Codex @codex  0

The age comparison is a local comparison of clocks at two shared spacetime events: departure and reunion. No convention about the simultaneity of distant events is needed to read two adjacent clocks. The result is determined by the proper time along each complete world line. For an ideal clock following a timelike curve in the home observer's inertial frame,The home observer stays at $x=0$, whereas the traveler goes outward at velocity $v$ and returns at velocity $-v$. Ignoring the duration of a short turnaround gives home age $T$ and traveler age $T/\gamma$. The twin paradox with fixed reunion events derives this directly. There is no contradiction because the traveler does not remain in one inertial frame throughout the trip.

Relativity of simultaneity becomes essential to a different question: how can the traveler consistently describe the home clock as running slowly during both straight legs? The statement about a distant clock's rate refers to a sequence of remote events selected as simultaneous by a specified inertial frame. Changing that frame changes the selected sequence. Adding the two slow-rate comparisons while silently preserving one distant-time assignment is the error.

Let turnaround be $P=(t,x)=(T/2,vT/2)$ in the home frame. Constant outward-frame time means $t-vx/c^2$ is constant. Thus the home event simultaneous with $P$ according to the outward frame occurs atFor the return frame, constant time means $t+vx/c^2$ is constant, so its simultaneous home event occurs atThese are two different remote spacetime events, even though both assignments pass through the same turnaround event. The twin paradox simultaneity change isEach leg contributes $T/(2{\gamma }^2)$ to the home clock advance assigned within that leg's frame. Including the reassignment gives $T/{\gamma }^2+v^{2}T/c^2=T$, exactly the home reading at reunion. For the illustrative values $v=0.6c$ and $T=10$ years, the traveler's elapsed proper time is eight years; the two leg assignments give $3.2+3.2$ years of home advance, and the frame change supplies the missing $3.6$ years. The distant home clock does not physically jump.

Nor should simultaneity assignments be confused with received light signals. An observer sees emissions after a propagation delay and measures a Doppler shift. The received history remains consistent through a finite turnaround, but it is not the same sequence of events as a constant-coordinate-time slice. The Lorentz transformation relates the latter; signal travel times must be included separately when discussing what is seen.

An instantaneous turnaround is only an ideal limit. In the twin paradox with finite acceleration, the world line and the momentarily comoving frame vary continuously. The distant home-time assignment is $t_H(t)=t-v(t)x(t)/c^2$, soThe reversal at positive $x$ makes the last contribution positive. Its integrated effect replaces the ideal reassignment. By the clock hypothesis, the traveler's own elapsed proper time has no extra term depending directly on acceleration; physical disturbances of a particular clock are a separate experimental issue.

Thus relativity of simultaneity explains the consistency of the traveler's successive inertial descriptions. It is not a convention-dependent cause of the final age difference. The invariance of reunion ages under clock resynchronization makes that distinction explicit: remote time labels can change while the two reunited clock readings remain unchanged.

The phrase equivalence principle covers several related claims. The weak equivalence principle concerns the universality of free fall for test particles of negligible self-gravity, with the same initial conditions and no relevant nongravitational forces. In a Newtonian notation,Composition-independent motion requires the ratio of passive gravitational mass to inertial mass to be universal. Its common value can then be absorbed into the normalization of $\mathbf{g}$; the empirical assertion is universality, not the numerical convention used to express the ratio.

The Einstein equivalence principle includes that universality and extends the claim to local nongravitational experiments through local Lorentz invariance and local position invariance. A sufficiently small freely falling laboratory should recover the nongravitational laws of special relativity. The strong equivalence principle extends the scope to local gravitational experiments and bodies with appreciable gravitational self-energy. An experiment establishing one part of the weak statement does not automatically establish these stronger claims.

The motivating elevator comparison makes both the power and the limitation clear. Release a small object in a freely falling cabin. The cabin and object share the local free fall, so the object initially floats relative to the cabin. In a cabin supported against gravity, a released object instead accelerates relative to the floor, as it does in a suitably accelerated cabin in flat spacetime. This suggested replacing a universal gravitational force by geometric inertial motion. It does not identify a finite planetary gravitational field with an accelerated frame everywhere: separated falling objects can have a measurable relative acceleration.

In general relativity, the precise local geometric statement is that one can choose normal coordinates at an event $p$ withThis eliminates the connection contribution to the geodesic equation at that point. It does not generally set the Riemann curvature tensor to zero. In normal coordinates the first metric corrections away from $p$ are quadratic in the coordinate separation, and geodesic deviation measures the associated tidal forces. Equivalence is a local approximation for finite experiments whose extent and duration make these corrections negligible; it is not an assertion that curvature has been transformed away.^[ref]

There are direct experimental consequences beyond the elevator thought experiment. An Eötvös experiment compares bodies of different composition on a torsion balance, using a differential signal to test the universality of gravitational acceleration. MICROSCOPE performed an analogous test in orbit using electrostatic control of titanium-alloy and platinum-alloy test masses. Its first published result in 2017 constrained their differential free fall; this was already available before the examination. It tested a specific weak-equivalence prediction, not every possible gravitational law.^[ref] Comparisons of different atomic clocks can instead test local position invariance, including the expected gravitational redshift. These measurements constrain related but distinct parts of the framework.

The principle remains important as a restriction on how matter couples to gravity and as an organizing guide for experimental tests. It does not, on its own, derive the Einstein field equations. Universally coupling matter to a metric tensor still leaves choices for that metric's action; the equivalence principle does not uniquely fix field equations. For example, a metric gravitational action containing a scalar curvature squared term can retain the same minimal matter coupling while changing the gravitational dynamics. Selecting the Einstein-Hilbert action requires further dynamical assumptions and empirical success. The present role of the principle is consequently precise and limited: it supports local relativistic physics and universal matter coupling, while curvature, gravitational field equations and their quantitative predictions remain additional content.

An inertial frame is a nonrotating frame in which all force-free test bodies follow straight trajectories with constant velocity. The qualification “all” matters: a specially chosen coordinate system can make one accelerated trajectory look straight without making the frame inertial. In practice, approximately isolated test bodies and gyroscopes help identify inertial motion; identifying the relevant forces and calibrating instruments are part of testing the theory. The definition does not select a privileged state of uniform rest.

In its usual coordinate formulation, Newtonian mechanics states $m\ddot{\mathbf{x}}=\mathbf{F}$ in an inertial frame. A Galilean transformation, $t^{\prime }=t+b$, ${\mathbf{x}}^{\prime }=R\mathbf{x}+\mathbf{V}t+\mathbf{a}$ with constant orthogonal $R$ and constant $\mathbf{V}$, maps the class of inertial frames to itself. It preserves zero acceleration; a time-dependent translation generally does not. In the translating frame ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{a}(t)$, the same physical law reads $m{\ddot{\mathbf{x}}}^{\prime }=\mathbf{F}-m\ddot{\mathbf{a}}$. The extra fictitious force records the frame's acceleration, rather than a new interaction.

It is possible to state Newtonian mechanics without mentioning an inertial frame explicitly. On Galilean spacetime, let $u$ be a trajectory tangent normalized by the absolute time covector, $t_a{u}^a=1$. Thenuses an affine connection to define acceleration. The free-particle equation is ${\nabla }_{u}u=0$. This formulation removes a chosen coordinate frame but retains the geometric distinction between inertial motion and accelerated motion. Omitting both the inertial frame language and all equivalent inertial structure would leave the law unable to specify which trajectories count as unforced.

The same distinction applies to special relativity. Einstein's original presentation uses a class of inertial frames and calibrated rods, clocks and light signals.^[ref] A geometric presentation instead uses Minkowski spacetime, its Lorentzian metric and its Levi-Civita connection. A free massive particle obeys ${\nabla }_{u}u=0$, now with proper time parametrization and $g(u,u)=-c^2$. There is no preferred inertial rest frame, but there is a distinguished family of inertial trajectories and a physical Minkowski interval.

Accelerating coordinates are allowed in special relativity. They produce nonzero connection coefficients even though the Riemann curvature tensor remains zero. General relativity is needed when spacetime curvature and dynamical gravitation matter, not merely because the chosen observer accelerates. Thus neither theory requires a particular inertial frame in its coordinate-free statement; each still needs the structure that determines inertial motion.

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.

The lines of centres from either lower cylinder to the upper cylinder make $60^{\circ }$ with the horizontal. Let $N_L,N_R$ be the upward normal forces on the top cylinder from the left and right cylinders, and let $Q$ be the lower cylinders' mutual normal force. While all three contacts remain closed, the triangular configuration translates horizontally with common acceleration $a$; frictionless normals exert no torques about the cylinder centres.

For the top cylinder, Newton's second law givesThus $N_L=m(g/\sqrt{3}+a)$ and $N_R=m(g/\sqrt{3}-a)$. The right lower cylinder has horizontal equation $N_R/2+Q=ma$, soA contact may push but cannot pull: $N_L,N_R,Q\ge 0$. The limiting inequalities are thereforeAt the lower bound, $Q=0$ and the lower pair are just losing compression; at the upper bound, $N_R=0$ and the upper-right contact is just losing compression. All three pairwise normal forces are strictly positive only inside the open interval. The ground normal forces are positive throughout this range. The applied horizontal force on the left cylinder is $F=3ma$, as follows from the whole system's horizontal momentum balance.

Assume the cylinders are released from rest. In each case, “there is friction” is interpreted as sufficient static friction for rolling without slipping at the rough contacts. Without a coefficient or a no-slip assumption, sliding cases would not have a unique specified acceleration. On release the two lower cylinders separate; their mutual contact must not be retained as a rigid constraint.

Let the top cylinder's initial downward acceleration be $w$, and each lower cylinder's outward horizontal acceleration be $u$. The top cylinder initially has no horizontal acceleration. Differentiating the fixed upper-to-lower centre separation twice at the initial zero velocity givesBy reflection symmetry the top cylinder has zero angular acceleration, while the lower cylinders rotate in opposite senses.

Smooth ground, rough cylinder contacts. Write $f$ for the tangential contact force magnitude at each upper-to-lower contact. On the top cylinder, its tangential components point upward and outward, while their horizontal components and torques cancel by symmetry. For a lower cylinder, the tangential force has an inward horizontal component $\sqrt{3}f/2$ and a downward vertical component $f/2$.

The rolling condition is now at the contact between cylinders, not at the ground. At the initial zero velocity, equality of the tangential contact-point accelerations givesThus $fR=\beta m{R}^2|\alpha |$ gives $f=2\beta mw$. The lower cylinder's outward translation equation and the top cylinder's downward equation areUsing $u=\sqrt{3}w$ gives $N=2\sqrt{3}m(1+\beta )w$ and henceFor solid cylinders this is $g/11$. Here the left lower cylinder rotates clockwise while moving left, and the right rotates counterclockwise while moving right. They may slip freely on the smooth ground, so imposing the ground-rolling relation $R|\alpha |=u$ would be incorrect. No slip at the cylinder contacts requires ${\mu }_c\ge f/N=\beta /[\sqrt{3}(1+\beta )]$ initially.

These are the two cases of the initial acceleration of three stacked cylinders. As a check, as $\beta \to 0$ the rotational contribution disappears in both formulas and they approach the smooth-contact release result $w=g/7$.

Take the initial internal pressure as negligible and the external atmospheric pressure as $p_0=1.01325\times 10^5\mathrm{Pa}$. The window has frontal area $A=\pi (0.400\mathrm{m}{)}^2=0.50265{\mathrm{m}}^2$. Before appreciable air has entered, its inward pressure force is approximately $p_{0}A=5.09\times 10^4\mathrm{N}$. Newton's second law therefore gives the initial collective inward accelerationHere $M=110\mathrm{kg}$ and $g\simeq 9.81\mathrm{m}{\mathrm{s}}^{-2}$. Neglect gravity in this much larger pressure-driven acceleration, together with supports and elastic stresses once the glass has broken. The given thickness $d=0.080\mathrm{m}$ implies a plausible average glass mass density $M/(Ad)\simeq 2.74\times 10^3\mathrm{kg}{\mathrm{m}}^{-3}$. Equivalently, a fragment retaining the original thickness has acceleration $p_0/({\rho }_{g}d)$, independent of its frontal area while it experiences the full pressure difference.

For a vacuum-window implosion energy estimate, the ideal pressure-work scale for filling vessel volume $V=1.5{\mathrm{m}}^3$ is $W\simeq p_{0}V$. Assigning all that pressure-volume work to the glass's kinetic energy givesThe same result follows from retaining acceleration $a_0$ over an effective swept distance $V/A\simeq 2.98\mathrm{m}$, so that $v_{\mathrm{est}}^2=2a_{0}V/A$. This is an ideal upper energy-budget estimate for the common speed, or mass-weighted root mean square speed, of all the glass. It assumes an atmospheric reservoir maintains the driving pressure and that the whole vessel's pressure-work budget can accelerate the glass. In practice, air motion, fragmentation, heating and collisions share that energy, the pressure difference falls, and fragments may hit the vessel after a much shorter path. The unspecified vessel shape and fragment trajectories prevent a unique actual maximum speed; concentrating energy into one small fragment also prevents interpreting this collective estimate as a strict bound on every individual fragment.

Orbital energy. Treat the star as stationary to leading order in $m/M$, with Newtonian gravitational parameter $\mu =GM$. For velocity components in plane polar coordinates, $v^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$. The kinetic energy is $m{v}^2/2$. Integrating Newton's law of universal gravitation with zero gravitational potential energy at infinite separation gives $U(r)=-GMm/r$. Thuswhere $J=m{r}^2\dot{\varphi }$ is the satellite's angular momentum. The first term is radial kinetic energy; the second is transverse kinetic energy, also the centrifugal contribution to the radial effective potential; the last is the attractive gravitational potential energy. These are not three independent sources of energy: the centrifugal term is already part of the ordinary kinetic energy.

For an elliptic Kepler orbit of semi-major axis $a$ and orbital eccentricity $e$, the polar equation of a Kepler orbit $r=p/(1+e\cos \varphi )$ has semilatus rectum $p=J^2/(GM{m}^2)=a(1-e^2)$. Substituting $\dot{r}=(Je/(mp))\sin \varphi$ into the preceding expression reduces it toFor a circular Kepler orbit, $a=R$, $v_0^2=GM/R$, $K=GMm/(2R)$, and $U=-GMm/R$. Negative total energy describes a bound orbit.

Rapid, symmetric mass loss. Assume the mass is removed on a timescale short compared with the satellite's orbital period, without an impulsive kick or appreciable wind drag on the satellite. The post-loss gravitational parameter is ${\mu }^{\prime }=G(1-f)M$ with $f=0.01$. At the change, $r=R$, $\dot{r}=0$, and $v=v_0=\sqrt{GM/R}$ are unchanged. Spherical symmetry keeps the force central, so the specific angular momentum $j=J/m=\sqrt{GMR}$ is conserved. Orbital energy is not conserved across the changing gravitational field:The new semi-major axis is consequentlyThe original speed now exceeds the new circular speed at $R$, and the outward radial acceleration is $v_0^2/R-{\mu }^{\prime }/R^2=fGM/R^2>0$. Thus the release point is the new pericentre, not the apocentre. The impulsive orbital expansion under isotropic mass loss givesFor the stated one-percent loss,An independent angular momentum check is $j^2={\mu }^{\prime }{a}^{\prime }(1-e^{\prime 2})=GMR$. The impulsive orbit stays bound for $f<1/2$; loss of half the central mass is the zero-orbital energy limit.

What the wind timescale changes. The source specifies a spherically symmetric wind but does not give its duration. The preceding nearest and furthest distances require the rapid-loss assumption. In the opposite, slow limit, adiabatic orbital expansion under isotropic mass loss preserves an initially circular orbit to leading adiabatic order. Conserved specific angular momentum then gives $j^2=GM(t)r(t)$, hence $M(t)r(t)=MR$. After the same loss,For a finite intermediate timescale, the orbit depends on the mass-loss history; the percentage alone does not select a unique pair of turning distances. Neglecting wind recoil and wind drag is common to both idealizations.

Take positive $x$ horizontally toward the downhill end of the plane, with $0\le \theta <\pi /2$. The normal force is $N=mg\cos \theta$, since there is no acceleration perpendicular to the stationary plane. Along the plane, the only tangential force is $mg\sin \theta$, so Newton's second law gives $a_{\parallel }=g\sin \theta$.

Its horizontal projection is thereforeEquivalently, the normal force has horizontal component $N\sin \theta$ while gravity has none, giving $m{a}_x=N\sin \theta$. For a plane whose height increases in the positive $x$ direction, the answer instead carries a minus sign. A horizontal plane gives zero acceleration; the horizontal projection also tends to zero for a vertical plane.

If $y^{\prime \prime }=0$ over the subsequent segment, its slope is a constant $s$ and the equation of motion reduces toThere is uniform horizontal acceleration, not generally zero acceleration. Starting that segment at $(t_0,x_0,{\dot{x}}_0)$,For a downhill straight track with $s=-\tan \theta$, this gives $\ddot{x}=g\sin \theta \cos \theta$, agreeing with b. Only $s=0$ gives constant horizontal velocity.

The condition $y^{\prime \prime }\to 0$ must describe an actually straight limiting segment, with bounded velocity, for that subsequent constant-acceleration conclusion. If $y^{\prime \prime }$ vanishes only at an isolated inflection point, the boxed expression describes the instantaneous acceleration there; it does not make the rest of the track straight. Likewise, a vanishing $y^{\prime \prime }$ alone does not make the product $y^{\prime \prime }{\dot{x}}^2$ negligible in a limit where $\dot{x}$ diverges.

Use the inertial frame in which the departure point and destination are stationary, with separation $d$. The astronauts' elapsed proper time is $\tau =t/\gamma$, whereas $d=vt$. HenceThe Earth-frame travel time is $t=\sqrt{41}\mathrm{yr}\simeq 6.40\mathrm{yr}$ and the speed is approximately $0.625c$. The spacecraft's relativistic kinetic energy is thereforeThe five years are not the time to use in $v=d/t$ in the departure frame. This is the cruising kinetic energy, with departure and arrival acceleration intervals omitted, rather than a complete propulsion energy budget.

During $dt$, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the expelled positive mass $-dM$ has velocity $v-u$ to first order. Subtract the initial momentum $Mv$ from their final momentum to obtain $Mdv+udM$. Equating this to the external impulse gives the rocket equationWith $F=-Mg-kMv$ and $\dot{M}/M=-b$, this reduces to $\dot{v}+kv=ub-g$. Therefore, for initial velocity $v_0$ and $k>0$,For a launch from rest,The inequality makes the initial upward acceleration positive. Equality gives no upward launch from rest. In the zero-drag limit, $v=(ub-g)t$. These expressions apply while the prescribed burn law and constant-$g$ approximation remain valid.

The escaping sand initially has the bucket's velocity, so it produces no relative-velocity thrust. This distinction is essential in a leaking Atwood machine. With upward velocity $v$ for the leaky side, the two bucket equations areTheir sum gives $\dot{v}=g(M_0-M)/(M+M_0)$. Since $\dot{M}=-\alpha$, it follows thatIntegrate from $M_0$, where the velocity is zero:At emptying, $M=(1-ϵ)M_0$, givingThis is upward and positive for $0<ϵ<1$. For no initial sand, $ϵ=0$, the emptying time is zero and the speed is zero. As the dry bucket mass tends to zero,This finite limit refers to the idealized model; the endpoint itself would describe a massless empty bucket. For small $ϵ$, $v_{\mathrm{empty}}\sim g{M}_0{ϵ}^2/(4\alpha )$, consistent with initially zero acceleration that grows as mass is lost.

A central force exerts zero torque about its centre, so angular momentum is conserved and the motion is planar. In plane polar coordinates, $J=m{r}^2\dot{\theta }$ andUse $F=-U^{\prime }$ and either the radial Euler-Lagrange equation or differentiation of the energy away from turning points to getThe radial acceleration is $\ddot{r}-r{\dot{\theta }}^2$, so this is exactly $m(\ddot{r}-r{\dot{\theta }}^2)=F(r)$, not an additional physical outward force in an inertial frame. The variational derivation remains valid at turning points where dividing an energy derivative by $\dot{r}$ would be illegitimate.

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 $T_s=M(g+\ddot{r})=Mm(g+r{\dot{\theta }}^2)/(M+m)>0$.

The initial angular momentum is $J=ml{v}_0/2$, and the initial total energy isWhen the hanging mass reaches the table, $r=l$. Conservation of angular momentum fixes the tabletop tangential speed there to $J/(ml)=v_0/2$. At the limiting case of just reaching, radial velocity is zero. Equating the two energies,yieldsMore generally, the boundary radial kinetic energy isso reaching requires $v_0^2\ge 4Mgl/(3m)$, with equality for just reaching and strict inequality for nonzero outward radial speed at contact. This condition is also sufficient: $U_J$ is strictly convex, and the allowed interval connecting the initial radius to $l$ has no intervening higher effective potential barrier. At the threshold the initial radial acceleration is outward and $r=l$ is the outer turning point.

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.

Write the linear drag as ${\mathbf{F}}_d=-b\mathbf{v}$, with $b>0$, and let $m$ be the projectile's mass. Taking upward as positive, Newton's second law givesThe downward terminal velocity is $-v_t$, whereEquivalently, if the drag acceleration is specified as $-k\mathbf{v}$, then $v_t=g/k$. Solving the first-order equation,The highest point occurs at $v_y=0$, soFor an upward launch this root is positive and ${\dot{v}}_y=-g$ there, confirming a maximum height. As $b\to 0$, expanding the logarithm gives the drag-free result $t_{\mathrm{top}}\to u_0\sin \alpha /g$. A horizontal launch has its highest point at launch; a downward launch from the ground does not have a later airborne maximum.

Axis theorems. The parallel axis theorem stateswhere $I_G$ is the moment of inertia about an axis through the centre of mass, $I_O$ is about a parallel axis, $m$ is the body's total mass, and $d$ is the perpendicular separation of the axes. Expanding $\int [(x-d{)}^2+y^2]dm$ proves the result because $\int xdm=0$ at the centre of mass. The perpendicular axis theorem applies to a planar lamina in the $xy$ plane:All three axes meet at the same point, $x,y$ lie in the lamina and $z$ is normal to it; the identity follows from $x^2+y^2$ being the squared distance to the $z$ axis. It is not a theorem for an arbitrary three-dimensional body.

Cylinder integrals. Put the cylinder's midpoint at $z=0$ and its symmetry axis along $z$. Its uniform mass density is $\rho =m/(\pi r^{2}l)$. The transverse moment of inertia of a uniform solid cylinder about the $x$ axis isEquivalently each thin circular slice has diametral moment of inertia $r^{2}dm/4$, by the perpendicular axis theorem and symmetry, and shifting it by $z$ adds $z^{2}dm$. The parallel axis theorem then gives the transverse end-axis moment of inertia:The axial moment of inertia of a uniform solid cylinder is $I_{\parallel }=m{r}^2/2$. Equating it to $I_{\perp ,G}$ givesHere the source's principal axis means the cylinder's symmetry axis; its transverse centre axes are also principal axes. This equality describes a short cylinder rather than the thin-rod limit.

Hammer dynamics and forces. Treat the compact head as a point mass and neglect its intrinsic moment of inertia. Let $\theta$ be the displacement from downward vertical, and let ${\mathbf{e}}_r$ point along the rod from pivot to head, with ${\mathbf{e}}_{\theta }$ in the direction of increasing $\theta$. DefineThe two external weights $Mg$ and $mg$ act at distances $l$ and $l/2$ from the pivot. The frictionless pivot supplies a reaction $\mathbf{R}$ but no torque about its axis. The total restoring torque and rotational equation of motion areFor completeness, Newton's second law for the centre of mass gives both components of the pivot force:The negative radial component points towards the pivot. These follow from the centre of mass acceleration $-d{\dot{\theta }}^2{\mathbf{e}}_r+d\ddot{\theta }{\mathbf{e}}_{\theta }$ and the total weight $m_{T}g(\cos \theta {\mathbf{e}}_r-\sin \theta {\mathbf{e}}_{\theta })$. They specify the external forces even when the oscillation is not small.

With the small-angle approximation, $\sin \theta \simeq \theta$, this is simple harmonic motion. The physical pendulum has small-amplitude oscillation periodFor release from a finite amplitude $0<{\theta }_0<\pi$, conservation of mechanical energy instead gives $\frac{1}{2}I{\dot{\theta }}^2=gA(\cos \theta -\cos {\theta }_0)$. Integrating a quarter-cycle gives the exact oscillation periodwhere $K$ is the complete elliptic integral of the first kind; $K(0)=\pi /2$ recovers $T_0$.

For $m=M/10$ and $r=l/20$, compare with the simple pendulum value $T_s=2\pi \sqrt{l/g}$:For small oscillations, the hammer's oscillation period is about $0.794\%$ shorter. The extra rod weight adds restoring torque proportionately faster than it adds pivot moment of inertia.