The planar pendulum as an ideal holonomic model: the multiplier, the tension and the vanishing reaction power of a scleronomic constraint

The datum is regular and scleronomic and its virtual displacements are the tangent line: $D_{x}G(t,x,y,z)=\left(\begin{array}{ccc}2x& 2y& 0\\ 0& 0& 1\end{array}\right)$ has rank $2$ at every point of $O$, the first row being nonzero because $(x,y)\mathrm{\ne }(0,0)$ there and the two rows being orthogonal; the locus is $\mathrm{\Sigma }=\mathbb{R}\times N$ with $N=\{(x,y,z):x^2+y^2=l^2,z=0\}$, so it is time-independent and the datum is scleronomic by [F2] (and then $G_t=0$ on $\mathrm{\Sigma }$); and $\delta \in \ker D_{x}G$ means $2x{\delta }_x+2y{\delta }_y=0$ and ${\delta }_z=0$, which at $X(\theta )=l(\sin \theta ,-\cos \theta ,0)$ is the line orthogonal to $u(\theta )$, that is, $\mathbb{R}{X}_{\theta }(\theta )$. Moreover $G(X(\theta ))=(l^2{\sin }^2\theta +l^2{\cos }^2\theta -l^2,0)=(0,0)$ and $D_{x}G(X(\theta ))X_{\theta }(\theta )=(2l^2(\sin \theta \cos \theta -\cos \theta \sin \theta ),0)=(0,0)$, so $X_{\theta }(\theta )$ is permitted for every $\theta$.

Every constrained $C^2$ motion admits a $C^2$ angular presentation on its interval $J$. Write its first two coordinates as $(a,b)$, so $a^2+b^2=l^2$, and set $\omega =(a{b}^{\mathrm{\prime }}-b{a}^{\mathrm{\prime }})\mathrm{/}{l}^2$, a $C^1$ function. Differentiating the circle identity gives $(a^{\mathrm{\prime }},b^{\mathrm{\prime }})=\omega (-b,a)$. Choose $t_{\ast }\in J$ and ${\theta }_{\ast }\in \mathbb{R}$ with $(a(t_{\ast }),b(t_{\ast }))=l(\sin {\theta }_{\ast },-\cos {\theta }_{\ast })$ by [F7], and put $\theta (t)={\theta }_{\ast }+{\int }_{t_{\ast }}^t\omega (s)ds$. The oriented FTC gives ${\theta }^{\mathrm{\prime }}=\omega$ and $\theta \in C^2$, with domain-relative endpoint derivatives where present. Both $(a,b)$ and $l(\sin \theta ,-\cos \theta )$ satisfy $w^{\mathrm{\prime }}=\omega (-w_2,w_1)$. Their difference has squared norm with derivative zero and value zero at $t_{\ast }$, so [F6] makes it zero on the interval. The third coordinate is zero by the constraint. Thus the motion is $X\circ \theta$; the following calculations apply to every constrained motion.

The multiplier of the balance theorem on this datum: with $\mathsf{M}=m{I}_3$ one has $C=D_{x}G{\mathsf{M}}^{-1}{D}_x{G}^{\mathsf{T}}=\frac{1}{m}\mathrm{diag}((2x,2y,0)\cdot (2x,2y,0),1)=\frac{1}{m}\mathrm{diag}(4(x^2+y^2),1)=\frac{1}{m}\mathrm{diag}(4l^2,1)$ on the locus, $D_{x}G{\mathsf{M}}^{-1}F=\frac{1}{m}(-2mgy,0)=(-2gy,0)$, $G_{tt}=0$, $G_{tq}=0$ and $G_{qq}[v,v]=(2(v_x^2+v_y^2),0)$; along $x(t)=X(\theta (t))$ the velocity is $\dot{x}=l{\theta }^{\mathrm{\prime }}(\cos \theta ,\sin \theta ,0)$, so $v_x^2+v_y^2=l^2{\theta }^{\mathrm{\prime }2}$, and hence $\mathrm{\Lambda }=-C^{-1}[(-2gy,0)+(2l^2{\theta }^{\mathrm{\prime }2},0)]=\left(\frac{m}{4l^2}\right(2gy-2l^2{\theta }^{\mathrm{\prime }2}),0)=(-\frac{m}{2l}(g\cos \theta +l{\theta }^{\mathrm{\prime }2}),0)$ at $y=-l\cos \theta$; this is the displayed multiplier.

Conversely, a solution of the pendulum equation gives an ideal constrained motion: let $\theta :J\to \mathbb{R}$ be $C^2$ with $m{l}^2{\theta }^{\mathrm{\prime }\mathrm{\prime }}+mgl\sin \theta =0$ and put $x:=X(\theta )$, with the model data and potential of [F1]; then $x$ is $C^2$ and constraint-preserving by step 1.1, and $\dot{x}=l{\theta }^{\mathrm{\prime }}(\cos \theta ,\sin \theta ,0)$, $\ddot{x}=l{\theta }^{\mathrm{\prime }\mathrm{\prime }}(\cos \theta ,\sin \theta ,0)+l{\theta }^{\mathrm{\prime }2}(-\sin \theta ,\cos \theta ,0)$. Set $\lambda :=(-\frac{m}{2l}(g\cos \theta +l{\theta }^{\mathrm{\prime }2}),0)$ and $R:=D_{x}G(x{)}^{\mathsf{T}}\lambda =(2x{\lambda }_1,2y{\lambda }_1,{\lambda }_2)=-m(g\cos \theta +l{\theta }^{\mathrm{\prime }2})u(\theta )$; the triple $(\cos \theta ,\sin \theta ,0)$, $u(\theta )$, ${\mathbf{e}}_3$ is an orthonormal basis of ${\mathbb{R}}^3$, and comparing components of $m\ddot{x}$ and $F+R$ in it: the tangent components are $ml{\theta }^{\mathrm{\prime }\mathrm{\prime }}$ and $-mg\sin \theta$, equal by the equation; the $u$-components are $-ml{\theta }^{\mathrm{\prime }2}$ and $mg\cos \theta -T=-ml{\theta }^{\mathrm{\prime }2}$ with $T=m(g\cos \theta +l{\theta }^{\mathrm{\prime }2})$; and the ${\mathbf{e}}_3$-components are $0$ and ${\lambda }_2=0$. Hence $\mathsf{M}\ddot{x}=F+R$ with $R=D_x{G}^{\mathsf{T}}\lambda$ annihilating the kernel, so $x$ is a constrained motion with ideal reaction and multiplier $\lambda$.

The tension and the reduced equation along a motion presented as $x=X(\theta )$: projecting $\mathsf{M}{x}^{\mathrm{\prime }\mathrm{\prime }}=F+R$ with $R=D_x{G}^{\mathsf{T}}\lambda$ of step 1.3 onto the tangent direction $(\cos \theta ,\sin \theta ,0)$ gives $ml{\theta }^{\mathrm{\prime }\mathrm{\prime }}=-mg\sin \theta$, that is $m{l}^2{\theta }^{\mathrm{\prime }\mathrm{\prime }}+mgl\sin \theta =0$; projecting onto $u(\theta )$ gives $-ml{\theta }^{\mathrm{\prime }2}=mg\cos \theta -T$ with $T:=-R\cdot u$, that is $T=m(g\cos \theta +l{\theta }^{\mathrm{\prime }2})$, the displayed tension; and projecting onto ${\mathbf{e}}_3$ gives $0={\lambda }_2$, so the plane constraint $z=0$ exerts no normal force on any constrained motion, whose motion and applied force already lie in the plane. The reaction on the bead is therefore $R=-Tu(\theta )$, a force along the radial line toward the origin when $T>0$. A load on any support requires additional assumptions about the support and force transmission; the one-particle equation determines only $R$.

Reaction power and energy: by step 2.2, $R=-Tu$ with $T\in \mathbb{R}$, and $x^{\mathrm{\prime }}=l{\theta }^{\mathrm{\prime }}(\cos \theta ,\sin \theta ,0)$ is orthogonal to $u$, so $R\cdot x^{\mathrm{\prime }}=0$ identically, in agreement with the scleronomic case of the power identity $R\cdot x^{\mathrm{\prime }}=-\lambda G_t$ of [F3] because $G_t=0$; with $U=mgy=-mgl\cos \theta$ static and no residual forces, [F4] gives $E=\frac{1}{2}m{l}^2{\theta }^{\mathrm{\prime }2}-mgl\cos \theta$ constant along the motion; independently, $\frac{d}{dt}E=m{l}^2{\theta }^{\mathrm{\prime }}{\theta }^{\mathrm{\prime }\mathrm{\prime }}+mgl\sin \theta {\theta }^{\mathrm{\prime }}=l{\theta }^{\mathrm{\prime }}(ml{\theta }^{\mathrm{\prime }\mathrm{\prime }}+mg\sin \theta )=0$ by the equation of step 2.2.

Energy expression of the tension and finite model data: substituting $l{\theta }^{\mathrm{\prime }2}=2E\mathrm{/}(ml)+2g\cos \theta$ (equivalent to the definition of $E$) into $T=m(g\cos \theta +l{\theta }^{\mathrm{\prime }2})$ gives $T=3mg\cos \theta +2E\mathrm{/}l$; with $E=\frac{1}{2}m{l}^2{\theta }_0^{\mathrm{\prime }2}-mgl\cos {\theta }_0$ this is $T=mg(3\cos \theta -2\cos {\theta }_0)+ml{\theta }_0^{\mathrm{\prime }2}$. For $m=1$, $l=1$, $g=10$, ${\theta }_0=\pi \mathrm{/}3$ and ${\theta }_0^{\mathrm{\prime }}=0$ at $t_0=0$, one has $E=-mgl\cos {\theta }_0=-5\mathrm{J}$ and $T(\theta )=30\cos \theta -10\mathrm{N}$, so $T({\theta }_0)=5\mathrm{N}$ and $T(0)=20\mathrm{N}$. Along this solution, conservation of $E$ from step 3.1 gives ${\theta }^{\mathrm{\prime }}(t{)}^2=(20\cos \theta (t)-10){\mathrm{s}}^{-2}\ge 0$, hence $\cos \theta (t)\ge \frac{1}{2}$ and $\theta (t)\in {\bigcup }_{k\in \mathbb{Z}}[2\pi k-\pi \mathrm{/}3,2\pi k+\pi \mathrm{/}3]$ for every $t$ in its interval $J\ni 0$. Since $J$ is connected and $\theta$ is continuous, $\theta (J)$ is connected; it contains $\theta (0)=\pi \mathrm{/}3$, so it lies in the component $[-\pi \mathrm{/}3,\pi \mathrm{/}3]$. Therefore $T\ge 5\mathrm{N}>0$ throughout this motion. The reaction power is zero at every point of the motion.

Clauses 1-5 of the Example are proved: the datum is a regular scleronomic holonomic constraint realizing the circle $x^2+y^2=l^2$ in the plane $z=0$, the multiplier has the displayed two components with the second vanishing along every constrained motion, the reaction is the displayed tension, the pendulum equation is the tangential projection of the constrained equation, the ideal reaction performs no work, the mechanical energy is conserved, and the finite data give a positive tension on the reachable arc.