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The particle moves on a horizontal circle of radiusIf $T$ is the tension in the light rod, vertical and radial balance giveDividing eliminates both $T$ and $m$, yielding the rotating conical pendulum with an offset pivot conditionThe ratio under the square root has units $T^{-2}$. The answer is independent of $m$; increasing $L$ increases the orbit radius and lowers the angular velocity required at fixed ${\varphi }_1$.

Let $x$ point downslope and $z$ point normally away from the plane, with $0\le z\le h$. Write the parallel velocity as $u(z)e_x$ and take $\tau >0$ to be the magnitude of the air's upslope stress. The steady equations arewith boundary conditionsThe free-surface condition makes $p_x=0$, and integration gives

The surface velocity, downslope shear exerted by the fluid on the plane, and volume flux per unit width areThus the inclined viscous film with opposing surface shear reverses in the three senses whenThe order of increasing required air stress is therefore

Writing the velocity as $(u,v)=(y,ax)$ givesso the flow is incompressible. With the conventiona stream function isThe streamlines are its level sets. For $a>0$, they are the hyperbolaswith separatrices $y=\pm \sqrt{a}x$ and a saddle at the origin. For $a<0$, they are concentric ellipsestraversed clockwise. This is the streamline classification of a planar linear saddle or centre.

The scalar vorticity isHence the flow is irrotational exactly whenThen $(u,v)=(y,x)=\nabla \varphi$, and a velocity potential is

Separation $y=T(t)X(x)$ and the fixed-end conditions givePutThe initial displacement and zero initial velocity then give the separated solution of the damped string equation

For the triangular initial displacement,Thus the even coefficients vanish and the odd ones alternate in sign.

LetThenOrthogonality of the sine and cosine modes, equivalently the Parseval identity, gives

The sign of its derivative is clearest directly from the equation. Integration by parts, using $y_t=0$ at the fixed endpoints, yieldsThis is the energy dissipation identity for a linearly damped string.

Take the sphere to move in the positive $z$-direction and use spherical coordinates centred on it. An axisymmetric harmonic potential that decays at infinity has the dipole form $A\cos \theta /r^2$. The no-penetration condition in the laboratory frame isIt fixes $A=-U{a}^3/2$, so the potential flow around a translating sphere isThe laboratory-frame velocity components are

The fluid kinetic energy iswhere $V=4\pi a^3/3$. Thus the added mass of a sphere is $\rho V/2$.

When the sphere falls, replacing fluid of density $\rho$ by material of density ${\rho }_s$ lowers the gravitational potential energy at rateThe total kinetic energy of sphere and fluid isConservation of total energy givesFor $U\ne 0$, the acceleration of a freely falling sphere with added mass isThe same formula holds at release by continuity.

For a general Lagrangian,Along an Euler-Lagrange trajectory,so energy is conserved when $L$ has no explicit time dependence.

For the given Lagrangian, write $v=|\dot{x}|$. At points where $v\ne 0$,The Euler-Lagrange equation is thereforeIn expanded form its left-hand side isThe momentum has fixed magnitude,so this is constant without needing to solve the equation of motion. The Hamiltonian is

The singular Legendre transform of a degree-one velocity Lagrangian applies here: $p$ determines only the direction of $\dot{x}$, not its magnitude, and the velocity Hessian is not invertible. Thus the usual inverse Legendre transform cannot reconstruct the Lagrangian from this Hamiltonian.

Take downward velocity as positive. Lubrication theory reduces the axial momentum equation to a constant pressure gradient, and the no-slip conditions are $u(0)=0$, $u(h)=U$. Twice integrating givesThe flux per unit circumference and the total gap flux areThe shear at the cylinder iswith the fluid traction on the falling cylinder opposing its motion.

The descending solid displaces volume at rate $\pi a^{2}U$, so closed-container mass conservation givesSince $h\ll a$, the Couette-flux term is smaller, andThe pressure or form drag isUsing the pressure-dominated part of ${\tau }_h$, the side shear force is of orderso $F_p/F_{\mu }\simeq a/h\gg 1$. Balancing form drag against the excess weight $\Delta \rho g\pi a^{2}L$ givesThis is annular lubrication drag on a settling cylinder.

For the outgoing wave, the pressure and velocity amplitudes are $-i\omega {\rho }_{+}T$ and $-i{k}_{+}T$. Its time-averaged acoustic energy flux is thereforeIf $\lambda =n\pi$, then $|T|=ϵc_{+}$ andIf $\lambda =(n+\frac{1}{2})\pi$, thenandWhen ${\rho }_{+}\ll {\rho }_{-}$ and $c_{+}\simeq c_{-}$, the half-integer family is strongly resonantly enhanced relative to the integer family by approximately $({\rho }_{-}/{\rho }_{+}{)}^2$ at equal $\omega$.

Use complex fields with time factor $e^{i\omega t}$ and put $k_{\pm }=\omega /c_{\pm }$. The outgoing potential in $x>0$ isContinuity of pressure and velocity at $x=0$ givesHenceThe piston condition ${\Phi }_{-}^{\prime }(-L)=i\omega ϵ$, with $\lambda =k_{-}L$, givesThis is the transmitted velocity potential from a piston through an acoustic interface.

The incompressible Stokes equations areTaking the curl gives ${\nabla }^2\omega =0$. For a planar stream function, $\omega =-{\nabla }^2\psi$, so the biharmonic stream function for planar Stokes flow satisfies

For $\psi =r^{2}f(\theta )$,The polar rate-of-strain components areHence

Sincebiharmonicity givesNo slip at $\theta =-\alpha$, no penetration at $\theta =0$, and the imposed tangential traction givePut $\Delta =\sin 2\alpha -2\alpha \cos 2\alpha$. Solving,The upper-surface velocity is radial and equals $r{f}^{\prime }(0)$. Therefore the similarity solution for tangentially forced Stokes wedge gives

Sincethe phase and group velocities are

Let $q=(q_1,q_2,q_3)$ and $n=(1,1,1)$. Sincethe Lagrangian depends on $q$ only through $|q{|}^2$ and $(q\cdot n{)}^2$, and on the velocity through $|\dot{q}{|}^2$. Rotations about the $n$-axis preserve all three quantities, proving the invariance.

The axially symmetric isotropic-mass Lagrangian therefore conserves the axial angular momentumup to an irrelevant normalization of $n$. Since $L$ has no explicit time dependence, conservation of energy from time-translation invariance gives the second conserved quantity

With streamwise and normal velocities $u,v$, zero exterior pressure gradient gives the steady boundary-layer equationsAt the organism,and matching to the quiescent exterior requires $u(x,y)\to 0$ as $y\to \infty$.

The interface is the material level set $F=r-R(\theta ,t)=0$. The condition $DF/Dt=0$ giveswhich is the stated equation.

Let the perturbation potentials inside and outside beLinearizing the kinematic condition at $r=a$ givesThe linearized interior Euler equations givewhereas the stationary exterior gives $p_o^{\prime }=-\rho \sigma {\varphi }_o$. Since the basic interior pressure has $d{p}_i/dr=\rho {\omega }^{2}r$, pressure continuity on the displaced boundary isEliminating $A$ and $B$ gives the circular vortex-sheet mode relationsoThe $k=1$ mode is a neutral displacement. Every $k>1$ mode has one exponentially growing branch, so the circular interface rolls up through a Kelvin--Helmholtz instability. Its pattern angular velocity is $\omega (k-1)/(2k)$, in the direction of the gyre but slower than the solid-body motion; the disturbances therefore propagate upstream relative to the rotating water.

At infinity the specific energy is $v^2/2$ and $l=bv$, soThe angle through which the velocity is deflected between the incoming and outgoing asymptotes is

Put $R=R_A+R_B$, $c=\sqrt{R^2-b^2}/R$, and $s=b/R$. At impact the unit vector from A to B is $n=(-c,s)$. The impulse is normal, so tangential velocity is unchanged, while the normal components undergo a one-dimensional elastic collision. Therefore

With $\omega =B/m$, magnetic force does no work, so $|\dot{x}|$ is constant; its zero $z$ component also makes $\dot{z}$ constant. For the stated data,This is a circle of radius $u/\omega$ centred at $(0,-u/\omega )$.

With $\alpha =\mu /m$, the complex velocity is $\dot{x}+i\dot{y}=u{e}^{-(\alpha +i\omega )t}$. HenceThe path spirals into $u(\alpha ,-\omega )/({\alpha }^2+{\omega }^2)$. As $u$ varies, this lies on $y=-(\omega /\alpha )x$, at distancefrom the origin.

Transverse force balance on a short element givesWith fixed ends,Here $A_n=0$. If $b_n$ are the sine coefficients of initial velocity, thenThe total deposited energy is $E=\rho v^2/2$, whileThe seventh mode is eliminated by striking at a node $l=jL/7$, $j=1,\dots ,6$. For a narrow hammer, each fixed-mode fraction is $O(\epsilon )$; beyond the inverse-width cutoff the envelope falls as $n^{-2}$, modulated by the two sine factors.

Taking curls givesBoth have zero divergence and curl, so they satisfy the static vacuum equations. Their upward fluxes through the loop areFor clockwise current $I$, the net force is zero in $B_1$, while in $B_2$

Put $K=2\pi b_2{r}^2$. Motion in $B_1$ changes no flux, so no current is induced and $z=z_0-g{t}^2/2$. In $B_2$, using counterclockwise current as positive,ThusWith $\Gamma =K^2/R$, the velocity tends to $-mg/\Gamma$ andAt terminal speed, $I^{2}R=mg|\dot{z}|$, so Joule loss equals gravitational-energy loss. For $R=0$, Ohm's quasistatic formula is singular: flux is conserved, persistent current stores magnetic energy, and ideal motion has no resistive terminal state (inductance must be retained).