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Write the electron electric charge as $-e$, with $e>0$ the elementary charge, and choose $\mathbf{B}=B\hat{\mathbf{z}}$ and initial velocity $u\hat{\mathbf{x}}$. In the nonrelativistic approximation, Newton's second law with the magnetic Lorentz force givesThe magnetic force is perpendicular to the velocity, so it does no work: $d(m_e|\mathbf{v}{|}^2/2)/dt=0$. Setting ${\omega }_c=eB/m_e$, the solution with $x(0)=y(0)=0$ isConsequently $x^2+(y-r{)}^2=r^2$, where $r=u/{\omega }_c$: this is a cyclotron orbit. Its frequency and cyclotron radius areThe accelerating voltage changes $u$ and $r$ but cancels out of $f$. Using $m_e=9.1094\times 10^{-31}\mathrm{kg}$ and $e=1.6022\times 10^{-19}\mathrm{C}$ givesOnly the magnitude of the magnetic field is determined; its direction is perpendicular to the orbital plane, with the sense of rotation fixed by the negative electric charge. The claimed voltage independence is a nonrelativistic result: exactly, $\gamma =1+eV/(m_e{c}^2)$ and $f=eB/(2\pi \gamma m_e)$. At $1\mathrm{kV}$ the correction is about $0.2\%$. The stated model also neglects the energy lost to emitted electromagnetic radiation.

Let $\mu$ be the instantaneous total mass of the rocket, including its remaining rocket propellant. During a short ejection it changes by $d\mu <0$. The expelled mass $-d\mu$ has velocity $v-v_0$ to first order in a fixed inertial frame. With no external force, momentum conservation readsDiscarding the second-order product gives $\mu dv=-v_{0}d\mu$. Integrate from initial total mass $m_0+M$ to rocket dry mass $m_0$:For constant mass flow rate $\alpha =-\dot{\mu }>0$, the same rocket equation gives $\mu (t)=m_0+M-\alpha t$ and $v(t)=v_0\ln [(m_0+M)/\mu (t)]$ until $t=M/\alpha$. The final result depends on the rocket mass ratio, not on how rapidly the fuel is ejected. These are Newtonian velocities; a very large rocket mass ratio would require a relativistic treatment even if the exhaust is slow in the instantaneous rocket frame.

Choose positive horizontal velocity to the right and positive spin angular velocity anticlockwise, as in the source drawing. At the bottom of the sphere, the rotational velocity is then to the right: the initial tangential contact-point velocity isFor the illustrated positive $v_1,{\Omega }_1$, the contact slips to the right, so the tangential contact force acts leftward. The upward normal force acts through the sphere centre and produces no spin torque. The tangential contact force acts at lever arm $a$ and produces a clockwise torque, reducing the positive spin. The weight acts vertically downward through the centre of mass.

During the short impact the contact impulses dominate the weight's impulse. In an ideal elastic rough contact, the tangential deformation can store and return energy; its contact force need not be a dissipative sliding-friction law throughout the collision. For other signed incoming states, reverse the initial tangential force if $w_1<0$; if $w_1=0$, no tangential impulse is required.

Take $x$ positive downward and put zero gravitational potential energy at the spring's support. The mass is then a distance $l+x$ below that reference, so $V_g=-mg(l+x)$. Hooke's law gives the elastic potential energy $k{x}^2/2$. ThereforeThe constant term $mgl$ fixes the requested reference but does not affect the motion. The Euler-Lagrange equation isThis is exactly Newton's second law with gravity downward and spring force $kx$ upward for positive extension. The static extension is $x_0=mg/k$.

Let $\theta$ be the displacement from the upward vertical and $v$ the speed of the ball's centre of mass. Its centre follows a circle of radius $R+r$. The rolling without slipping constraint is $v=r\omega$, with $\omega$ the ball's spin angular velocity. The static contact friction does no work on the ball because the instantaneous contact point is stationary. Thus conservation of energy givesProject Newton's second law toward the hemisphere's centre. The inward component of gravity is $Mg\cos \theta$ and the normal force $N$ points outward, soAt contact loss $N=0$. Eliminating $v$ gives the ideal sphere rolling off a hemisphere resultThe radius cancels, so replacing $R+r$ by $R$ under the stated small-ball approximation leaves the same answer.

Finite-friction qualification. This is an ideal rolling-constraint result. It does not describe uninterrupted pure rolling to detachment with any finite coefficient of static friction. The tangential force and torque equations giveso $f=(2/7)Mg\sin \theta$ acts uphill, whereasConsequently the required ratio isFor a concrete example, a coefficient of static friction of one allows rolling only up to about $47.54^{\circ }$, where $17\cos \theta -10=2\sin \theta$, before sliding begins. A subsequent physical contact-loss angle needs a sliding-friction model. The boxed $53.97^{\circ }$ answers the source's expressly imposed no-slip idealization.

For a differentiable central potential, the force is $\mathbf{F}=-V^{\prime }(r)\hat{\mathbf{r}}$. The angular momentum is $\mathbf{J}=m\mathbf{r}\times \dot{\mathbf{r}}$, soThis proves conservation of angular momentum. For $J\ne 0$, the trajectory lies in the fixed plane perpendicular to $\mathbf{J}$. In plane polar coordinates, $J=m{r}^2\dot{\varphi }$. Substituting this into the kinetic energy gives the radial energy equationThe second term is the centrifugal contribution to the effective potential. Its radial derivative reproduces $m\ddot{r}=-V^{\prime }(r)+J^2/(m{r}^3)$. Purely radial motion is the $J=0$ case.

For the satellite, write $\mu =GM$, with $M$ Earth's mass and $m$ the satellite's mass. At apogee and perigee, the radial velocity is zero, so angular momentum conservation and conservation of energy giveEliminating $v_{\mathrm{P}}$ yieldsFor an elliptic orbit, $r_{\mathrm{A}}+r_{\mathrm{P}}=2a$. Therefore the vis-viva equation at apogee isSubstitution into the orbital energy gives $E=-GMm/(2a)$. The circular case follows by continuity, or directly from the radial force balance.

An instantaneous tangential retrograde burn at apogee leaves the position unchanged. For the speed change $v\mapsto (1-ϵ)v$, the orbital energy change isDifferentiating $E=-\mu m/(2a)$ gives $\Delta E=\mu m\Delta a/(2a^2)$ to first order. Since $r_{\mathrm{A}}=a(1+e)$, the vis-viva equation implies $v^2=\mu (1-e)/[a(1+e)]$. HenceThe factor of two distinguishes the major axis from the semimajor axis. As a finite-change check, the new semimajor axis is exactlywithin the instantaneous-burn, unchanged-mass model. Expanding this expression gives the stated result.

The massless inextensible string has hanging length $y=l-r$, measured downward from the table. Thus $\dot{y}=-\dot{r}$. In polar coordinates, the tabletop mass has squared speed ${\dot{r}}^2+r^2{\dot{\theta }}^2$. Adding the two kinetic energies and setting the hanging gravitational potential energy to $-Mgy$ givesand the Lagrangian isThe canonical momenta are $p_r=(M+m)\dot{r}$ and $p_{\theta }=m{r}^2\dot{\theta }$. Performing the Legendre transform gives the HamiltonianIt equals the total mechanical energy and is conserved because $L$ has no explicit time dependence. Also $\theta$ is a cyclic coordinate, so $J=p_{\theta }=m{r}^2\dot{\theta }$ is conserved angular momentum about the vertical through the hole. The table and the string guide supply external forces; conservation of the system's total linear momentum does not follow. These are the two independent generic integrals needed for the reduced tabletop mass coupled to a hanging mass.

The retained mass at time $t$ is $M(t)=M_0+kt$. Incoming rain has zero horizontal momentum in the ground reference frame. During $dt$, horizontal momentum conservation for the carriage and the rain about to enter givessoThis is a variable-mass system: the entering mass must be accelerated horizontally, even though no external horizontal force acts on the carriage-plus-rain system. Integrating the velocity,For $u_0\ge 0$ this is also the distance travelled. Its $k\to 0$ limit is $u_{0}t$. Vertical impact can dissipate kinetic energy; horizontal momentum conservation does not require kinetic energy conservation.

Take downward distance $s$ and downward speed $v=ds/dt\ge 0$, with constant gravitational acceleration $g>0$ and quadratic drag coefficient $k>0$. The printed vector square has no standard vector meaning; the directionally consistent drag force is ${\mathbf{F}}_d=-mk|\mathbf{v}|\mathbf{v}$. For this downward fall, Newton's second law becomesThe terminal velocity is obtained by setting the net force to zero:From rest, the speed increases monotonically towards this limiting value. In fact $v(t)=v_{\infty }\tanh (\sqrt{gk}t)$; thus it is a supremum approached at infinite time, and a tower of finite height need not permit it to be reached.

For the required distance, use the chain rule $dv/dt=vdv/ds$. Writing $w=v^2$ gives a linear ordinary differential equation:Consequently,Here $k$ has units of inverse length, so the result has the required units of distance and is independent of the falling mass.

First take $a\ne 0$, put $A=|a|$, and suppose $k>0$. The force is the negative derivative of the potential energy:The potential energy is an odd function, passes through $U(0)=0$, and approaches zero from above as $x\to -\infty$ and from below as $x\to +\infty$. Its stationary points are $x=\pm A$:Thus the sole stable equilibrium is the local minimum at $x=A$; the local maximum at $x=-A$ is an unstable equilibrium. The requested values at the stable equilibrium areThe force is an even function, positive for $|x|<A$, negative for $|x|>A$, and zero at $x=\pm A$. Its central maximum is $F(0)=k/A^2$. Sinceits two minima occur at $x=\pm \sqrt{3}A$, with $F=-k/(8A^2)$, and $F\to 0^{-}$ at both ends. These values fix the shapes of both requested sketches.

In one dimension, attraction towards the origin means $xF(x)<0$, while repulsion from the origin means $xF(x)>0$. With that explicitly stated convention,A positive force alone means rightward, not necessarily outward. At the origin $F(0)\ne 0$, so it is not an equilibrium; the inward/outward distinction is undefined exactly there.

The sign assumptions matter because the paper calls $a,k$ merely constants. For $k<0$ both curves reverse sign, the stable equilibrium is $x=-A$, its potential energy is $-|k|/(2A)$, and the attraction/repulsion intervals interchange. For $k=0$ the force vanishes everywhere on the nonsingular domain and there is no isolated potential energy minimum. If $a=0$ and $k\ne 0$, then $U=-k/x$ for $x\ne 0$ and $F=-k/x^2$, with no finite equilibrium.

Axis theorems. The parallel axis theorem stateswhere $I_G$ is the moment of inertia about an axis through the centre of mass, $I_O$ is about a parallel axis, $m$ is the body's total mass, and $d$ is the perpendicular separation of the axes. Expanding $\int [(x-d{)}^2+y^2]dm$ proves the result because $\int xdm=0$ at the centre of mass. The perpendicular axis theorem applies to a planar lamina in the $xy$ plane:All three axes meet at the same point, $x,y$ lie in the lamina and $z$ is normal to it; the identity follows from $x^2+y^2$ being the squared distance to the $z$ axis. It is not a theorem for an arbitrary three-dimensional body.

Cylinder integrals. Put the cylinder's midpoint at $z=0$ and its symmetry axis along $z$. Its uniform mass density is $\rho =m/(\pi r^{2}l)$. The transverse moment of inertia of a uniform solid cylinder about the $x$ axis isEquivalently each thin circular slice has diametral moment of inertia $r^{2}dm/4$, by the perpendicular axis theorem and symmetry, and shifting it by $z$ adds $z^{2}dm$. The parallel axis theorem then gives the transverse end-axis moment of inertia:The axial moment of inertia of a uniform solid cylinder is $I_{\parallel }=m{r}^2/2$. Equating it to $I_{\perp ,G}$ givesHere the source's principal axis means the cylinder's symmetry axis; its transverse centre axes are also principal axes. This equality describes a short cylinder rather than the thin-rod limit.

Hammer dynamics and forces. Treat the compact head as a point mass and neglect its intrinsic moment of inertia. Let $\theta$ be the displacement from downward vertical, and let ${\mathbf{e}}_r$ point along the rod from pivot to head, with ${\mathbf{e}}_{\theta }$ in the direction of increasing $\theta$. DefineThe two external weights $Mg$ and $mg$ act at distances $l$ and $l/2$ from the pivot. The frictionless pivot supplies a reaction $\mathbf{R}$ but no torque about its axis. The total restoring torque and rotational equation of motion areFor completeness, Newton's second law for the centre of mass gives both components of the pivot force:The negative radial component points towards the pivot. These follow from the centre of mass acceleration $-d{\dot{\theta }}^2{\mathbf{e}}_r+d\ddot{\theta }{\mathbf{e}}_{\theta }$ and the total weight $m_{T}g(\cos \theta {\mathbf{e}}_r-\sin \theta {\mathbf{e}}_{\theta })$. They specify the external forces even when the oscillation is not small.

With the small-angle approximation, $\sin \theta \simeq \theta$, this is simple harmonic motion. The physical pendulum has small-amplitude oscillation periodFor release from a finite amplitude $0<{\theta }_0<\pi$, conservation of mechanical energy instead gives $\frac{1}{2}I{\dot{\theta }}^2=gA(\cos \theta -\cos {\theta }_0)$. Integrating a quarter-cycle gives the exact oscillation periodwhere $K$ is the complete elliptic integral of the first kind; $K(0)=\pi /2$ recovers $T_0$.

For $m=M/10$ and $r=l/20$, compare with the simple pendulum value $T_s=2\pi \sqrt{l/g}$:For small oscillations, the hammer's oscillation period is about $0.794\%$ shorter. The extra rod weight adds restoring torque proportionately faster than it adds pivot moment of inertia.

Let $D$ be the separation of the Earth and Moon, and let $x$ be the spacecraft's distance from the Earth's centre towards the Moon. On $0<x<D$, the two gravitational forces oppose each other. Their equality, by Newton's law of universal gravitation, isTaking the positive square root givesThe centre of mass of the two bodies is at $x_G=D{M}_M/(M_E+M_M)$ from the Earth's centre. Thus the gravitational field null between two point masses lies a distancefrom their centre of mass, on the Moon's side. With $M_E/M_M=81$, $x=9D/10$ and $x_G=D/82$, soThe point is $3.84\times 10^7\mathrm{m}$ from the Moon. It is an instantaneous zero of the physical gravitational field, not a co-rotating Lagrange point: a spacecraft remaining fixed relative to the Earth-Moon line would need a nonzero centripetal acceleration. With zero net force it cannot follow that circular co-rotation.

Let $a$ be the center of mass acceleration downhill and $F$ the uphill static friction. Translation and rotation about the center of mass obeyThe rolling without slipping constraint is $a=r\alpha$. For the coin's moment of inertia $I=m{r}^2/2$, this gives $F=ma/2$, soThe normal force is $N=mg\cos \theta$. Static friction can enforce rolling without slipping only if $F\le \mu N$, givingThe static friction acts uphill to provide the torque; it is not automatically equal to $\mu N$, attaining that limiting value only at the threshold. For finite $\mu \ge 0$ the result lies between $0$ and $\pi /2$.

Define $f(r)>0$ as the magnitude of the inward central force, so $\mathbf{F}=-f(r)\hat{\mathbf{r}}$ and $V^{\prime }(r)=f(r)$. The conserved angular momentum is $J=m{r}^2\dot{\theta }$. Eliminating $\dot{\theta }$ from the mechanical energy yieldsThe added term in the effective potential is the centrifugal barrier. If instead $f$ denotes the signed radial force component, the convention becomes $V^{\prime }=-f$.

For the power law central potential $V=\beta r^k$, differentiate the effective potential:The circular orbit condition is $m\beta k{r}_0^{k+2}=J^2$. Thus, when $J\ne 0$, $\beta k>0$ and $k\ne -2$,The sign condition $\beta k>0$ expresses attraction. At $k=0$ the force vanishes and there is no nontrivial circular orbit. At $k=-2$, attraction requires $\beta <0$ and the condition reduces to $J^2=-2m\beta$. In that exceptional case the effective potential is identically zero and every $r_0>0$ is a possible circular radius for the tuned $J$; there is no isolated preferred radius. A small radial velocity then gives $\ddot{r}=0$, rather than an oscillation.

Put $r=r_0+\eta$, where $|\eta |\ll r_0$. The Taylor expansion at a circular orbit isThe constant $V_{\mathrm{eff}}(r_0)$ has no effect on the force. The paper's displayed harmonic potential should be read as this expansion after subtracting that constant and neglecting higher orders; it is not an equality for the absolute effective potential in general. For example, when $m=J=\beta =1$ and $k=2$, $r_0=2^{-1/4}$ and $V_{\mathrm{eff}}(r_0)=\sqrt{2}$, whereas the printed quadratic expression is zero at $r_0$.

Differentiating again and then using $\beta k{r}_0^{k-2}=J^2/(m{r}_0^4)$ givesThe linearized radial equation of motion is $m\ddot{\eta }=-V_{\mathrm{eff}}^{\prime \prime }(r_0)\eta$, so the radial frequency in a power-law central potential isFor $k<-2$ the circular orbit is unstable and radial disturbances grow exponentially, with growth rate $\sqrt{-(k+2)}|J|/(m{r}_0^2)$. At $k=-2$ the linear restoring force vanishes, as described in part (c). Thus an attractive central force alone does not guarantee small oscillations.

Use $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$ for the magnetic Lorentz force on each wire segment. Let $+x$ point right, $+y$ upward and $+z$ out of the page, so $\mathbf{B}=-B\hat{\mathbf{z}}$. With clockwise electric current, the four magnetic forces areThe upper side carries electric current to the right and experiences an upward force, opposing the fall. The bottom side lies outside the magnetic field and has no magnetic force. On the right side the electric current is downward and the force is outward to the right; on the left it is upward and the force is outward to the left. Each side's in-field vertical length is $h$, so these horizontal forces cancel in the net force on the loop.

The wire's total mass is $M=4a{\rho }_m{A}_w$. Uniform wire gives each side a downward weight $Mg/4$. Internal mechanical forces at the corners transmit the unequal loads and maintain the assumed square shape; they cancel for the loop as a whole. The total external vertical force is therefore $Mg-IaB$ downward.

At terminal velocity, the upward magnetic force balances the weight. Substitute the electric current from the preceding solution:The geometric factors cancel, givingHere $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $B=1\mathrm{T}$, ${\rho }_m=8960\mathrm{kg}{\mathrm{m}}^{-3}$ and ${\rho }_c=1.7\times 10^{-8}\mathrm{\Omega }\mathrm{m}$. As numerical checks, the specified $a=0.08\mathrm{m}$ and $d_w=0.001\mathrm{m}$ give $M=2.252\times 10^{-3}\mathrm{kg}$, $R=6.926\times 10^{-3}\mathrm{\Omega }$ and $I_s=0.276\mathrm{A}$.

The gravitational power lost, $Mg{v}_s$, equals the Joule heating $I_s^{2}R$, consistent with energy conservation. This terminal velocity applies only while the top arm remains in the magnetic field; after it exits, the magnetic braking ceases, apart from any brief inductive transient.

Choose the entry point as $(0,0)$, the initial velocity as $v\hat{\mathbf{x}}$, and the slit as $(d,a)$, with $a,d>0$. A magnetic field perpendicular to the page changes the direction of the velocity, while the Lorentz force does no work. For positive $q$, an upward initial force requires the magnetic flux density to point into the page. The resulting circular motion has centre $(0,r)$, since its initial tangent is horizontal.

The slit must lie on the circle, giving $d^2+(a-r{)}^2=r^2$ and hence $r=(d^2+a^2)/(2a)$. Equating the centripetal force to the magnetic part of the Lorentz force, $m{v}^2/r=|q|vB$, givesA negative electric charge requires the opposite direction of magnetic flux density. These expressions assume nonrelativistic motion and negligible energy loss through electromagnetic radiation.

For an opaque screen, reaching the slit must be the first encounter with $x=d$. The velocity there has $v_x/v=(d^2-a^2)/(d^2+a^2)$, so passage from left to right requires $a<d$, as in the drawn arrangement. At $a=d$ the trajectory merely touches the screen. If $a>d$, the same circle first meets the screen at height $d^2/a<a$, where the particle is blocked; its later passage through $(d,a)$ does not solve the opaque-screen problem. This qualification distinguishes a geometric circle through two points from an accessible slit trajectory.

By symmetry, the resultant Coulomb force on each charge points along the line from the centroid to its vertex. Their accelerations and subsequent speeds are equal, so the centroid remains fixed and the equilateral triangle expands without rotating. Each charge follows a straight radial trajectory; the separation grows and the repulsive force decreases.

For example, if the instantaneous side is $s$ and the centroid-to-charge distance is $r=s/\sqrt{3}$, the two pair forces have a $60^{\circ }$ angle between them. Their resultant is $\sqrt{3}{k}_e{q}^2/s^2=k_e{q}^2/(\sqrt{3}{r}^2)$, along the outward radius. This confirms that the symmetric shape persists.

Assuming release from rest, nonrelativistic motion and negligible radiation losses, energy conservation givesThe three velocity vectors sum to zero, consistent with momentum conservation. Their magnitudes approach the displayed value as the remaining electrostatic potential energy tends to zero.