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Assume a slender homogeneous elastic beam obeying linear elasticity, with plane sections remaining plane and no resultant longitudinal force. A length element of its neutral axis subtending a small angle $d\varphi$ has length $Rd\varphi$. A neighbouring fibre at distance $z$ from that axis has a length change of magnitude $zd\varphi$, giving longitudinal strain $\epsilon =z/R$ with signs chosen consistently for tensile and compressed fibres. By Hooke's law, its longitudinal stress is $\sigma =E\epsilon =Ez/R$.

The torque of these longitudinal forces about the neutral axis is thereforeHere $E$ is Young's modulus and $I$ the second moment of area. The displayed relation uses a signed radius and bending moment; an unsigned geometrical radius is $|R|=EI/|B|$. For consistency with the negative bending moments in the later parts, take the deflection $y$ positive downward and the signed curvature as $1/R\simeq y^{\prime \prime }$. Then the small-slope Euler-Bernoulli beam equation reads $EI{y}^{\prime \prime }=B$. Reversing the vertical coordinate reverses the corresponding sign relation but leaves the physical deflection unchanged.

By vertical force balance and reflection symmetry, each support exerts an upward force $W/2$. Use the paper's convention in which the sagging bending moment is negative. For a cut at $0<x<L/2$, the left support's moment about that cut givesFor $L/2<x<L$, include the central downward force when taking moments of all loads to the left:ThusThe bending moment is continuous at the load, with $B(L/2)=-WL/4$, and vanishes at both freely rotating supports. The jump is in its first derivative, reflecting the concentrated transverse force.

The uniform weight per unit length is $q=\rho abg$, where $\rho$ is the mass density and $g$ the gravitational acceleration. The total weight is $qL$, so each support provides an upward force $qL/2$. The weight on the portion from $0$ to $x$ is $qx$ acting at that portion's midpoint, a distance $x/2$ from the cut. In the same bending moment convention as part (c),The units are $(\mathrm{N}{\mathrm{m}}^{-1}){\mathrm{m}}^2=\mathrm{N}\mathrm{m}$. Its greatest magnitude occurs at $x=L/2$, where $B=-q{L}^2/8$.

Newton's second law takes its familiar form $m\mathbf{a}=\mathbf{F}$ in an inertial frame. Axes fixed to a turning apparatus or to the Earth form a rotating reference frame. Extra inertial forces account for the acceleration of these coordinates. Their values depend on the chosen reference frame; physical interactions such as gravity, tension and contact forces must still be included separately.

Let the rotating origin have inertial acceleration $\mathbf{A}$, and let its axes have angular velocity $\mathbf{\Omega }(t)$. For any vector $\mathbf{q}$, the rotating-frame derivative formula isEach rotating basis vector changes by $\mathbf{\Omega }\times {\mathbf{e}}_i$, which proves this rule by expanding $\mathbf{q}$ in that basis. Apply it twice to the position relative to the moving origin. With ${\mathbf{v}}^{\prime }$ and ${\mathbf{a}}^{\prime }$ measured in the rotating axes,Consequently the equation in the rotating reference frame isThe additional terms are the translational inertial force, Coriolis force, Euler force and centrifugal force, respectively. The last two simplify when the angular velocity is constant: the Euler force vanishes, while the other rotation terms remain.

The centrifugal force points away from the rotation axis. If ${\mathbf{r}}_{\perp }$ is the perpendicular displacement from that axis, thenFor uniform rotation it has potential energyA mass at rest relative to a turntable can therefore be described as having an inward tension or friction balancing its outward centrifugal force. In an inertial frame, the inward physical force instead produces the required centripetal acceleration. These are two descriptions of the same motion; one must not insert the rotating-frame outward term into the inertial equation as an additional physical interaction.

A rotating liquid gives a useful quantitative example. For a liquid in hydrostatic equilibrium relative to a vessel rotating uniformly about the vertical axis,Along the free surface the pressure is constant, so ${\Omega }^{2}sds-gdz=0$ andThe surface is a elliptic paraboloid. This is the balance of gravity and the centrifugal force in the rotating description. On the Earth, the same outward acceleration reduces effective gravity, most strongly at the equator, and contributes to the planet's equatorial bulge.

The Coriolis force depends on relative motion:It is perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}^{\prime }$, vanishes for motion along the rotation axis, and reverses when the relative velocity reverses. The dot product givesThus the Coriolis force does no work in the rotating reference frame: it changes the direction of motion rather than directly changing the rotating-frame kinetic energy. For example, on a horizontal turntable rotating anticlockwise as viewed from above, an outward-moving slider has a clockwise Coriolis acceleration, whereas an inward-moving slider has an anticlockwise one. The deflection can be observed by comparing a laboratory camera with a camera rotating with the table.

For slow horizontal motion on the Earth at latitude $\lambda$, keep the vertical component ${\Omega }_E\sin \lambda$ of its angular velocity. In east-north coordinates $(u,v)$ the horizontal Coriolis acceleration isIt deflects motion to the right in the northern hemisphere and to the left in the southern hemisphere. In large-scale atmospheric and oceanic motion it can balance a horizontal pressure gradient, giving geostrophic balance. With $\hat{\mathbf{z}}$ upward, the steady horizontal equation without friction isThe resulting velocity runs along lines of equal pressure. This approximation requires rotation to dominate local inertial changes; it is inappropriate close to the equator where $f$ tends to zero.

The Foucault pendulum gives another terrestrial example. In the small-angle approximation, its horizontal coordinates satisfyTo first order in the small ratio ${\Omega }_E/{\omega }_0$, the oscillation plane precesses relative to the ground at angular rate $-{\Omega }_E\sin \lambda$: clockwise in the northern hemisphere, zero at the equator, and one revolution per sidereal day at a pole. The Coriolis force explains this changing orientation without requiring energy input. Together these examples show why inertial forces are useful whenever the apparatus or environment supplies naturally rotating coordinates.

The method of images solves certain electrostatics problems by replacing a conductor by fictitious electric charges outside the region where the solution is required. The replacement must satisfy the actual boundary values of the electric potential as well as the prescribed sources. The uniqueness of Poisson equation then establishes that the constructed electric potential equals the physical one in that region. The image locations are mathematical devices; they are not real charges inside the metal.

For a point charge $q$ at $(0,0,a)$ above an infinite grounded conducting plane $z=0$, put an image charge $-q$ at $(0,0,-a)$. In a cylindrical coordinate system with coordinates $(s,z)$ the potential in the physical region isThis vanishes at $z=0$, has the required singularity at the real charge, and tends to zero at infinity. With electric field $\mathbf{E}=-\nabla \varphi$, the induced surface charge density follows from the normal field immediately above the conductor:The force on $q$ is the field of the image evaluated at the real charge:The electrostatic energy relative to infinite separation is $U=\frac{1}{2}q{\varphi }_{\mathrm{image}}=-q^2/(16\pi {ϵ}_{0}a)$. The factor $1/2$ accounts for the fact that the induced charge changes as the real charge is assembled; treating the image as an independently fixed physical charge would give the wrong energy.

A second standard geometry is a grounded conducting sphere of radius $b$ with a charge $q$ a distance $d>b$ from its centre. Its exterior field is generated by an image $q^{\prime }=-qb/d$ at distance $b^2/d$ from the centre on the line towards the real charge. On the sphere the distances to the real and image charges are in the ratio $d/b$, making their potentials cancel. For an isolated neutral sphere, an additional central image $+qb/d$ makes the total induced charge zero and leaves the surface an equipotential. The construction is powerful for suitable symmetric boundaries, but arbitrary boundaries do not in general admit a finite image system.

For point masses, Newton's force is attractive along their separation with magnitude $G{m}_1{m}_2/r^2$. Treat the narrow rod as uniform, with linear mass density $\mu =M/(2a)$; a mass profile would otherwise be needed beyond total mass and length.

For axial and transverse gravity of a uniform rod, on the axis at $b>a$, integrating the element force givesdirected toward the rod. On the perpendicular bisector, axial components cancel and transverse components add:At $r=b\gg a$,Both approach the point-mass force, but $F_1/F_2=1+3a^2/(2b^2)+O(a^4/b^4)$, so the axial attraction is slightly larger.

For a circular orbit perpendicular to a uniform rod, rotational symmetry about the rod makes a centred circular orbit possible in its perpendicular-bisector plane. With $m\ll M$, centrifugal balance givesFor $r\gg a$, it is Keplerian, $v\sim \sqrt{GM/r}$. For rod thickness $\ll r\ll a$, the force approximates that of an infinite line and $v\to \sqrt{GM/a}$, a flat speed limit rather than a divergence.

The instantaneous mechanical power supplied by a force iswhere v is the velocity of its point of application in the chosen frame.

The net force equals the time derivative of momentum in an inertial frame:

For a mass change in a viscously damped spring oscillator, use linear viscous damping $m\ddot{x}+b\dot{x}+kx=mg$, holding the spring constant $k$ and damping coefficient $b$ fixed. Measuring $x$ from $mg/k$ removes the constant force. The envelope is $e^{-\gamma t}$, whereDoubling $m$ halves both $k/m$ and $\gamma$. ThereforeNeglecting damping in the frequency gives $0.5/\sqrt{2}=0.35355\mathrm{Hz}$, indistinguishable at the supplied precision. The release condition sets the phase and new initial amplitude, not these envelope ratios. A different mass-dependent drag law would require its damping coefficient to be specified.

For a fixed set of bodies in an inertial frame, total linear momentum is constant if the resultant external force vanishes. Summing Newton's second law cancels internal force pairs by Newton's third law. For an open subsystem, momentum carried across its boundary must also be included.

For an equal-intake-fuel mass-loading rocket, enclose the original craft and the stationary material collected during $dt$. Its initial momentum is $mu$. At the end, the craft has mass $m-\mu dt$ and the exhaust has mass $2\mu dt$. To first order,The first term is rocket thrust; the second is the momentum cost of collecting stationary matter. It vanishes in the conventional rocket equation, which therefore has no analogous fixed exhaust-speed ceiling in Newtonian mechanics. Here the acceleration vanishes at $u_{\max }=2u_0$; from rest it stays positive below that value while fuel remains. Relativistic limits are outside this model.

For constant $\mu$, mass conservation gives $m=m_i-\mu t$. Separating variables,ThereforeThe formal limiting speed would require $t=m_i/\mu$ and zero remaining craft mass. A positive rocket dry mass and finite carried fuel terminate the burn earlier. Maintaining nonzero collection rate even at rest is an idealization explicitly prescribed by the problem.

The bob experiences weight $Mg$ vertically downward and a rod force towards the pivot. The tangential component of weight is $-Mg\sin \theta$, soThe rod force supplies the radial acceleration and contributes no tangential torque about the pivot.

To the same small-angle order, writing $\chi ={\omega }_{0}t+\varphi$,Thus $K+U=MgL{\theta }_0^2/2$ is constant. Put $b=\dot{\theta }(0)$, a signed angular velocity. ThenIf the printed angular speed $\beta$ is intended as the signed rate, set $b=\beta$. If it means magnitude only, $b=\pm \beta$ gives two possible phases; the amplitude is unique. At zero amplitude the phase is arbitrary.

For a finite-radius spherical pendulum bob, the PDF dimension $L$ runs from pivot to centre of mass. The parallel axis theorem gives $I_P=M{L}^2+2M{a}^2/5$. The restoring torque remains $-MgL\theta$, henceWith $a=[3M/(4\pi \rho ){]}^{1/3}=0.02605\mathrm{m}$,The finite bob increases the inertia without changing the gravitational lever arm.

An electric dipole consists of opposite charges separated by a small displacement, with electric dipole moment $\mathbf{p}=q\mathbf{d}$ directed from negative to positive. At distances large compared with its size,The net charge term vanishes; the field decays faster than a monopole field. In an imposed slowly varying electric field,for a permanent fixed dipole. A uniform field produces a torque but no net force. For a linear induced dipole $\mathbf{p}=\alpha \mathbf{E}$, the interaction energy is instead $-\alpha E^2/2$, because building the dipole also costs energy.

A small planar current loop has magnetic dipole moment $\mathbf{\mu }=IA\hat{\mathbf{n}}$, with the normal fixed by the right-hand rule. Outside the source,Magnetic field lines form closed loops; a current-loop dipole does not require magnetic monopoles. Dipole moments describe the leading directional field of a neutral localized source, and their torques explain alignment in external fields.

In electrostatic equilibrium a conductor has $\mathbf{E}=0$ throughout its bulk and constant electric potential on each connected component. Otherwise free charge would continue moving. Gauss's law then excludes bulk excess charge; it resides on surfaces. Just outside a conductor the tangential field vanishes and the normal jump isfor a vacuum interface with outward normal $\hat{\mathbf{n}}$. Surface charge rearranges in an external field to enforce these boundary conditions. Greater curvature often concentrates charge, but the actual distribution is a global boundary-value problem.

An empty closed cavity inside a conductor has constant potential and zero field, even in an external electrostatic field. If a charge $q$ is placed in the cavity, a Gaussian surface in the metal shows that the cavity wall carries total induced charge $-q$; its detailed distribution depends on geometry. An initially neutral isolated conductor then has $+q$ on its exterior surface. A grounded conductor can exchange charge with Earth, so this isolation condition no longer applies.

A useful example is a charge $q$ at height $d$ above an infinite grounded plane. In the physical half-space, the method of images replaces the induced charges by an image $-q$ at height $-d$, giving the attractive force $q^2/(16\pi {ϵ}_0{d}^2)$. The image is a mathematical construction, not an actual charge behind the conductor. Equipotential surfaces and charge conservation determine conductor fields together.

In the centre-of-momentum frame the two equal masses have opposite velocities. Conservation of kinetic energy in an elastic collision fixes their common speed, so scattering changes their directions only. For incoming laboratory velocities $\mathbf{v}$ and $0$, the frame velocity is $\mathbf{V}=\mathbf{v}/2$ and the incoming relative velocities are $\pm \mathbf{v}/2$. If the outgoing relative velocities are $\pm \mathbf{w}$, then $|\mathbf{w}|=v/2$ and the laboratory velocities are $\mathbf{V}\pm \mathbf{w}$. Their dot product is zero: nonzero outgoing laboratory velocities are perpendicular.

For this smooth equal-disc oblique collision, let $\mathbf{n}$ point from A to B at contact. The impulse acts only along $\mathbf{n}$; the tangential velocity of A is unchanged. With $\mathbf{v}\cdot \mathbf{n}=v\cos \theta$, equal masses exchange their normal velocity components:ThereforeNo rotation about a disc centre is generated by these central impulses. At exactly head-on impact A stops, so its outgoing direction is undefined; the angle formula then denotes a limiting direction.

For this adhesive collision of two equal discs, use the parallel axis theorem about the midpoint of the touching centres:Its magnitude is $mav\sin \theta$. Linear and angular momentum conservation giveThe rotation has the sign of ${\mathbf{L}}_{\mathrm{CM}}$, clockwise for the printed geometry. The energy loss is $m{v}^2(1/4-{\sin }^2\theta /6)>0$. After impact the midpoint translates uniformly and the rigid pair rotates uniformly about it: there are no external horizontal forces or axial torques.

Rough unglued edges can transmit a tangential impulse and generate angular momentum about the individual disc centres. The centre-of-mass velocity and total angular momentum still have the values above, but individual translational speeds, angular velocities and energy loss require a friction/contact law and normal restitution. Roughness alone does not imply permanent sticking or the adhesive result.

For a fixed rotation axis with constant moment of inertia, the axial torque obeys $\tau =I\ddot{\theta }$, with $\tau =\sum (\mathbf{r}\times \mathbf{F})\cdot {\hat{\mathbf{e}}}_{\mathrm{axis}}$. For a point mass at radius $l$, the tangential equation is $F_t=ml\ddot{\theta }$; multiplying by $l$ gives $l{F}_t=m{l}^2\ddot{\theta }$, as required. Radial forces have zero moment about the axis.

Take increasing $\theta$ in the direction of the clapper's initial fall and let the contact force $N$ oppose that tangential motion. Initial equilibrium of the clapper givesWhile contact persists the bell and clapper rotate together. Their respective offsets from the upward vertical are $\theta -\beta$ and $\theta$. Summing external gravitational moments eliminates their internal contact force:The clapper alone satisfies $m{l}^2\ddot{\theta }=mgl\sin \theta -Nl$. Hence for $m\ll M$,A nonadhesive contact cannot exert tension. The first zero of this expression on the released trajectory is the loss-of-contact angle. Write $q=lh/k^2=0.9375$. ThenThe second-quadrant root is reached from $\theta =\beta =30^{\circ }$; selecting the principal negative arctangent would choose the wrong branch. Indeed the exact contact force before taking $m/M\to 0$ is $mMg[k^2\sin \theta -lh\sin (\theta -\beta )]/(M{k}^2+m{l}^2)$, so its zero is unchanged.

Measure $x$ from the static equilibrium position, so the constant gravitational force has already cancelled the equilibrium spring force. With stationary oil, Newton's second law gives $m\ddot{x}=-b\dot{x}-kx$. The characteristic exponents are $-\gamma \pm i{\omega }_1$, whereFor the stipulated phase of the damped cosine,At $t_n=2\pi n/{\omega }_1$, the velocity is not zero: $\dot{x}=-\gamma A{e}^{-\gamma t_n}$. ThusExpanding $e^{-z}$ at small $z=4\pi \gamma /{\omega }_1$ gives the requested light-damping loss $z$. The exact instantaneous energy also satisfies $dE/dt=-b{\dot{x}}^2\le 0$.

From the static extension, $k=mg/(0.20\mathrm{m})=44.145\mathrm{N}{\mathrm{m}}^{-1}$; the instantaneous drag measurement gives $b=0.10\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$. Consequently $\gamma =0.05556{\mathrm{s}}^{-1}$ and ${\omega }_1=7.003{\mathrm{s}}^{-1}$. Two successive downward equilibrium crossings are one damped period apart, so the same energy ratio applies. At the first crossing $E=m{v}^2/2=1.125\times 10^{-3}\mathrm{J}$. ThereforeThe speed has decreased by $2.43\times 10^{-3}\mathrm{m}{\mathrm{s}}^{-1}=0.243\mathrm{cm}{\mathrm{s}}^{-1}$. These values use the exact decay factor, not the first-order approximation.

In magnetostatics Ampère's circuital law is $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{\mathrm{enc}}$. Its general vacuum version adds ${\mu }_0{ϵ}_{0}d{\Phi }_E/dt$ on the right. Faraday's law is $\mathcal{E}=\oint \mathbf{E}\cdot d\mathbf{l}=-d{\Phi }_B/dt$ for a fixed loop. A circular Amperian path around a long straight wire giveswith direction given by the right-hand rule. The force on a wire element is $d\mathbf{F}=Id\mathbf{l}\times \mathbf{B}$; its magnitude is $IBdl\sin \chi$ and its direction follows this cross product.

Choose the straight-wire current rightward and the square below it, as printed. Its positive magnetic field there is into the page. With an into-page loop normal, the magnetic flux linkage isHere $\mathcal{M}$ denotes mutual inductance, not a mass. Choose positive coil current to produce an out-of-page field. Then its induced current in the resistance-dominated approximation isThis reproduces the specified sign with the declared orientation. It neglects coil self-inductance, requiring $\omega L_{\mathrm{coil}}\ll R$, as implicit in the requested expression.

The nearest side contains N parallel current segments and is at a uniform distance $s$, so its force magnitude isThe two sides perpendicular to the wire have opposite currents and equal magnetic fields at corresponding distances; their forces cancel pairwise. The far side opposes the near-side force and has the smaller field. ThusWhen $I$ and $i$ have the same sign under the chosen conventions, the near-side coil current opposes the straight-wire current and the total force is away from the wire; otherwise it is toward it. In particular the signed force alternates with $\sin (\omega t)\cos (\omega t)$. The coil is fixed, so these forces do no mechanical work on it. The mean Joule heating is

A gravitational field is force per unit test mass, $\mathbf{g}=\mathbf{F}/m$. Newton's law of universal gravitation gives the field of a point mass M:The potential is potential energy per unit mass, with zero at infinity for an isolated mass. Fields and potentials superpose. The potential difference fixes the work per unit mass, and a conservative gravitational field has zero circulation around a closed path.

For a continuous mass distribution, the gravitational Gauss law is $\oint \mathbf{g}\cdot d\mathbf{S}=-4\pi G{M}_{\mathrm{enc}}$. A spherical body therefore has the same exterior field as a point mass at its centre. Inside a uniform sphere of density $\rho$,The field is continuous at its surface and vanishes at its centre. An empty spherical shell has zero interior field but a nonzero constant interior potential. Near a small region of Earth's surface one may approximate $\mathbf{g}$ as uniform and the potential change as $g\Delta h$. Farther away its spatial gradient produces tidal relative accelerations. Escape to infinity from radius R requires speed $\sqrt{2GM/R}$ when rotation and other bodies are neglected.

The total ring mass is $M=2\pi r{\lambda }_m=1.257\times 10^7\mathrm{kg}$. For the axial gravitational field of a uniform ring, transverse force components cancel by symmetry; each axial component has projection factor $z/\sqrt{r^2+z^2}$. ThusThe force points towards the ring's centre along its axis.

Take $q=\ddot{x}$ positive when the hanging mass descends and the other mass moves uphill. For the sliding block, Newton's second law gives $mg-T=mq$ and $T-Mg\sin \theta =Mq$. ThereforeFor the pulley coupled to a rolling disc, tension and weight act through its centre; the contact normal is perpendicular to the slope. Static friction provides the torque required for rolling without slipping. Define signed $F$ down the slope, so $F>0$ for uphill acceleration.

The rotational equation and rolling constraint give $Fa=Iq/a$, hence $F=Mq/2$. The translational equations become $T-Mg\sin \theta -F=Mq$ and $mg-T=mq$, givingThus the threshold remains $m>M\sin \theta$: the moment of inertia increases the effective mass but does not change the sign of the driving force. For downhill acceleration $F$ reverses. The normal force is $Mg\cos \theta$; static friction does no work at the stationary contact point and sufficient roughness is assumed.