Take the natural release interpretation: both block and bob are initially at rest, the block starts at $x_0$, and the bob has small horizontal displacement $\delta =l{\theta }_0$ relative to the downward line through the pivot. Then $P=0$, $\dot{\theta }(0)=0$ andThe bob's horizontal position isThe block and bob therefore oscillate in opposite phase about the fixed horizontal centre of mass $x_0+m\delta /(M+m)$. Their horizontal oscillation amplitudes are $m\delta /(M+m)$ and $M\delta /(M+m)$ respectively. They pass the centre-of-mass line together when the rod is vertical. If the bob were also given an unspecified initial velocity, the additional sine and translational terms in part (d) would be needed; the source's displacement-release case selects the cosine solution.

When the block is free, $x$ becomes a dynamical generalized coordinate and its kinetic energy $M{\dot{x}}^2/2$ must be included. Expanding to quadratic order about the downward stationary state and dropping a constant givesThe coordinate $x$ is cyclic. Its conserved canonical momentum is the total horizontal momentum,The two linear Euler-Lagrange equations areEliminating $\ddot{x}$ givesThus the pendulum with a freely sliding support has a zero-frequency translational mode and an oscillatory mode. With four independent constants $X_c,V,C,D$, the general solution isHere $V=P/(M+m)$ and $X_c+Vt$ is the horizontal centre of mass coordinate to first order, since $X_{\mathrm{CM}}=x+[ml/(M+m)]\theta$. The oscillation is faster than that of a pendulum with a fixed pivot; the limit $M\to \infty$ restores $\Omega \to \sqrt{g/l}$.

As $\omega$ approaches ${\omega }_0=\sqrt{g/l}$, the driving motion couples resonantly to the pendulum. The particular-solution amplitude has denominator ${\omega }_0^2-{\omega }^2$, so a steady harmonic-response formula alone is singular. For a specified finite initial state, the near-resonant motion instead develops beats, and its exact resonant limit has a growing envelope.

For example, with the vertical zero-angular-velocity initial condition above, taking the limit givesThe ideal undamped linear theory predicts amplitude growing linearly with time, not an infinite displacement at the first instant. In reality the motion leaves the small-angle regime, and damping and nonlinear changes of the pendulum's oscillation period alter the response. Below and above resonance, the sign of the harmonic response coefficient changes, corresponding to a phase reversal.

The Euler-Lagrange equation for $\theta$ isFor small angles, set $\cos \theta \simeq 1$ and $\sin \theta \simeq \theta$. With $x=A\cos \omega t$ and ${\omega }_0^2=g/l$, the harmonic forcing of a horizontally driven pendulum obeysFor $\omega \ne {\omega }_0$, its general solution isThe constants $C,D$ set the initial angle and angular velocity. If the bob starts vertical with zero angular velocity at $t=0$, this reduces toThe solution remains an approximation only while the resulting angle is small.

Let the pivot's prescribed horizontal position be $x(t)$ and measure $\theta$ from the downward vertical. The bob's coordinates are $X=x+l\sin \theta$ and $Y=-l\cos \theta$. ThereforeWith zero gravitational potential energy at pivot height, the bob's Lagrangian isHere $x(t)$ is a prescribed input, not yet a free coordinate. Terms depending only on that prescribed time function can be dropped without changing the $\theta$ equation. The moving block's own kinetic energy becomes relevant when its motion is free in part (d).

The air fall lastsRetain the underwater deceleration rather than replacing the whole water trajectory with motion at terminal velocity. DefineSeparation of the descent equation gives the hyperbolic cotangent solution and its integrated depth:where $t=0$ now means water entry. The derivative of the logarithm verifies $\dot{z}=v$, while $v(0)=v_t\coth \chi =u$. At water depth $H_w=5\mathrm{m}$, inversion using the inverse hyperbolic sine yieldsThe total elapsed time is thereforeFor an estimate without evaluating the exact expression, $b{H}_w\gg 1$ permits $\mathrm{arsinh}x\simeq \ln (2x)$ here, givingThis gives the same $4.43\mathrm{s}$ to the quoted precision. The simpler $H_w/v_t\simeq 4.72\mathrm{s}$ overestimates the underwater time because the pebble initially moves faster than $v_t$.

For superterminal descent under quadratic drag, use $\dot{v}=vdv/dz$ in $\dot{v}=b(v_t^2-v^2)$. Equivalently,The speed approaches $v_t$ asymptotically from above. There is no finite exact distance at which it first equals the limiting speed. A practical estimate must specify an accuracy. Define “reached” as $v\le (1+ϵ)v_t$. ThenFor one-percent accuracy, $ϵ=0.01$ givesA five-percent criterion gives $z_{5\%}\simeq 0.642\mathrm{m}$. The characteristic decay length of the excess squared speed is $1/(2b)\simeq 0.0952\mathrm{m}$, but several such lengths are needed because the entry speed is so far above the limiting value. These estimates depend on the stated water-entry idealization; the source gives no finite tolerance or entry-loss model.

Fixed-angular-momentum form. Eliminating $\dot{\varphi }$ using the conserved angular momentum $h$ givesEquivalently, the polar motion has effective potentialFor the steady conical pendulum at ${\theta }_0$, $\ddot{\theta }=0$, soThe sign of $h$ agrees with the sign of $\Omega$; substituting $h=m{l}^2{\sin }^2{\theta }_0\Omega$ verifies the constant-angle result in part (d).

Period of small polar oscillations. Put $\theta ={\theta }_0+\alpha (t)$, with $|\alpha |\ll 1$ and $h$ held fixed. This is the appropriate small nutation about a conical pendulum; keeping $\dot{\varphi }$ fixed instead would violate the angular momentum constraint during the oscillation. Ifthen $f({\theta }_0)=0$ andThe linearization is $\ddot{\alpha }+{\omega }_{\theta }^2\alpha =0$, givingSince $0<{\theta }_0<\pi /2$, the coefficient is positive and these small polar oscillations are stable. To first order, $\dot{\varphi }=\Omega (1-2\alpha \cot {\theta }_0)$, making the compensating change in azimuthal angular velocity explicit. The limit ${\omega }_{\theta }/|\Omega |\to 2$ as ${\theta }_0\to 0$ concerns radial modulation of a small nonzero circular orbit; the exactly planar $h=0$ pendulum is a different, degenerate limit of these polar coordinates.

Constant azimuth. If $\varphi$ is constant, $\dot{\varphi }=0$ and the motion is a planar simple pendulum:For the small-angle approximation, $\sin \theta \simeq \theta$, giving simple harmonic motion withThe harmonic angular frequency requires small amplitude. At finite libration amplitude ${\theta }_{\max }<\pi$, the exact oscillation period is $4\sqrt{l/g}K(\sin ({\theta }_{\max }/2))$, where $K$ is the complete elliptic integral of the first kind; the nonlinear pendulum does not have the same amplitude-independent period.

Constant polar angle. If $\theta ={\theta }_0$ is nonzero and constant, the bob traces a horizontal circle: this is a conical pendulum. Setting $\ddot{\theta }=0$ and dividing by $\sin {\theta }_0$ givesThe sign distinguishes the two senses of rotation. The exactly downward configuration ${\theta }_0=0$ is stationary; its azimuth is undefined and it must not be obtained by dividing by $\sin {\theta }_0$. A nontrivial constant-angle orbit of this gravity-only system needs $\cos {\theta }_0>0$ even though the supporting rod can also push.

The $\theta$ Euler-Lagrange equation readsHenceThe first term reflects the coupling to azimuthal motion; the second is the gravitational restoring contribution for an angle measured from the downward vertical.

The azimuth $\varphi$ is a cyclic coordinate, since $\partial \mathcal{L}/\partial \varphi =0$. Its Euler-Lagrange equation gives conservation of the vertical component of angular momentum:This $h$ is the actual angular momentum, not specific angular momentum. The Lagrangian also has no explicit time dependence. The conserved energy isAlong solutions, $\dot{E}=-\partial \mathcal{L}/\partial t=0$. These are the conserved quantities associated with rotational symmetry about the vertical and time-translation symmetry, respectively.

Use spherical coordinates about the fixed pivot, with fixed rod length $l$, polar angle $\theta$ measured from the downward vertical, and azimuth $\varphi$. The bob's position iswhen the Cartesian $z$ axis points upward. Differentiation gives kinetic energy $T=m{l}^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2)/2$. Taking zero gravitational potential energy at pivot height gives $V=-mgl\cos \theta$. The spherical pendulum's Lagrangian is thereforeThe rod is light, rigid and of fixed length; pivot friction and air drag are neglected. A rigid rod, unlike a string, may sustain either tensile or compressive constraint force.

The supplied PDF labels this equation clarified from original for tutorial purposes. Use the model of an eccentric wheel with missing-mass kinetic energy neglected: neglect the missing disc's contribution to kinetic energy, while retaining its subtraction from gravitational potential energy. Do not silently replace the gravitational mass $W$ by the full-disc mass $D$ as well.

Choose $\theta$ as the clockwise angle of $\overrightarrow{CH}$ from the upward vertical, and set $s=R\theta$. The rolling without slipping constraint makes changes in $s$ equal changes in the distance C travels down the slope. Choose its additive origin so $s=0$ when the hole is vertically above C. Then, up to an irrelevant height constant,The approximate kinetic energy is that of the complete uniform disc rolling at $\dot{s}=R\dot{\theta }$:Subtracting the missing disc's gravitational potential energy givesApply the Euler-Lagrange equation to $\mathcal{L}=T-U$:ThereforeHere $K_1,K_2$ have dimensions of acceleration and $\gamma$ has dimensions of inverse length. The initial conditions are $s(0)=\dot{s}(0)=0$ under the vertical-above interpretation. If an initial orientation is instead specified relative to the slope normal, the same coordinate convention produces a different initial $s$, not different constants. With $r_o\to 0$ the equation becomes $\ddot{s}=(2/3)g\sin \alpha$, the familiar solid-disc result. Retaining the hole's exact kinetic energy would introduce coordinate-dependent inertia and extra ${\dot{s}}^2$ terms; the printed neglect is what permits the requested constant-coefficient form.

For a freely rolling rigid body, decompose the kinetic energy into translation of its centre of mass G and rotation about G. The centre of mass after removing material lies on the side opposite H:Another use of the parallel axis theorem givesThus at the same angular speed,This compares the rotational component, not the total rolling kinetic energy. The total also contains $\frac{1}{2}W{v}_G^2$ and need not be smaller than $T_H$. Describing all rolling motion as instantaneous rotation about its contact point includes that translational contribution; it is not the rotational component intended by this comparison.

For the eccentric circular-hole disk inertia, let $D=\pi \rho R^2$ be the mass of the complete disc before removing material, $h=\pi \rho r_o^2$ the missing mass, and $W=D-h$ the wheel's remaining mass. Denote the complete disc's centre by C and the axle/hole centre by H; $CH=r_o$. Assume the hole fits inside the disc, $0<r_o<R/2$. The parallel axis theorem gives the full disc's moment of inertia about H as $\frac{1}{2}D{R}^2+D{r}_o^2$. Subtract the absent little disc about its own centre H:The axle is a support, not part of the rotating wheel's stated mass. Its frictionless bearing does not alter this moment of inertia. For fixed-axis angular velocity $\omega$,The missing material must be subtracted with its actual displacement relative to the axis; treating the wheel as a concentric annulus would give the wrong result.

Put $m_A=M=m$ and $m_B=2m$. At reversal the transferred mass is $q_r=0.45m$, leaving A with $1.55m$ and, after capture of that exhaust, B with $2.45m$. The first burn givesAfter reversal A emits its remaining exhaust away from B, so B receives no further fuel and eventually coasts at this velocity. Applying the rocket equation with reversed thrust to A givesBoth now move toward negative x, but A moves faster in that direction. The asymptotic closing speed isThis conclusion does not require a burn rate or an initial separation: after a finite burn, the positive eventual closing speed closes any finite remaining gap. Those additional inputs would be required to calculate the meeting time. Any pre-reversal exhaust still in transit affects the timing, not B's eventual momentum after collecting that fixed amount.

Neglecting exhaust flight time, no momentum is stored in transit andSince the velocities have opposite signs, equal nonzero speeds imply equality of the instantaneous masses. For $m_B=2m_A$ and $M=5m_A$, this occurs at $6m_A-q=2m_A+q$, or $q=2m_A$. ThereforeThe common-speed event occurs after $40\%$ of A's original fuel has been transferred, rather than when either rocket has its final mass. At this event the rearward exhaust still overtakes B: $v_A-u_0=-0.5945u_0<v_B=-0.4055u_0$. Thus the interception restriction in b does not invalidate this event even though extending complete ballistic capture over the much later burn would need separate checking.

Let $q$ be the transferred fuel mass. Before any engine reversal, rocket A has mass $a=m_A+M-q$ and velocityFor exhaust collection by a second rocket, under the stipulated eventual capture of all exhaust, final conservation of momentum for the complete initially stationary system gives $m_A{v}_{Af}+(m_B+M)v_{Bf}=0$. ThusCapture is inelastic: no kinetic energy conservation for B is assumed. Finite exhaust flight time changes the intermediate sharing of momentum but not this endpoint balance if the stated complete capture actually occurs.

The ballistic exhaust interception condition is a physical restriction on that idealization. For purely ballistic exhaust to reach a receding collector to the rear, its laboratory velocity must be more negative than B's while interception occurs. In the negligible-flight-time approximation this requires $v_A-u_0<v_B$. Very large fuel ratios need not meet that condition throughout a complete burn; the boxed endpoint formula is conditional on the prompt's complete-collection stipulation, not an unconditional prediction for every ballistic two-rocket geometry.

Choose positive x along the rocket's motion. If its instantaneous total mass is $\mu$ and it ejects a positive amount $d\eta =-d\mu$, the exhaust velocity in the original inertial frame is $v-u_0$ to first order. Conservation of momentum over that increment givesIntegrating the rocket equation from $\mu =m+M$ to $\mu =m$ givesThe constant exhaust speed is relative to the rocket, not the initial frame. Kinetic energy of rocket plus exhaust is not conserved because stored fuel energy is released. The calculation is Newtonian and requires resulting speeds small compared with the speed of light.

Take the light to travel along positive x, while the water moves with signed velocity $-v$. In the water's rest frame its light phase velocity is $c/n$. The relativistic velocity addition law from a gives the laboratory phase velocityExpanding to first order in $v/c$, with $nv/c\ll 1$, yieldsEquivalently $w\simeq c/n-v(1-1/n^2)$: the counterflow reduces the forward light speed. This is the nondispersive Fresnel drag coefficient. The refractive index is taken as constant over the relevant frequencies; dispersion and its frequency-shift corrections are outside this approximation.