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Write the position of a mass element relative to the centre of mass as $r$. Expand its squared velocity and integrate:The mixed term vanishes because $\int rdm=0$. Alsousing the scalar triple-product identity. ThereforeIn principal axes the inertia tensor is diagonal, so the second term is $\frac{1}{2}(A{\omega }_1^2+B{\omega }_2^2+C{\omega }_3^2)$. Substituting the three Euler-angle components given in the source gives its displayed kinetic-energy expression directly.

For $A=B$, the first two component squares have opposite mixed terms and sum to ${\dot{\theta }}^2+{\dot{\varphi }}^2{\sin }^2\theta$. Put $n=\dot{\psi }+\dot{\varphi }\cos \theta$. Using the orientation-dependent potential printed in the source therefore gives the formal modelThe cyclic coordinates $x,y,\psi$ have conserved momenta $M\dot{x},M\dot{y},Cn$. Consequently $\dot{x},\dot{y},n$ are constant. The coordinate $\varphi$ is also cyclic, withFor imposed fixed tilt $\theta =\alpha \in (0,\pi /2)$, conservation of $n$ and $p_{\varphi }$ forces $\dot{\varphi }=\Omega$ constant. Impose the fixed-height condition $z=h$ in this formal constrained model. Its $\theta$ equation isSet the tilt constant and divide by its nonzero sine. This givesA real polynomial root requires its quadratic discriminant to be nonnegative:When this condition holds the two algebraic rates are $(Cn\pm \sqrt{C^2{n}^2-4AMgh\cos \alpha })/(2A\cos \alpha )$.

There is a physical inconsistency in the printed data that must be kept distinct from this algebra. If $(x,y,z)$ are literally Cartesian coordinates of the centre of mass in uniform gravity, the gravitational potential is $Mgz$, not $Mgz\cos \theta$. The printed Lagrangian consequently does not follow from that Cartesian-height interpretation. Moreover its unconstrained $z$ equation is $\ddot{z}=-g\cos \theta$, so constant height at the indicated tilt requires a constraint reaction. The requested rate and discriminant follow from the printed orientation-dependent potential with fixed height imposed; a standard heavy-top interpretation instead uses a height proportional to an axis distance times $\cos \theta$, with the reference point and corresponding inertia specified consistently. The kinetic-energy derivation is independent of this source inconsistency.

Use coordinates with the downward vertical as reference. The first rod's center is at $(a/2)(\sin \theta ,-\cos \theta )$, so its translational plus rotational kinetic energy is $m{a}^2{\dot{\theta }}^2/6$. The second center isIts squared speed is $a^2{\dot{\theta }}^2+a^2{\dot{\varphi }}^2/4+a^2\dot{\theta }\dot{\varphi }\cos (\theta -\varphi )$. Adding its center-of-mass rotation, using moment of inertia $m{a}^2/12$, gives the totalThe two center heights give $V=-mga(3\cos \theta +\cos \varphi )/2$. Hence $L=T-V$ is exactly the stated Lagrangian.

Small stable oscillations are about the downward equilibrium $\theta =\varphi =0$. Retaining quadratic terms gives $T_2=\frac{1}{2}{\dot{q}}^{T}M\dot{q}$ and $V_2=\frac{1}{2}{q}^{T}Kq$, whereA normal mode $q=v{e}^{i\omega t}$ satisfies $(K-{\omega }^{2}M)v=0$. With $\kappa =a{\omega }^2/g$, the determinant equation isThus the angular frequencies areBoth squared frequencies are positive. The amplitude ratio is $\varphi /\theta =(9-8\kappa )/(3\kappa )$: it is positive in the lower-frequency mode and negative in the higher-frequency mode. The rods therefore swing in phase in the first mode and in opposite phase in the second.

Write $q=(x,X{)}^T$, $M=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$ and $K=\left(\begin{array}{cc}\alpha & \beta \\ \beta & \gamma \end{array}\right)$. The linear equations are $M\ddot{q}+Kq=0$. A normal mode has all coordinates oscillating with one frequency and fixed relative amplitudes, $q=v\cos (\omega t+\phi )$. Its squared frequency solvesand its amplitude vector is a nonzero null vector of $K-{\omega }^{2}M$. The assumed distinct positive roots give two normal frequencies, their positive square roots.

For the rods let ${\theta }_1,{\theta }_2$ be their small inclinations from the downward vertical. To first order, ${\theta }_1=x/a$ and ${\theta }_2=(X-x)/(2a)$. The first rod has kinetic energy $m{\dot{x}}^2/6$. The second has centre-of-mass horizontal velocity $(\dot{x}+\dot{X})/2$ and angular velocity $(\dot{X}-\dot{x})/(2a)$, givingTogether these give $T=m(3{\dot{x}}^2+2\dot{x}\dot{X}+2{\dot{X}}^2)/6$. The centre-of-mass height changes giveproving the displayed quadratic Lagrangian.

Here $M=(m/3)\left(\begin{array}{cc}3& 1\\ 1& 2\end{array}\right)$ and $K=(mg/(2a))\left(\begin{array}{cc}6& -1\\ -1& 1\end{array}\right)$. Set $\lambda =2a{\omega }^2/(3g)$. The determinant equation reduces to $5{\lambda }^2-17\lambda +5=0$, soThe first row of the mode equation gives $X/x=(6-3\lambda )/(1+\lambda )$. The actual rod-angle ratio is thereforeIt is positive for the smaller root and negative for the larger root. Thus the low-frequency mode has the rods in phase, and the high-frequency mode has them out of phase. Using the rod-angle ratio is necessary because $X$ is the displacement of the end of the second rod, not that rod's inclination.

The autonomous Lagrangian has the Beltrami identityHere $L$ is homogeneous of degree two in the velocities, so $\dot{u}{L}_{\dot{u}}+\dot{v}{L}_{\dot{v}}=2L$. Thus $L$ itself is constant, say $C\ge 0$.

For $C>0$, put $\ell =\sqrt{L}$. Its Euler-Lagrange expression along the curve isThe first term vanishes by stationarity for $L$, and the second by constancy of $L$. Thus the same path is stationary for its length functional $\int \sqrt{L}dt$.

On a regular segment described as a graph $v=v(u)$, reparametrize by $u$. Since $\dot{v}=v_u\dot{u}$,Choose the orientation with increasing $u$; reversing it does not change stationarity. Invariance of length under regular reparametrization therefore givesThis is the principle of length extremals from a constant quadratic energy. The stated nonnegativity also permits the degenerate case $C=0$. Then the length is zero, its absolute minimum since the integrand is nonnegative, so it is an extremal in the minimizing sense. The differentiated square-root Euler-Lagrange equation above is only used where $L>0$; it would not be legitimate to divide by $\sqrt{L}$ on a null curve. For a genuine surface metric and a regular nonconstant path, $C>0$ automatically.

For the surface of revolution $r(\eta ,\zeta )=(\eta ,f(\eta )\cos \zeta ,f(\eta )\sin \zeta )$, the coordinate tangent vectors have squared lengths $1+f^{\prime 2}$ and $f^2$, and their dot product is zero. Hence its first fundamental form isUse the energy Lagrangian $L=A{\dot{\eta }}^2+f^2{\dot{\zeta }}^2$. Its Euler-Lagrange equations areFor a meridian $\zeta ={\zeta }_0$, choose arc length as parameter, so $\dot{\eta }=1/\sqrt{A}$. Differentiating gives $2A\ddot{\eta }+A^{\prime }{\dot{\eta }}^2=0$, while the second equation is automatic. Part (b) makes this path an extremal of length. Thus every meridian is a geodesic.

For a nonconstant parallel $\eta ={\eta }_0$, choose constant nonzero $\dot{\zeta }$. The second equation holds, and the first reduces to $f({\eta }_0)f^{\prime }({\eta }_0){\dot{\zeta }}^2=0$. Since $f>0$, this is equivalent toConversely that condition makes both equations hold at constant speed, hence supplies a length extremal. Geometrically, a parallel is geodesic exactly at a stationary radius of the surface.

Use the usual Feynman rule convention in which the diagonal quadratic terms of the free Lagrangian, including the physical W boson and Z boson masses, are incorporated into their propagators. Then there areA line joining two ends is a propagator, not an interaction vertex. Gauge boson self-interactions begin with cubic terms in the Yang-Mills theory Lagrangian. The question's $g,W,Z,\gamma$ name four families: they include eight gluon colour components and the two charged W bosons $W^{\pm }$.

There is a convention to distinguish here: if a calculation expands around massless free fields, quadratic $m_W^2{W}_{\mu }^{+}{W}^{-\mu }$ and $m_Z^2{Z}_{\mu }{Z}^{\mu }/2$ terms may be drawn as two-leg mass insertions. Those two insertion types belong to that bookkeeping convention. Similarly, two-point self-energy corrections occur in loop diagrams. Neither is an additional fundamental interaction in the standard physical-mass, tree-level rules used for the count above.

There are three vertex types, with all legs labeled in the following original drawings:

The gluon cubic interaction vertex follows from the non-Abelian gauge field strength of Quantum chromodynamics. The electroweak Yang-Mills theory Lagrangian, after expressing the fields in the physical charged and neutral basis, produces the two charged-W boson vertices. In particular there are no fundamental $\gamma \gamma \gamma$, $ZZZ$, $ZZ\gamma$, or $Z\gamma \gamma$ vertices, and the separate colour and electroweak gauge factors do not generate a mixed $gg\gamma$ or $ggZ$ vertex. The count is by the boson labels requested, not by separate assignments of gluon colour indices.

There areFor a Yang-Mills theory, schematically $F=\partial A+gAA$. The gauge kinetic Lagrangian is quadratic in $F$, so its terms are at most quartic in the gauge fields:Thus there is no elementary five-leg pure-gauge interaction vertex in the renormalizable Standard Model. A five-boson scattering amplitude can still be assembled from several lower-valence vertices, or generated by loop diagrams. The number of external legs of a complete Feynman diagram is not the valence of a fundamental vertex.

The Lagrangian has no explicit time dependence, so its energy function is constant. With total speed $v^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$, division by $M_p$ givesConsequently, between the initial point $(r_i,v_i)$ and any other point,This energy relation holds for nonradial motion too. Radially outward motion is the stipulated approximation for the subsequent probe calculation; angular momentum is then zero.

A cyclic coordinate $q_j$ is absent explicitly from the Lagrangian: $\partial L/\partial q_j=0$, even though ${\dot{q}}_j$ may occur. The Euler-Lagrange equation then givesThus a continuous coordinate-translation symmetry gives a conserved conjugate momentum. The conserved quantity need not be ordinary linear momentum: an absent azimuthal angle can give conserved angular momentum. The derivative is taken holding the other coordinates and velocities fixed.

Define the energy function $\mathcal{E}={\sum }_i{\dot{q}}_i{p}_i-L$. Along solutions of the Euler-Lagrange equations,ConsequentlyFor an ordinary isolated mechanical system with $L=T-U$, a time-independent quadratic kinetic energy and a velocity-independent potential energy, ${\sum }_i{\dot{q}}_i{p}_i=2T$, so $\mathcal{E}=T+U$. For a more general Lagrangian, the conserved quantity is the displayed energy function; identifying it with $T+U$ requires these assumptions.

Use the angle $\theta$ from the vertical as the generalized coordinate. For a uniform ladder, its centre of mass is $(l\sin \theta ,l\cos \theta )$, so $v_G^2=l^2{\dot{\theta }}^2$. Its central moment of inertia is $I_G=m(2l{)}^2/12=m{l}^2/3$. With ideal contact constraints that do no work, the Lagrangian, conjugate momentum, and Euler-Lagrange equation areRelease from rest at ${\theta }_0$ also gives ${\dot{\theta }}^2=3g(\cos {\theta }_0-\cos \theta )/(2l)$ by conservation of energy. These results hold while both contacts persist. The ideal wall reaction is normal to the wall; an unspecified tangential wall friction would require an additional generalized force and would prevent determination from the supplied data alone.

Measure the vertical displacement $x$ downward from the spring's unstretched position. The kinetic energy is $m{\dot{x}}^2/2$, while the spring and gravitational potential energies are $k{x}^2/2$ and $-mgx$. Thus the Lagrangian isApplying the Euler-Lagrange equation, $d(\partial \mathcal{L}/\partial \dot{x})/dt-\partial \mathcal{L}/\partial x=0$, givesGravity shifts the equilibrium to $x_{\ast }=mg/k$. For displacement $\xi =x-x_{\ast }$ from that equilibrium, the equation of motion becomes $\ddot{\xi }+(k/m)\xi =0$, giving simple harmonic motion with angular frequency $\omega =\sqrt{k/m}$. Choosing an upward coordinate instead reverses the gravitational term, but leaves the equilibrium-relative angular frequency unchanged.

For the general Lagrangian just obtained, the Euler-Lagrange equation givesUsing the small-angle approximation $\sin \theta \simeq \theta$, this is simple harmonic motion with ${\omega }_0^2=A/I$. ThusThe endpoint mass cancels in the light-rod limit. A concrete comparison shows why the finite rod cannot silently be included: the same specified endpoint $m$ with $M_r=m$ gives $T=2\pi \sqrt{8l/(9g)}$, distinct from the negligible-rod result. This is a physical pendulum when the rod contributes, and a simple pendulum in the light-rod limit. Both expressions require small oscillation amplitude; the full sine equation remains nonlinear.

Allowing azimuthal motion makes the endpoint speed satisfy $v^2=l^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\phi }}^2)$. The light-rod spherical pendulum therefore hasThe azimuth $\phi$ is a cyclic coordinate, since the Lagrangian is independent of it. Its conjugate canonical momentum is the conserved vertical angular momentum:For a finite uniform rod, replace $m{l}^2$ by $I$ and $mgl$ by $A$ from i. The same azimuthal symmetry then gives $p_{\phi }=I{\sin }^2\theta \dot{\phi }$.

The Hamiltonian is the Legendre transform of this Lagrangian. With $p_{\theta }=I\dot{\theta }$,There is no explicit time dependence, so $dH/dt=\partial H/\partial t=0$. Here $H$ also equals the total mechanical energy: the constraint is stationary, the kinetic energy is quadratic in generalized velocities, and gravity is conservative. The apparent singularity of the momentum expression at $\sin \theta =0$ is a spherical-coordinate singularity; trajectories reaching a pole must be described consistently, and nonzero $J_z$ prevents reaching it at finite energy.

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).

A central force confines nonradial motion to a plane. In plane polar coordinates, the Lagrangian for $V(r)=\beta r^k$ isThe azimuth $\varphi$ is a cyclic coordinate, so $L=m{r}^2\dot{\varphi }$ is conserved angular momentum. The Lagrangian has no explicit time dependence, giving conserved energyAt fixed $L$, the radial equation of motion is $m\ddot{r}=-V_{\mathrm{eff}}^{\prime }(r)$. The effective potential packages the transverse kinetic energy as a centrifugal term alongside the actual potential energy. It gives a one-dimensional radial problem without introducing an extra physical interaction.

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 parallel axis theorem gives $I=I_{\mathrm{CM}}+M{d}^2$ for parallel axes separated by $d$, with the reference axis through the centre of mass. For a plane lamina, the perpendicular axis theorem gives $I_z=I_x+I_y$ for mutually perpendicular axes meeting at one point, with $x,y$ in the plane.

Treat the removed material as negative mass. With the hole's centre at $b\hat{\mathbf{x}}$, its moment of inertia about $O$ is $m(a^2/2+b^2)$. The remaining moment of inertia and centre of mass areFor a uniform original disk, $m/M=a^2/R^2$. Write $d_1=mb/(M-m)$ and measure $\varphi$ from the downward equilibrium direction of its centre of mass. The physical pendulum then has LagrangianThe Euler-Lagrange equation is $I_1\ddot{\varphi }+mbg\sin \varphi =0$. For small oscillations, $\sin \varphi \simeq \varphi$, givingNow add the removed small disk on the opposite side, with its centre a distance $c=R-a$ from $O$ so that it is internally tangent to the large disk. The total mass returns to $M$, while its first moment about $O$ is $-m(b+c)\hat{\mathbf{x}}$. ThusReplacing both the inertia and gravitational restoring moment givesThe same first-moment method applies to a general perforated-disc pendulum. The added disk is attached to the lamina and shares its rotation; treating it as a massless patch or updating only the inertia would omit its restoring torque.

Let $x$ be the downward displacement of $M$ from its initial position just beyond the table edge. The smaller mass moves towards the edge by the same distance, and the hanging chain length is $x$. All the chain and both endpoint masses move at speed $\dot{x}$, so their total moving mass is $\mathcal{M}=M+m+\rho L$. The hanging chain's gravitational potential energy, relative to the initial configuration, is $-\rho g{x}^2/2$. Thus the Lagrangian for the chain sliding over a table edge isThe Euler-Lagrange equation gives $\mathcal{M}\ddot{x}=g(M+\rho x)$. Put $\Omega =\sqrt{\rho g/\mathcal{M}}$. With $x(0)=\dot{x}(0)=0$, the solution isThese are also the smaller mass's displacement and speed towards the edge. The formula applies until $M$ reaches the floor, at $t_f={\Omega }^{-1}\mathrm{arcosh}(1+\rho y/M)$. Since $y<L$, the smaller mass has not yet reached the edge during this interval. As $\rho \to 0$, the result tends to $x=Mg{t}^2/[2(M+m)]$, as expected for a massless connector.

Let the spring's spring constant be $k$, and write its extension as $e_0+x$, allowing for an initial extension $e_0$. The new Lagrangian isThe specified equilibrium is $x_e=y-y_1$, with $k(e_0+x_e)=g(M+\rho x_e)$. Put $\xi =x-x_e$; the equation becomesThe spring-stabilized chain over a table edge oscillates only when $k>\rho g$. More hanging chain increases its own gravitational pull, reducing the effective restoring stiffness. The equilibrium height alone does not determine $k$ without information about the spring's natural length or preload.