Exercises — Derivation of Euler-Lagrange equations from D'Alembert's principle

Pick the pieces. Only one coordinate, $y$. The velocity is $\dot{y}$, so the kinetic energy is $T=\frac{1}{2}m{\dot{y}}^2.$ Gravity pulls down; potential energy grows with height: $V=mgy\Rightarrow L=T-V=\frac{1}{2}m{\dot{y}}^2-mgy.$ Turn the crank. The two ingredients of the Euler-Lagrange machine: $\frac{\partial L}{\partial \dot{y}}=m\dot{y}\Rightarrow \frac{d}{dt}\left(m\dot{y}\right)=m\ddot{y},\frac{\partial L}{\partial y}=-mg.$ Subtract: $m\ddot{y}-(-mg)=0\Rightarrow m\ddot{y}+mg=0\Rightarrow \boxed{\ddot{y}=-g}.$ The acceleration is $g$ downward — exactly Newton. ✓

$T=\frac{1}{2}m{\dot{x}}^2,V=\frac{1}{2}k{x}^2,L=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{x}^2.$ $\frac{\partial L}{\partial \dot{x}}=m\dot{x}\Rightarrow \frac{d}{dt}(m\dot{x})=m\ddot{x},\frac{\partial L}{\partial x}=-kx.$ Euler-Lagrange: $m\ddot{x}+kx=0\Rightarrow \boxed{\ddot{x}=-\frac{k}{m}x}.$ Simple harmonic motion with $\omega =\sqrt{k/m}$. ✓

With $x=\ell \sin \theta ,y=-\ell \cos \theta$, the speed is $\ell \dot{\theta }$, so $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2,V=-mg\ell \cos \theta ,L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+mg\ell \cos \theta .$ $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }\Rightarrow \frac{d}{dt}=m{\ell }^2\ddot{\theta },\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta .$ EL: $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0\Rightarrow \ddot{\theta }=-\frac{g}{\ell }\sin \theta .$ Small angles: $\sin \theta \approx \theta$, so $\ddot{\theta }=-\frac{g}{\ell }\theta$. That is SHM with $\boxed{\omega =\sqrt{\frac{g}{\ell }}},T_{\text{period}}=2\pi \sqrt{\frac{\ell }{g}}.$

Position (angle from bottom): $x=R\sin \theta ,y=-R\cos \theta$ (bottom is $y=-R$). This is identical in form to the pendulum with $\ell \to R$ — the hoop provides the same rigid constraint the rod did, and the normal force points along the radius, perpendicular to the allowed (tangential) motion, so it does no virtual work and never appears. $T=\frac{1}{2}m{R}^2{\dot{\theta }}^2,V=-mgR\cos \theta .$ $\boxed{\ddot{\theta }=-\frac{g}{R}\sin \theta }.$ Same physics, new geometry. ✓

The bead's velocity has a radial part $\dot{r}$ and a tangential part $r\omega$ (from the wire spinning). These are perpendicular, so $T=\frac{1}{2}m({\dot{r}}^2+r^2{\omega }^2),V=0.$ Here $r^2{\omega }^2$ depends on $r$, so $\partial T/\partial r\ne 0$ — this is the whole point. $\frac{\partial T}{\partial \dot{r}}=m\dot{r}\Rightarrow \frac{d}{dt}=m\ddot{r},\frac{\partial T}{\partial r}=mr{\omega }^2.$ EL ($Q_r=0$): $m\ddot{r}-mr{\omega }^2=0\Rightarrow \boxed{\ddot{r}={\omega }^{2}r}$. The positive sign means the bead flies outward — the centrifugal effect emerged automatically from $\partial T/\partial r$. Solution $r(t)=A{e}^{\omega t}+B{e}^{-\omega t}$: exponential escape. ✓

If $m_1$ descends by $x$, then $m_2$ rises by $x$ (string is inextensible — the constraint). Both move at speed $\dot{x}$: $T=\frac{1}{2}(m_1+m_2){\dot{x}}^2.$ Heights: $m_1$ at $-x$, $m_2$ at $-(\ell -x)=x-\ell$. So $V=-m_{1}gx+m_{2}g(x-\ell )=(m_2-m_1)gx-m_{2}g\ell .$ $\frac{\partial L}{\partial \dot{x}}=(m_1+m_2)\dot{x}\Rightarrow \frac{d}{dt}=(m_1+m_2)\ddot{x},\frac{\partial L}{\partial x}=-(m_2-m_1)g=(m_1-m_2)g.$ EL: $(m_1+m_2)\ddot{x}-(m_1-m_2)g=0\Rightarrow \boxed{\ddot{x}=\frac{(m_1-m_2)g}{m_1+m_2}}.$ The tension never appeared — it is the internal constraint force, killed by choosing the single coordinate $x$. ✓

$T=\frac{1}{2}M{\dot{X}}^2+\frac{1}{2}m{\dot{x}}^2,V=\frac{1}{2}k(x-X{)}^2.$ $L=\frac{1}{2}M{\dot{X}}^2+\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k(x-X{)}^2.$ Equation for $X$: $\frac{\partial L}{\partial \dot{X}}=M\dot{X},\frac{\partial L}{\partial X}=-\frac{1}{2}k\cdot 2(x-X)\cdot (-1)=k(x-X).$ $\Rightarrow M\ddot{X}-k(x-X)=0\Rightarrow \boxed{M\ddot{X}=k(x-X)}.$ Equation for $x$: $\frac{\partial L}{\partial \dot{x}}=m\dot{x},\frac{\partial L}{\partial x}=-k(x-X).$ $\Rightarrow \boxed{m\ddot{x}=-k(x-X)}.$ Newton's third law is visible: the two forces are equal and opposite. ✓

Bob position (pivot at height $y_p$): $x=\ell \sin \theta ,y=y_p-\ell \cos \theta .$ Velocities: $\dot{x}=\ell \dot{\theta }\cos \theta$, $\dot{y}={\dot{y}}_p+\ell \dot{\theta }\sin \theta$. ${\dot{x}}^2+{\dot{y}}^2={\ell }^2{\dot{\theta }}^2+{\dot{y}}_p^2+2\ell \dot{\theta }{\dot{y}}_p\sin \theta .$ $T=\frac{1}{2}m({\ell }^2{\dot{\theta }}^2+{\dot{y}}_p^2+2\ell \dot{\theta }{\dot{y}}_p\sin \theta ),V=mgy=mg(y_p-\ell \cos \theta ).$ The ${\dot{y}}_p^2$ and $mg{y}_p$ pieces depend only on time, not on $\theta$ or $\dot{\theta }$, so they drop out of the EL derivatives. Keep the relevant parts: $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }+m\ell {\dot{y}}_p\sin \theta .$ $\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\ddot{\theta }+m\ell {\ddot{y}}_p\sin \theta +m\ell {\dot{y}}_p\dot{\theta }\cos \theta .$ $\frac{\partial L}{\partial \theta }=m\ell \dot{\theta }{\dot{y}}_p\cos \theta -mg\ell \sin \theta .$ Subtract; the $m\ell {\dot{y}}_p\dot{\theta }\cos \theta$ terms cancel: $m{\ell }^2\ddot{\theta }+m\ell {\ddot{y}}_p\sin \theta +mg\ell \sin \theta =0.$ With ${\ddot{y}}_p=-a{\Omega }^2\cos (\Omega t)$: $\boxed{\ddot{\theta }=-\frac{1}{\ell }(g-a{\Omega }^2\cos \Omega t)\sin \theta }.$ The drive acts like a time-varying effective gravity $g_{\text{eff}}(t)=g-a{\Omega }^2\cos \Omega t$. (This is the seed of the famous inverted-pendulum stabilization.) ✓

No potential, so use $\frac{d}{dt}\frac{\partial T}{\partial \dot{x}}-\frac{\partial T}{\partial x}=Q_x$. $T=\frac{1}{2}m{\dot{x}}^2,Q_x=\vec{F}\cdot \frac{\partial \vec{r}}{\partial x}=F(t)\cdot 1=F_0\sin (\Omega t).$ $m\ddot{x}=F_0\sin (\Omega t)\Rightarrow \ddot{x}=\frac{F_0}{m}\sin (\Omega t).$ Integrate once ($\dot{x}(0)=0$): $\dot{x}=\frac{F_0}{m\Omega }(1-\cos \Omega t).$ Integrate again ($x(0)=0$): $\boxed{x(t)=\frac{F_0}{m\Omega }\left(t-\frac{\sin \Omega t}{\Omega }\right)}.$ The bead drifts (the $t$ term) while wiggling — a net secular motion from an oscillating force. ✓

$\varphi$ is cyclic means $L$ does not contain $\varphi$ itself (only $\dot{\varphi }$). Indeed $\partial L/\partial \varphi =0$. The $\varphi$ Euler-Lagrange equation is then $\frac{d}{dt}\frac{\partial L}{\partial \dot{\varphi }}=0\Rightarrow \frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }=\text{const}\equiv p_{\varphi }.$ So the angular momentum $p_{\varphi }=m{r}^2\dot{\varphi }$ is conserved — a symmetry (rotational invariance of $L$) yielding a conservation law, exactly Noether's theorem in miniature. Radial equation: $\frac{\partial L}{\partial \dot{r}}=m\dot{r}\Rightarrow \frac{d}{dt}=m\ddot{r},\frac{\partial L}{\partial r}=mr{\dot{\varphi }}^2-kr.$ $m\ddot{r}-mr{\dot{\varphi }}^2+kr=0.$ Eliminate $\dot{\varphi }=p_{\varphi }/(m{r}^2)$: $\boxed{m\ddot{r}=\frac{p_{\varphi }^2}{m{r}^3}-kr}.$ The first term is the centrifugal barrier, the second the spring pull. ✓

$T=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}I{\dot{\varphi }}^2,V=0(\text{level ground}).$ The constraint $\dot{x}-R\dot{\varphi }=0$ is non-holonomic (it involves velocities); the $\delta q_i$ are not independent, so we append multiplier terms. The modified equations are $\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_i}-\frac{\partial L}{\partial q_i}=\lambda a_i$, where $a_x=1,a_{\varphi }=-R$ come from the constraint $a_x\dot{x}+a_{\varphi }\dot{\varphi }=0$. $x$ equation: $m\ddot{x}=\lambda$. $\varphi$ equation: $I\ddot{\varphi }=-\lambda R\Rightarrow \frac{1}{2}m{R}^2\ddot{\varphi }=-\lambda R\Rightarrow \frac{1}{2}mR\ddot{\varphi }=-\lambda .$ Constraint (differentiated): $\ddot{x}=R\ddot{\varphi }$. Substitute $\ddot{\varphi }=\ddot{x}/R$ into the $\varphi$ equation: $\frac{1}{2}mR\cdot \ddot{x}/R=-\lambda \Rightarrow \frac{1}{2}m\ddot{x}=-\lambda$. Combine with $m\ddot{x}=\lambda$: $\frac{1}{2}(\lambda )=-\lambda \Rightarrow \frac{3}{2}\lambda =0\Rightarrow \boxed{\lambda =0,\ddot{x}=0}.$ On level ground with no driving, the friction (constraint) force $\lambda$ is zero and the hoop rolls at constant speed — friction is only needed when something tries to change the rolling. The multiplier $\lambda$ is that friction force. See Lagrange multipliers for non-holonomic constraints. ✓

Constraint force wanted explicitly, non-holonomic ::: Lagrange multiplier $\lambda$ (raw form with $\lambda a_i$) Force is $-\partial V/\partial q$, holonomic ::: standard $L=T-V$ Euler-Lagrange Explicit time-dependent driving force, no potential ::: raw form with generalized force $Q_i$ $L$ has no $q_i$ (only ${\dot{q}}_i$) ::: cyclic coordinate, $\partial L/\partial {\dot{q}}_i$ conserved