + \beta(1/s)(-1/s) = 1 - \beta/s^2 = 0\) ✓. The Euclidean product is \(1 - 1/s^2 = 1 - 25/9 = -16/9 \ne 0\).

Check. \(\beta = \tfrac12\) (equal masses) gives \(\lambda = 1/(1 \pm 0.70711) = 0.58579, 3.41421 = 2 \mp \sqrt2\), the deck's values, and mode ratio \(\pm\sqrt2\) ✓. \(\beta \to 0\): both \(\lambda \to 1\), both periods \(2\pi\sqrt{l/g} = 2.006\) s ✓. Dimensions: \(\lambda\) is a pure number, so \(\omega^2 = \lambda g/l\) is \(\mathrm{s^{-2}}\) ✓.

Idea. Three coordinates, one Lagrangian; the forces in the string appear only as \(ke_i\), and never need a name. The equations are linear, so the motion is a uniformly accelerated "rigid Atwood" solution plus normal modes, one of which has zero frequency because the rigid motion costs no spring energy.

1. With \(e_1 = y_1 + z - c_1\), \(e_2 = y_2 - z - c_2\) (depths measured downwards, so gravity's potential energy is \(-m_igy_i\)): \[ L = \tfrac12 m_1\dot y_1^2 + \tfrac12 m_2\dot y_2^2 + \tfrac12 J\dot z^2 + m_1gy_1 + m_2gy_2 - \tfrac12 ke_1^2 - \tfrac12 ke_2^2 . \] Since \(\partial e_1/\partial y_1 = 1\), \(\partial e_1/\partial z = 1\), \(\partial e_2/\partial y_2 = 1\), \(\partial e_2/\partial z = -1\): \[ m_1\ddot y_1 = m_1g - ke_1,\qquad m_2\ddot y_2 = m_2g - ke_2,\qquad J\ddot z = -ke_1 + ke_2 . \] (The kinetic energy of the pulley is \(\tfrac12 I\dot\varphi^2 = \tfrac12 J\dot z^2\).)

2. Constant stretches: \(\ddot y_1 = -\ddot z\), \(\ddot y_2 = \ddot z\). Write \(A = \ddot z\). The first two equations give \(ke_1 = m_1(g + A)\), \(ke_2 = m_2(g - A)\), and the third \(JA = m_2(g - A) - m_1(g + A)\), so \[ A = \frac{(m_2 - m_1)\,g}{m_1 + m_2 + J} = \frac{0.1 \times 9.81}{1.35} = \mathbf{0.7267\ m\,s^{-2}} , \] the textbook Atwood result with the pulley's inertia, and independent of \(k\). Stretches: \(e_1 = 0.5 \times 10.537/200 = \mathbf{26.34\ mm}\), \(e_2 = 0.6 \times 9.083/200 = \mathbf{27.25\ mm}\).

3. The equations are linear with constant coefficients, so the difference \(u\) of any two solutions solves the homogeneous system. Subtracting the particular solution, \(M\ddot u + Ku = 0\) with \(u = (u_1, u_2, u_z)\), \[ M = \begin{pmatrix}m_1 & 0 & 0\\ 0 & m_2 & 0\\ 0 & 0 & J\end{pmatrix},\qquad K = k\begin{pmatrix}1 & 0 & 1\\ 0 & 1 & -1\\ 1 & -1 & 2\end{pmatrix}, \] \(K\) being the Hessian of \(\tfrac12 k(u_1 + u_z)^2 + \tfrac12 k(u_2 - u_z)^2\). \(K(1, -1, -1)^{\top} = 0\): a zero mode, \(\omega = 0\), whatever the masses. It is the rigid motion — mass 1 down by one unit, mass 2 up by one, the rim turning so the left side goes down — which changes neither stretch. The reason is a symmetry: the spring energy is unchanged by that shift, and the corresponding direction in configuration space costs no restoring force. (The zero-mode solution is \(u = (1, -1, -1)(c_0 + c_1t)\): uniform motion of the rigid Atwood machine, superposed on the accelerating one.)

4. With \(m_1 = m_2 = m\) and \(J = \beta m\) (here \(\beta = 0.5\)): try \((1, 1, 0)\): \(K(1, 1, 0)^{\top} = k(1, 1, 0)^{\top}\), \(M(1,1,0)^{\top} = m(1, 1, 0)^{\top}\), so \(\omega^2 = k/m\): the two masses bounce in phase on their springs, the pulley still. Then \((1, -1, c)\): \(K(1, -1, c)^{\top} = k(1 + c,\ -1 - c,\ 2 + 2c)^{\top}\) and \(M(1, -1, c)^{\top} = m(1, -1, \beta c)^{\top}\). Matching the first components, \(\omega^2 = k(1 + c)/m\); the third then needs \(2k(1 + c) = \omega^2m\beta c = k(1 + c)\beta c\), so \(c = 2/\beta\) (or \(c = -1\), the zero mode). Hence \(\omega^2 = (k/m)(1 + 2/\beta)\), with mode \((1, -1, 2/\beta)\): masses in antiphase, the pulley swinging against them. Numbers: \[ \omega_1 = \sqrt{200/0.5} = \mathbf{20.0\ s^{-1}}\ (\text{period } 0.314\ \mathrm s),\qquad \omega_2 = \sqrt{400 \times 5} = \mathbf{44.72\ s^{-1}}\ (\text{period } 0.140\ \mathrm s). \] Limits. \(J \to \infty\) (\(\beta \to \infty\)): \(\omega_2^2 \to k/m\), and both modes become each mass bouncing on its own spring from a fixed pulley ✓. \(k \to \infty\): \(\omega_1, \omega_2 \to \infty\), the oscillations freeze out, and only the zero mode and the accelerating solution remain — the rigid Atwood machine of problem P1 ✓.

Check. Dimensions: \(A\) in \(\mathrm{m\,s^{-2}}\) ✓, \(e = m(g \pm A)/k\) in metres ✓. Part 2's numbers satisfy the third equation: \(JA = 0.1817\) N and \(k(e_2 - e_1) = 0.1817\) N ✓. The modes are \(M\)-orthogonal: \((1,1,0)\cdot M(1,-1,4)^{\top} = m - m = 0\), and each is \(M\)-orthogonal to \((1,-1,-1)\): \(m - m = 0\) and \(m + m - \beta m\cdot4\cdot1 = 2m - 2m = 0\) ✓.