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Use a translating accelerated reference frame; its axes do not rotate. If $\mathbf{R}=(\xi ,\eta ,\zeta )$ is the block's position relative to the table, then its inertial position is $\mathbf{r}=\mathbf{R}+\mathbf{A}\sin \omega t$. Its relative velocity is $\dot{\mathbf{R}}$ and its inertial acceleration is $\ddot{\mathbf{R}}-{\omega }^2\mathbf{A}\sin \omega t$. Let $N$ be the upward normal force. During contact $\zeta =\dot{\zeta }=\ddot{\zeta }=0$, and the linear tangential resistance givesThese are Newton's second law transformed to the table's reference frame; the extra terms arise from the table's acceleration. The last equation gives $N=m(g-A_z{\omega }^2\sin \omega t)$ while contact persists. If contact is lost, both the contact resistance and $N$ vanish, and free flight must be described by the corresponding gravity-only equations.

For $A_z=0$, assume $\lambda >0$ and put $\gamma =\lambda /m$. Seek a periodic particular solution ${\xi }_p=b\sin \omega t+c\cos \omega t$. Substitution into the first differential equation givesSolving yieldsThe homogeneous solutions are a constant and a multiple of $e^{-\gamma t}$. Choose the phase $0<\theta <\pi /2$ with $\tan \theta =\omega /\gamma =m\omega /\lambda$. Then $b=-A_x{\sin }^2\theta$ and $c=-A_x\sin \theta \cos \theta$, soAlso $\eta (t)={\eta }_{\infty }+C_{\eta }{e}^{-\lambda t/m}$. The decaying transients leave the stated steady orbit. Positive damping is essential to this late-time assertion; for $\lambda =0$ an additional constant drift may persist.

No attractive contact force is available, so $N$ cannot be negative. Over a whole oscillation its minimum is $m(g-{\omega }^2|A_z|)$. Thus strictly positive contact loading throughout requires $\boxed{{\omega }^2|A_z|<g}$, or the printed form with $A_z$ taken as a nonnegative vertical amplitude. If ${\omega }^2|A_z|>g$, contact cannot persist: it would require $N<0$ during part of the cycle. At equality, $N$ vanishes only at isolated instants and the ideal constrained motion can still remain in contact. Hence the strict inequality characterizes positive loading; mere nonnegative contact permits the marginal equality case.

Label the initially left-hand rod $A$ and the right-hand rod $B$. At impact their centres are $(0,\ell )$ and $(0,-\ell )$, and their contact points are both the origin. Each rod has total mass $2m$ and moment of inertiaLet the impulse on $B$ be $+J{\mathbf{e}}_x$, so the impulse on $A$ is $-J{\mathbf{e}}_x$. The impulsive form of Newton's second law givesThe contact lever arms from the centres are $(0,-\ell )$ for $A$ and $(0,+\ell )$ for $B$. Both angular impulses are $-\ell J{\mathbf{e}}_z$. Thus, taking counterclockwise rotation as positive,Equal and opposite linear impulses therefore produce equal clockwise rotations, not opposite rotations.

The restitution law uses the velocities of the contacting endpoints, including the rotational terms. For $A$ that velocity isand for $B$ it is $u_B=-V+J/m$. The approach speed before impact is $2V$. The coefficient of restitution $e$ therefore givesSubstituting into the centre and angular velocities yields the endpoint collision of two freely rotating equal rods:For the first, perfectly elastic case, $e=1$, soAll initial translational energy has become rotational energy. Indeed the total initial kinetic energy is $2m{V}^2$, while the final energy for general $e$ isThe loss is $m{V}^2(1-e^2)$, zero for an elastic collision. Conservation of total angular momentum about the origin provides another check: initially it is $-4m\ell V$; after collision the centre-motion contribution is $-2m\ell V(1-e)$ and the spins contribute $-2m\ell V(1+e)$, with the same total. The initial distance $d$ determines when the collision occurs but does not enter the impulsive velocities.

Let $m$ be the car's mass and let $v(t)$ be its forward speed. Initially the driven contact surfaces slip backward relative to the ground because $v<r\Omega$. Thus kinetic friction on the car acts forward. On the horizontal surface the total normal force is $mg$, so Newton's second law gives $m\dot{v}=\mu mg$. While slipping persists,The contact slip speed is $r\Omega -v(t)$. Its first zero is the onset of rolling without slipping, henceThe engine maintains the prescribed wheel speed, so wheel spin-down is not part of this calculation.

On a slope of angle $\alpha >0$, directed uphill, the normal force is $mg\cos \alpha$ and gravity contributes $-mg\sin \alpha$ along the slope. Therefore the uphill acceleration while slipping is $g(\mu \cos \alpha -\sin \alpha )$. If it is positive, the onset of rolling without slipping occurs atThe inequality follows because $\mu \cos \alpha -\sin \alpha <\mu$. If $\mu \cos \alpha \le \sin \alpha$, the car does not accelerate uphill to $r\Omega$, so there is no finite catch-up time in this sliding-friction model. The finite-time conclusion presumes positive traction sufficient to climb the slope.

Let ${\mathbf{e}}_r,{\mathbf{e}}_{\varphi },{\mathbf{e}}_z$ be the cylindrical unit vectors. The surface has an upward normal vector proportional to ${\mathbf{e}}_z-h^{\prime }(r){\mathbf{e}}_r$. If $N_z$ is the vertical component of the normal force, its radial component is $-h^{\prime }(r)N_z$. In the stated approximation, vertical acceleration and the vertical driving-force component are neglected, so vertical balance gives $N_z\simeq Mg$. The horizontal contact force is consequently $-Mg{h}^{\prime }(r){\mathbf{e}}_r$. The projected velocity and its speed areThe driving force is parallel to this velocity, with components $F\dot{r}/u$ and $Fr\dot{\varphi }/u$. Use the acceleration in polar coordinates and Newton's second law to obtainThese equations apply where $u\ne 0$; at rest the direction of an exactly velocity-aligned drive is not determined.

For a prescribed path, write $r^{\prime }=dr/d\varphi$, $r^{\prime \prime }=d^{2}r/d{\varphi }^2$. The chain rule gives $\dot{r}=r^{\prime }\dot{\varphi }$ and $\ddot{r}=r^{\prime \prime }{\dot{\varphi }}^2+r^{\prime }\ddot{\varphi }$. Subtract $r^{\prime }/r$ times the azimuthal equation from the radial equation. The driving force and $\ddot{\varphi }$ both cancel:Thus the constraint for tangentially driven motion on an axisymmetric surface isThe denominator must give a nonnegative squared speed, and the calculation uses $r>0$. At the polar origin a limiting Cartesian description is appropriate.

For the paraboloid and Archimedean spiral, putThen, for $\varphi >0$,The branch tending to $\varphi =\infty$ has positive angular velocity, henceSince $u=ϵ\dot{\varphi }\sqrt{1+{\varphi }^2}$, solve the azimuthal equation for the required force:This is positive for $\varphi >0$, consistent with a forward drive. Taking the limit gives

These are the large-time limits of the printed projected equations. Their mechanical approximation has a precise range limitation: on the exact surface, $z=h(r)$ gives $\dot{z}=(2r/\ell )\dot{r}$, which along this solution grows like $2{ϵ}^2\alpha \varphi /\ell$. At fixed positive $ϵ$, it is not small as $\varphi \to \infty$. Thus the requested limits are formal limits of that reduced model; neglect of vertical motion is not a uniform approximation to the full three-dimensional motion all the way to infinity.

Dot Newton's second law with $\dot{\mathbf{r}}$. The magnetic cross-product term is perpendicular to this velocity, soThusorThe two halves are the kinetic energy and the isotropic spring's potential energy. The expression in the question is twice their sum. The Lorentz force changes the direction of motion but supplies no power, because it is perpendicular to velocity; this explains why the magnetic field does not appear in the conserved mechanical energy. The sign corresponds to charge $-e$.

For a fixed central gravitational source, the radial and transverse components of Newton's second law in polar coordinates areThe second equation gives $r^2\dot{\theta }=h$, the constant specific angular momentum. Take $h>0$ by choosing the orientation of $\theta$, and put $u=1/r$. Writing $u^{\prime }$ and $u^{\prime \prime }$ for derivatives with respect to $\theta$,Substituting into the radial equation and dividing by $-h^2{u}^2$ proves the Binet equationAt perihelion the velocity is tangential. The escape velocity is $v_0=\sqrt{2GM/r_0}$, so $h=r_0{v}_0$ and $h^2=2GM{r}_0$. Consequently $k=1/(2r_0)$. The initial conditions are $u(0)=1/r_0$ and $u^{\prime }(0)=0$. Solving the linear ordinary differential equation gives the parabolic Kepler orbitUsing angular momentum conservation again,On the outgoing branch $r=2r_0$ is first reached at $\theta =\pi /2$. Integrate from release at perihelion, using $w=\tan (\theta /2)$:More generally the same integral gives $t=(2r_0^2/h)(w+w^3/3)$, the Barker equation for this parabolic orbit.

For constant particle masses define the centre of mass by $M\mathbf{X}={\sum }_i{m}_i{\mathbf{r}}_i$, with $M={\sum }_i{m}_i$. Summing Newton's second law over all particles givesEach pair of internal forces cancels by Newton's third law. HenceOnly the action-reaction condition is needed for this translation law; centrality of the internal forces is needed for the angular momentum result in part (ii).

Let $r=|{\mathbf{r}}_1-{\mathbf{r}}_2|$ and $\hat{\mathbf{r}}=({\mathbf{r}}_1-{\mathbf{r}}_2)/r$. A differentiable central potential energy gives the internal forcesThere are no external forces. Differentiating the total kinetic energy and using Newton's second law therefore givesSince $\dot{r}=\hat{\mathbf{r}}\cdot ({\mathbf{v}}_1-{\mathbf{v}}_2)$, the last expression is $-U^{\prime }(r)\dot{r}=-dU/dt$. ThusThis conserves total mechanical energy, including the kinetic energy of centre-of-mass motion, not merely the relative kinetic energy.

Take upward velocity as positive, and consider the closed material system consisting of the rocket and the fuel it ejects during a short interval $dt$. Initially its mass and momentum are $m$ and $mv$. Let $\delta m>0$ be the ejected mass, so the rocket's mass change is $dm=-\delta m$. At the end its velocity is $v+dv$, while the ejected gas has inertial velocity $v-u$ to first order; its downward relative velocity need not be downward in the inertial frame.

The final total momentum, to first order, isNeglect air resistance and take the gravitational acceleration $g$ constant. The external gravitational impulse on this material system is $-mgdt$ to first order. Newton's second law therefore gives $mdv-u\delta m=-mgdt$, orThe positive thrust is $-u\dot{m}$ because fuel loss makes $\dot{m}<0$. Applying $d(mv)/dt$ to the rocket alone without accounting for the escaping momentum would give the wrong equation.

For constant $u$ and positive mass, divide by $m$ and integrate from release at $t=0$:Thus the gravity-corrected rocket equation is equivalent to

The potential factors as $V=x^2(3-2x)$, so its zeros are $0$ and $3/2$, with a double zero at the origin. Its derivatives are $V^{\prime }=6x(1-x)$ and $V^{\prime \prime }=6-12x$. Thus there is a local minimum $(0,0)$, a local maximum $(1,1)$ and an inflection at $(1/2,1/2)$. The potential tends to $+\infty$ on the far left and $-\infty$ on the far right. These features determine the left-hand sketch.

Put $v=\dot{x}$. Newton's second law gives the phase plane system $\dot{x}=v$, $\dot{v}=-V^{\prime }=6x(x-1)$. Its conserved mechanical energy and trajectories areThe linearization at $(0,0)$ has eigenvalues $\pm i\sqrt{6}$, and the surrounding energy curves are closed, so it is a center equilibrium. At $(1,0)$ the eigenvalues are $\pm \sqrt{6}$, giving a saddle equilibrium. The figure shows the cubic potential barrier phase portrait; arrows point right when $v>0$ and left when $v<0$.

For $0<E<1$, there are three turning-point roots of $V(x)=E$. The two leftmost enclose a closed orbit around the origin: the particle oscillates between them. There is also a separate open component to the right of the third root, where an incoming particle turns and escapes to $+\infty$. The closed orbits run clockwise in the $(x,v)$ plane.

For $E=1$, the factorization $1-V(x)=(x-1{)}^2(2x+1)$ gives the separatrixThe branch with $-1/2\le x<1$ is a homoclinic orbit: it leaves the saddle asymptotically, turns at $x=-1/2$ and returns asymptotically. On $x>1$, one branch approaches the saddle and the other departs towards infinity. Approaching the saddle takes infinite time. The saddle itself is also an equilibrium trajectory.

For $E>1$, the particle can cross the barrier. There is one left turning point below $-1/2$; an incoming particle passes through the well, turns on the left and escapes to the right. For $E<0$, only a right-hand open component exists, with its turning point beyond $3/2$. At $E=0$, the stationary centre coexists with a right-hand open trajectory turning at $3/2$. Thus negative energy does not imply trapping here: the cubic potential is unbounded below on the right.

At the initial instant the circulation around every closed curve is zero because the fluid is at rest. A smooth material-flow map carries closed curves bijectively between times. Conservation of circulation around every material curve therefore makes the line integral of the velocity zero around every closed curve at the current time, not just small contractible loops. Fix a reference point in a connected fluid region and defineThe vanishing closed-curve integrals make this independent of the path. Thus $\mathbf{u}=\nabla \varphi$: the fluid has a globally defined velocity potential. Its incompressible flow condition is $\nabla \cdot \mathbf{u}=0$, hence ${\nabla }^2\varphi =0$. This argument establishes the global potential without silently assuming the fluid region is simply connected.

Use the laboratory frame, with the sphere instantaneously centred at the origin and moving in the $\theta =0$ direction. Impermeability equates fluid and sphere normal velocities, while the fluid is at rest at infinity:Choose the additive constant so that $\varphi \to 0$ at infinity. Substitution of $\varphi =f(r)\cos \theta$ into the Laplace equation in spherical polar coordinates givesThe far-field condition gives $A=0$, and $f^{\prime }(a)=U$ gives $B=-U{a}^3/2$. ThereforeThe radial and polar velocity components areFor fluid density $\rho$, its kinetic energy isThus $T_f=\frac{1}{2}{m}_a{U}^2$ with added mass of a sphere $m_a=2\pi \rho a^3/3$, half the mass of displaced fluid.

For the initial acceleration after release, let $\dot{U}$ denote the upward acceleration, $\mathcal{V}=4\pi a^3/3$, and $m_b={\rho }_b\mathcal{V}$. At the instant $U=0$, the velocity-squared term in the Unsteady Bernoulli equation vanishes. The time derivative of the potential at the sphere is ${\varphi }_t(a,\theta )=-a\dot{U}\cos \theta /2$; centre-motion corrections also vanish at this instant. Relative to hydrostatic pressure, the additional pressure is consequentlyIts upward force on the sphere isAdding buoyancy and weight, Newton's second law gives $(m_b+m_a)\dot{U}=(\rho -{\rho }_b)\mathcal{V}g$. ThereforeThis is the acceleration of a freely falling sphere with added mass, with upward chosen positive. If ${\rho }_b>\rho$, the negative value represents a downward acceleration.

Put $M=m_1+m_2$ and let $\mathbf{R}=(m_1{\mathbf{r}}_1+m_2{\mathbf{r}}_2)/M$ be the center of mass. Applying Newton's second law separately to the two point masses givesThus the center of mass has constant velocity, $\mathbf{R}(t)=\mathbf{R}(0)+t\dot{\mathbf{R}}(0)$. For the relative position $\mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2$, subtraction givesHere $\mu$ is the reduced mass; it converts the relative motion into a one-body equation of motion.

For the given linear force, the relative position obeys simple harmonic motion, $\ddot{\mathbf{r}}+{\Omega }^2\mathbf{r}=0$, where ${\Omega }^2=k/\mu$. The initial rest condition givesFor $d>0$, collision first occurs when the relative position vanishes, namely when $\Omega t=\pi /2$. ThereforeThe separation $d$ cancels because the frequency of this simple harmonic motion is independent of its amplitude.

Newton's second law with the computed Lorentz force gives $m\dot{v}=-(B^2{\ell }^2/R)v$. Separating variables and imposing $v(0)=v_0$ yieldsThe velocity vector is this scalar times $\hat{\mathbf{x}}$, and the characteristic decay time is positive for the stated parameters.

Work in an inertial frame, take all masses to be constant, and assume the internal forces obey the pairwise relation from Newton's third law $F_{ij}=-F_{ji}$. For the angular momentum result, also assume these pair forces are central, so $F_{ij}$ is parallel to $r_i-r_j$ and its pair torque vanishes. The equations from Newton's second law areSumming them and pairing the internal forces gives

About a fixed point $a$, define the total angular momentum and external torque byDifferentiation produces no velocity term because ${\dot{r}}_i\times m_i{\dot{r}}_i=0$. Grouping internal terms into pairs givesEach final cross product is zero by centrality, proving $\boxed{d{L}_a/dt=G_a}$. Equal and opposite internal forces give momentum conservation when the total external force is zero, but do not in general make their internal torques cancel; the central-force assumption supplies that extra step.

Let $M={\sum }_i{m}_i$ and let the centre of mass be $R=M^{-1}{\sum }_i{m}_i{r}_i$, so $P=M\dot{R}$. For an arbitrary moving reference point $a(t)$, the same differentiation givesTaking $a=R$, the extra term is $-\dot{R}\times M\dot{R}=0$. Therefore the corresponding angular momentum about the centre of mass satisfiesIts definition using velocities relative to $\dot{R}$ agrees with the one above, since ${\sum }_i{m}_i(r_i-R)=0$ makes ${\sum }_i(r_i-R)\times m_i\dot{R}=0$. Thus this result remains true even when the centre of mass accelerates.

Finally take the common mass to be $m$ and $F_i=-k{\dot{r}}_i$, with constant $k$. About either a fixed point or the centre of mass,The angular-momentum equation becomes $\dot{L}=-(k/m)L$. Multiplying by the integrating factor $e^{kt/m}$ gives $\frac{d}{dt}(e^{kt/m}L)=0$, hence the angular momentum decay under uniform linear drag isThe common mass and common drag coefficient ensure that the same decay factor applies to every term.

With no electric field, the Lorentz force is $F=q\dot{r}\times B\hat{z}$. The equation from Newton's second law and its first integral areThe cyclotron frequency is $|\omega |$; the sign of $\omega$ specifies the sense of rotation. Since $\hat{x}\times \hat{z}=-\hat{y}$, the initial data giveThe component equations are $\dot{x}=\omega y$, $\dot{y}+\omega x=u+a\omega$ and $\dot{z}=v$. For $\omega \ne 0$, differentiating the first givesSolving this linear differential equation and then using $y=\dot{x}/\omega$ givesThe perpendicular projection is a circle of radius $|u/\omega |$ centred at $(a+u/\omega ,0)$, the guiding centre. The velocity parallel to the magnetic field stays constant, producing helical motion in a uniform magnetic field.

When $a\omega +u=0$, this simplifies toThus the trajectory is a helix about the $z$ axis, with radius $|a|$ and axial advance $2\pi v/|\omega |$ per complete revolution in increasing time. Its projection moves clockwise when viewed from positive $z$ for $\omega >0$. If $v=0$ it is a circle; if $a=0$ it is a straight line along the axis.

For $\omega =0$, the Lorentz force vanishes and the solution is instead $r(t)=a\hat{x}+ut\hat{y}+vt\hat{z}$, also the limit of the general formula. The special condition then forces $u=0$.

For the moment of inertia of a uniform solid sphere, write the mass density as $\rho =3m/(4\pi a^3)$. In spherical coordinates about the chosen axis, the squared perpendicular distance is $R^2{\sin }^2\vartheta$. The volume element is $R^2\sin \vartheta dRd\vartheta d\phi$, so

Take $0<\alpha <\pi /2$ and $\ell >0$, with a fixed plane. The normal force has magnitude $N=mg\cos \alpha$ and is perpendicular to the plane. On the perfectly smooth plane there is no friction. Both gravity and the normal force have zero torque about the marble's centre, so a marble released from rest does not rotate. The component of Newton's second law down the plane is $m{A}_1=mg\sin \alpha$, henceThe normal force does no work because the velocity is tangent to the plane. Gravity is conservative, so mechanical energy, the sum of translational kinetic energy and gravitational potential energy, is conserved.

For rolling without slipping, let the static friction force have magnitude $f$, pointing up the plane. It supplies the torque that spins the marble while reducing its translational acceleration. If $\Omega$ is the angular velocity and $A_2$ the acceleration down the plane, the rolling constraint is $a\Omega =\dot{s}$, hence $a\dot{\Omega }=A_2$. The translation and rotation equations areThus $f=I{A}_2/a^2$ andThese are the equations for rolling acceleration with rotational inertia; the plane must be rough enough to supply this static friction.

The point of contact is instantaneously at rest relative to the fixed plane, so static friction does no total work on the rolling marble. Equivalently, its translational power is $-f\dot{s}$ and its rotational power is $fa\Omega =f\dot{s}$, which cancel. The normal force also does no work. Consequently mechanical energy is conserved here too, now with rotational kinetic energy included:where $H$ is the centre's height. No energy loss is implied by the presence of ideal static friction. Constant acceleration from rest gives

For the hollow marble, put $I=\lambda m{a}^2$ in the same equations. The smooth-plane time is unchanged, while the rolling acceleration becomes $g\sin \alpha /(1+\lambda )$. ThereforeThis again assumes rolling without slipping and enough static friction; the ratio depends on the mass distribution through its moment of inertia.

Apply the rotating-frame derivative formula first to position and then to its inertial velocity:The angular velocity is constant, so there is no term involving $\dot{\mathbf{\omega }}$. By Newton's second law in the inertial frame, the inertial acceleration is $\mathbf{F}/m$. Rearrangement gives the equation of motion in a rotating frame:The two additional accelerations are the Coriolis acceleration and centrifugal acceleration.

Take the scalar product of Newton's second law with the velocity $\dot{\mathbf{r}}$. The cross product is perpendicular to its first factor, soConsequentlyThis is conservation of energy: the sum of kinetic energy and potential energy stays fixed. The magnetic-type Lorentz force $\dot{\mathbf{r}}\times \mathbf{B}$ does no work, because it is perpendicular to the instantaneous velocity. It may turn the trajectory but cannot directly change its speed, regardless of the spatial variation of $\mathbf{B}$.

Apply the rotating-frame derivative formula first to $r$, then to $v^{\prime }=[dr/dt{]}_{S^{\prime }}$. Since the angular velocity is constant in either frame,The two apparent forces in the rotating equation are the Coriolis force and the centrifugal acceleration multiplied by mass. ThusSet ${\varphi }_{\mathrm{eff}}=\varphi -\frac{m}{2}|\omega \times r{|}^2$. The supplied gradient identity gives $\nabla {\varphi }_{\mathrm{eff}}=\nabla \varphi +m\omega \times (\omega \times r)$. Taking the scalar product with $v^{\prime }$ removes $G$, since it is orthogonal to $v^{\prime }$, and removes the Coriolis force. Therefore the energy balance in a uniformly rotating frame isThe displayed quantity is conserved when the potential is stationary in the rotating coordinates. This includes a static inertial potential invariant under rotations about the chosen axis, such as vertical gravity in part (b). If the word conservative only means a time-independent inertial potential, that extra rotational invariance is needed; a conservative force alone does not imply the claimed conservation.