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The free-body diagram shows a meridional section of the right circular cone. On the first particle, gravity acts vertically downwards, tension points along the straight line to the apex, and the normal reaction is perpendicular to the surface. The smooth mechanical constraint supplies no tangential friction. The second particle has only its downward gravitational weight and the upward tension.

Adding the forces on each particle and applying Newton's second law givesAt $z_1=-r_1$, a normal vector is ${\mathbf{e}}_r+{\mathbf{e}}_z$. Its sign depends on the supporting face of the cone; either sign has zero dot product with every permitted velocity. The diagram chooses the reaction towards the upper side of the surface.

The massless inextensible string has a common tension magnitude $T$. Since ${\mathbf{r}}_1=r_1({\mathbf{e}}_r-{\mathbf{e}}_z)$, the unit vector from the first particle towards the apex is $(-{\mathbf{e}}_r+{\mathbf{e}}_z)/\sqrt{2}$. The second particle lies below the apex, soThe string length constraint isThe velocity of the first particle in cylindrical coordinates is ${\dot{r}}_1{\mathbf{e}}_r+r_1\dot{\theta }{\mathbf{e}}_{\theta }+{\dot{z}}_1{\mathbf{e}}_z$. Therefore the powers delivered by the two tension forces areThusThe tension can transfer energy between the particles, but its total power on the pair is zero.

All forces on the first particle lie in the plane spanned by ${\mathbf{e}}_r$ and ${\mathbf{e}}_z$: this holds for gravity, tension, and the normal reaction. Hence the azimuthal component of Newton's second law vanishes:Using the product rule givesStrictly, $h$ is the axial specific angular momentum; the actual axial angular momentum is $m_{1}h$, which is conserved as well.

Differentiate the stated total energy using the chain rule and then substitute the two Newton's second law equations:The gravitational potential energy terms have canceled the power of gravity. Also $\mathbf{N}$ is proportional to ${\mathbf{e}}_r+{\mathbf{e}}_z$, whose dot product with ${\dot{\mathbf{r}}}_1$ is ${\dot{r}}_1+{\dot{z}}_1=0$. Thus the stationary smooth mechanical constraint does no work. The remaining tension powers cancel by part (b), proving $\dot{E}=0$ and conservation of total energy while the massless inextensible string remains taut and contact is maintained.

Use the ideal transverse-motion model of the question: constant tension, small slopes, no bending stiffness, and no transverse body-force term. On a short segment away from $x=0$, the vertical components of the endpoint tensions are, to first order in $y_x$, $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$, since $\rho$ is linear mass density. Newton's second law givesDividing by $\Delta x$ and taking its limit proves the wave equation on a string on each side of the mass:The string on the right pulls the point mass with vertical component $T{y}_x(0^{+},t)$, while the string on the left contributes $-T{y}_x(0^{-},t)$. HenceThe fixed supports and attachment impose the boundary conditionsTogether with the mass equation, these are the endpoint and matching conditions. The slope can jump at $x=0$: the jump supplies the mass's vertical acceleration. Requiring the slope to be continuous would incorrectly remove its dynamics.

For a short string element of length $\Delta x$, the small-slope approximation makes the upward components of tension at its two ends $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$. Newton's second law givesDivide by $\Delta x$ and take the limit to obtain the wave equation on a string,Here $\rho$ is linear mass density, and the derivation assumes small transverse slopes, uniform tension, and a flexible string.

At the right endpoint the string pulls the ring toward the interior. Its vertical force is therefore $-T{y}_x(L,t)$ to first order; the frictionless wire supplies no vertical force. The ring obeysThis is the dynamic endpoint law for a string with an endpoint mass.

For a separated oscillatory solution, write $y=X(x)q(t)$ with $q^{\prime \prime }=-{\omega }^{2}q$ and $X^{\prime \prime }+(\omega /c{)}^{2}X=0$. The fixed left end removes the spatial cosine, so the nonzero normal modes haveThe endpoint law gives $M{\omega }_n^2\sin ({\omega }_{n}L/c)=T({\omega }_n/c)\cos ({\omega }_{n}L/c)$. Since $\sin ({\omega }_{n}L/c)=0$ would also force the cosine to vanish, it can be divided out for a nonzero mode. Using $T=\rho c^2$ givesFor physical $M\ge 0$, these give all positive modal frequencies. There is no nonzero static linear mode: the left value and right slope conditions would force it to vanish. Positivity also follows by integration by parts from

For $M=0$, the right endpoint is free, with the Neumann boundary condition $y_x(L,t)=0$. At $L=\pi$, the allowed wave numbers are $k_n=n+1/2$, $n\ge 0$, and ${\omega }_n=c{k}_n$. Zero initial displacement removes every cosine-in-time coefficient. The initial velocity contains only the modes $k=1/2$ and $k=5/2$, each with coefficient one. Dividing those velocity coefficients by their angular frequencies givesAt $x=0$ each term vanishes, at $x=\pi$ each spatial derivative vanishes, and at $t=0$ the displacement and its time derivative reproduce the two initial conditions.

For the uniform unloaded space-elevator cable, let the constant linear mass density be $\lambda$ and the outer endpoint be $b={\eta }^{\prime }R$. In the co-rotating frame the outward effective force per unit length is $\lambda ({\omega }^{2}r-G{M}_E/r^2)$. The cable is just clear of the surface, with neither a ground load nor an endpoint counterweight, so both ends have zero applied tension. The integrated effective force must vanish:Using $G{M}_E={\omega }^2{\eta }^3{R}^3$, this becomesThe root ${\eta }^{\prime }=1$ is a zero-length cable. The physical nonzero cable hasIt extends well beyond the geostationary orbit so its outer centrifugal force balances the inner segment's weight. If a cable is anchored with a nonzero endpoint load or has a counterweight, that load changes the integrated balance; its length is not fixed by the zero-end-load formula.

An element of the rotating cable satisfiesWith $T(R)=0$, integration givesIt increases below $r_g$, decreases above $r_g$, and returns to zero at $b$. HenceThe tension is greatest at the geostationary orbit, where local effective gravity is zero. A zero local effective force does not mean zero tension: that section transmits the loads of the entire inner and outer cable portions.

In the rim-mass pulley coupled to a hanging mass, the printed figure shows the smaller pulley fixed overhead, and the mechanical string leaves the right-hand tangent of the larger pulley. Let $\theta$ increase counterclockwise from the downward vertical to the radius carrying $M$. Then the hanging mass $m$ moves downward by $R\theta$. The string is taken to grip the rim without slipping. Its tension must be solved dynamically, rather than set equal to $mg$ during motion.

The kinetic energy and potential energy areThe Euler-Lagrange equation and the condition for equilibrium areOne can also eliminate the tension from $M{R}^2\ddot{\theta }=TR-MgR\sin \theta$ and $mR\ddot{\theta }=mg-T$. For $0<m<M$, the solution with $\cos {\theta }_e>0$ is stable; the other solution in a full turn is unstable.

The radial and angular Euler-Lagrange equations areWriting the conserved angular momentum aswe obtainA positive constant radius requires zero radial acceleration. Hence the circular motion hasprovided $J\ne 0$ and $0<r_0<l$. Its angular velocity is ${\dot{\theta }}_0=J/(m{r}_0^2)$. No positive circular radius exists for $J=0$ with $M,g>0$.

For later use, the conserved radial energy isThe effective potential has $U_J^{\prime \prime }(r)=3J^2/(m{r}^4)>0$, so its stationary radius is a stable minimum. The tension remains positive in the ideal allowed region: the hanging Newton's second law gives $T_s=M(g+\ddot{r})=Mm(g+r{\dot{\theta }}^2)/(M+m)>0$.

Keep the signs of part i. The tension has horizontal component $T\cos \theta$ but still has counterclockwise torque $Tr$, since it is tangent to the axle. ThusSolving these equations gives the critical pulling angle of a rolling spool lawThe zero-acceleration angle is thereforeAt this angle the line of action of the pull passes through the ground contact, so its torque about that point vanishes. A spool initially at rest stays at rest while the contact constraints hold; one already rolling has zero acceleration, rather than being brought to rest.

For equilibrium the vertical normal force is $N=Mg-T\sin {\theta }_c$ and $|f|=T\cos {\theta }_c$. The static friction inequality $|f|\le \mu N$ becomesHence, for $0<r<R$,At this limit the normal force is still nonnegative; for finite $\mu$ and positive $r$ it is positive. For ${\theta }_c<\theta \le \pi /2$, the numerator of $a$ is negative. Released from rest, the spool rolls leftward, away from the pulling hand, with counterclockwise rotation, provided the assumed rolling without slipping and ground contact remain possible. The nonzero acceleration is not obtained by assuming that friction simply opposes the centre's motion.

Take rightward translation as positive and counterclockwise angular acceleration $\alpha$ as positive. Let $f$ denote signed ground friction, positive to the right. The source's lower-tangent winding makes the tension exert counterclockwise torque $Tr$. For a horizontal pull, translation and rotation obeyThe final relation is the rolling without slipping constraint. Eliminating $f$ and $\alpha$ givesThus the requested magnitude of static friction isIt acts leftward. Released from rest, the spool accelerates rightward, toward the pulling hand, and rotates clockwise. There is no contradiction between leftward friction and rightward motion: static friction prevents slipping of the contact point, and the horizontal tension is larger than $|f|$ because $r<R$.

Write $k=1/(4\pi {ϵ}_0)$. The separation of the two point charges is $2l\sin \varphi$, so Coulomb's law gives the horizontal repulsive force $F=k{Q}^2/(4l^2{\sin }^2\varphi )$. Resolving the rod's tension vertically and horizontally gives $T\cos \varphi =mg$ and $T\sin \varphi =F$. HenceThe left side increases from zero to infinity for $0<\varphi <\pi /2$, so this determines a unique outward equilibrium. For a small angle, ${\varphi }^3\simeq Q^2/(16\pi {ϵ}_{0}mg{l}^2)$. With $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, direct numerical solution gives

A large electrical conductor between the point charges provides electrostatic shielding: one must not retain their original mutual Coulomb force through an ideal infinite plate. In the usual broad-plate approximation, let $\sigma$ denote the unperturbed surface charge density on each face. The electrostatic boundary condition gives an outward electric field $E_0=\sigma /{ϵ}_0$. Neglecting local electrostatic induction, the rod's tension then satisfies $mg\tan \varphi =Q{E}_0$, givingThere are two physical qualifications. First, if $\sigma$ means the total charge per projected plate area, shared equally between the two faces, then $E_0=\sigma /(2{ϵ}_0)$ and the same approximation gives $\varphi \simeq 0.989^{\circ }$. The factor of two is a convention about the specified surface charge density, not a different Gauss's law.

Second, a point charge near an electrical conductor makes its surface charge density nonuniform. For an infinite plane with prescribed background electric field $E_0$, the method of images adds an attractive force $k{Q}^2/(4x^2)$, where $x=l\sin \varphi$. The more accurate equilibrium equation isIts larger, stable root is approximately $1.96^{\circ }$ for the per-face convention, or $0.877^{\circ }$ for the total-area convention. This ideal-plane refinement is not an exact answer for an unspecified finite plate: its dimensions and the meaning of the quoted surface charge density would be needed. In particular, adding a uniform-sheet electric field to the original mutual repulsion would describe a different, nonshielding model.

Let the common tension be T, the linear mass densities be ${\mu }_1<{\mu }_2$ and the speeds be $c_j=\sqrt{T/{\mu }_j}$, so $c_2<c_1$. Displacement and transverse-force continuity give the displacement coefficients at a string junctionThe reflected triangular pulse is inverted, has smaller height $|r|H$, unchanged spatial width and unchanged speed, and travels left. Its left-to-right asymmetric profile is reversed as well as vertically inverted. The transmitted pulse remains upright, has smaller height tH, travels more slowly, and has width $W{c}_2/c_1$. The string-pulse width mapping at a junction leaves its temporal duration unchanged. The transmitted slopes are steeper by $2c_1/(c_1+c_2)$; the reflected slope magnitudes are smaller by $|r|$.

The density ratio is unspecified, so only these inequalities are determined. The original plot below illustrates ${\mu }_2/{\mu }_1=4$, giving $r=-1/3$, $t=2/3$ and half-width transmission; it is not a numerical reading of the examination figure.

The string speed is $c=\sqrt{T/\rho }$. The characteristic impedance is the ratio of the forward-wave transverse force to transverse velocity, $Z=T/c=\sqrt{T\rho }$. With peak displacement amplitude $A$, the mean power isDisplacement and transverse force are continuous at a massless junction. Since $Z_2=2Z_1$, the displacement amplitudes are $r=(Z_1-Z_2)/(Z_1+Z_2)=-1/3$ and $t=2Z_1/(Z_1+Z_2)=2/3$. Taking the incident amplitude real, the light-string motion isIn this standing-and-travelling decomposition of a reflected string wave, the first term travels and the second stands; the decomposition is not a separation into independent incident-power fractions because these fields interfere. The reflected wave has a phase reversal. power coefficients and standing-wave ratio are

For a uniform inserted matching section, unequal exterior impedances require a quarter-wave impedance transformer: $Z_M=\sqrt{Z_1{Z}_2}$ and $k_{M}a=(2j+1)\pi /2$. At equal tension this means middle density $2{\rho }_0$ and an odd-quarter-wavelength length in that section. Reflections at its two ends then cancel at the design frequency. An arbitrary density or an arbitrary half-wavelength segment cannot perfectly match these unequal exterior strings; the ideal match is frequency-selective.

amplitude scaling along a slowly tapered string instead follows from matching over an adiabatic frequency range. The source fixes frequency, while conserved power and the final characteristic impedance determine amplitude. In the heavy string,Its wave speed halves. The no-reflection approximation needs small fractional characteristic impedance change over each local wavelength, rather than length alone for a taper containing abrupt features.

time average of harmonic power. Expand the two real parts:The terms proportional to $e^{\pm 2i\omega t}$ average to zero over complete periods, while $F{u}^{\ast }$ is time-independent. For real peak amplitudes this is $F_0{u}_0\cos \varphi /2$.

String dynamics. A short element of length $dx$ experiences transverse force $T[{\psi }_x(x+dx)-{\psi }_x(x)]$. Newton's second law therefore gives $\rho dx{\psi }_{tt}=Tdx{\psi }_{xx}$, orThe approximation is a flexible, lossless string with small slope and constant tension; $\rho$ is its linear mass density.

energy transport. Across a cut toward increasing $x$, the transverse force is $F_y=-T{\psi }_x$ and the transverse velocity is $u={\psi }_t$. For a rightgoing wave these have complex amplitudes $ikTA$ and $i\omega A$. HenceThis real mechanical impedance is force divided by transverse velocity for a travelling wave, with units $\mathrm{kg}{\mathrm{s}}^{-1}$.

Matching and power ratios. At an ordinary massless junction without an external transverse force, displacement is continuous and transverse tension is balanced: ${\psi }_1={\psi }_2$, $T_1{\psi }_{1x}=T_2{\psi }_{2x}$. For the same tension, the slope is therefore continuous even if density changes. With real positive impedances,Their sum equals $P_i$; the transmitted power includes the impedance ratio and is not simply $t_{\psi }^2{P}_i$.

Let $T$ be the common tension and ${\mu }_j$ the linear mass densities. The characteristic impedance of a stretched string is $Z_j=T/c_j=\sqrt{T{\mu }_j}$. Continuity of displacement and transverse force at a massless junction gives $1+r=t$ and $Z_1(1-r)=Z_{2}t$. ConsequentlyThe reflected power is $7.18\%$. The minus sign of $r$ represents a phase reversal, not negative power. Both incident and reflected waves travel in the first string, so their power ratio is simply the squared displacement-amplitude ratio. As a check, $\mathcal{T}=(Z_2/Z_1)|t{|}^2=0.92820$ and $\mathcal{R}+\mathcal{T}=1$.

The symbol $\rho$ here is linear mass density, with units $\mathrm{kg}{\mathrm{m}}^{-1}$. A short element $[x,x+\delta x]$ has transverse forcebecause small slopes give $\sin \theta \simeq \tan \theta ={\Psi }_x$. Its mass is $\rho \delta x$. Applying Newton's second law, dividing by $\delta x$, and taking the limit gives the wave equation on a string,The derivation assumes constant positive tension, small transverse displacements and slopes, a flexible string with negligible bending stiffness, and negligible damping. Variations of tension due to the motion enter beyond the linear approximation.

At an ordinary point or a massless join, the string is continuous, soWith no concentrated transverse force or point mass, transverse force balance requiresFor the common tension in this problem, the slope $\partial \Psi /\partial x$ is therefore continuous as well. A jump in linear mass density changes wave speed, but introduces no finite inertial force at a zero-length join. A point mass, external support force, or unequal tensions would require modified boundary conditions.

There is no reflected wave when the characteristic mechanical impedances match:for the same tension. Then $r=0$, $C/A=1$ and $\mathcal{T}=1$. If tensions were independently adjustable, the general condition would be $T_1{\rho }_1=T_2{\rho }_2$, together with transverse-force matching; equality of densities alone would no longer be the universal criterion.