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

Pieces chunno. Sirf ek coordinate hai, $y$. Velocity $\dot{y}$ hai, toh kinetic energy hai $T=\frac{1}{2}m{\dot{y}}^2.$ Gravity neeche pull karti hai; potential energy height ke saath badhti hai: $V=mgy\Rightarrow L=T-V=\frac{1}{2}m{\dot{y}}^2-mgy.$ Machine chalao. Euler-Lagrange machine ke do ingredients: $\frac{\partial L}{\partial \dot{y}}=m\dot{y}\Rightarrow \frac{d}{dt}\left(m\dot{y}\right)=m\ddot{y},\frac{\partial L}{\partial y}=-mg.$ Subtract karo: $m\ddot{y}-(-mg)=0\Rightarrow m\ddot{y}+mg=0\Rightarrow \boxed{\ddot{y}=-g}.$ Acceleration $g$ neeche ki taraf hai — bilkul Newton ki tarah. ✓

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

$x=\ell \sin \theta ,y=-\ell \cos \theta$ ke saath, speed $\ell \dot{\theta }$ hai, toh $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2,V=-mg\ell \cos \theta ,L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+mg\ell \cos \theta .$ $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }\Rightarrow \frac{d}{dt}=m{\ell }^2\ddot{\theta },\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta .$ EL: $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0\Rightarrow \ddot{\theta }=-\frac{g}{\ell }\sin \theta .$ Small angles: $\sin \theta \approx \theta$, toh $\ddot{\theta }=-\frac{g}{\ell }\theta$. Ye SHM hai 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 $y=-R$ par hai). Ye form mein pendulum se identical hai $\ell \to R$ ke saath — hoop wahi rigid constraint provide karta hai jo rod karta tha, aur normal force radius ke along point karti hai, allowed (tangential) motion ke perpendicular, isliye ye no virtual work karti hai aur kabhi appear nahi hoti. $T=\frac{1}{2}m{R}^2{\dot{\theta }}^2,V=-mgR\cos \theta .$ $\boxed{\ddot{\theta }=-\frac{g}{R}\sin \theta }.$ Same physics, nayi geometry. ✓

Bead ki velocity mein ek radial part $\dot{r}$ aur ek tangential part $r\omega$ (wire ke spinning se) hai. Ye perpendicular hain, toh $T=\frac{1}{2}m({\dot{r}}^2+r^2{\omega }^2),V=0.$ Yahan $r^2{\omega }^2$ depends on $r$ karta hai, toh $\partial T/\partial r\ne 0$ — yahi poora point hai. $\frac{\partial T}{\partial \dot{r}}=m\dot{r}\Rightarrow \frac{d}{dt}=m\ddot{r},\frac{\partial T}{\partial r}=mr{\omega }^2.$ EL ($Q_r=0$): $m\ddot{r}-mr{\omega }^2=0\Rightarrow \boxed{\ddot{r}={\omega }^{2}r}$. Positive sign ka matlab hai bead baahir ki taraf jaati hai — centrifugal effect automatically $\partial T/\partial r$ se nikal aaya. Solution $r(t)=A{e}^{\omega t}+B{e}^{-\omega t}$: exponential escape. ✓

Agar $m_1$ $x$ se neeche jaata hai, toh $m_2$ $x$ se upar jaata hai (string inextensible hai — ye constraint hai). Dono $\dot{x}$ speed se chalte hain: $T=\frac{1}{2}(m_1+m_2){\dot{x}}^2.$ Heights: $m_1$ at $-x$, $m_2$ at $-(\ell -x)=x-\ell$. Toh $V=-m_{1}gx+m_{2}g(x-\ell )=(m_2-m_1)gx-m_{2}g\ell .$ $\frac{\partial L}{\partial \dot{x}}=(m_1+m_2)\dot{x}\Rightarrow \frac{d}{dt}=(m_1+m_2)\ddot{x},\frac{\partial L}{\partial x}=-(m_2-m_1)g=(m_1-m_2)g.$ EL: $(m_1+m_2)\ddot{x}-(m_1-m_2)g=0\Rightarrow \boxed{\ddot{x}=\frac{(m_1-m_2)g}{m_1+m_2}}.$ Tension kabhi appear nahi hui — ye internal constraint force hai, single coordinate $x$ choose karne se khatam ho gayi. ✓

$T=\frac{1}{2}M{\dot{X}}^2+\frac{1}{2}m{\dot{x}}^2,V=\frac{1}{2}k(x-X{)}^2.$ $L=\frac{1}{2}M{\dot{X}}^2+\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k(x-X{)}^2.$ $X$ ke liye equation: $\frac{\partial L}{\partial \dot{X}}=M\dot{X},\frac{\partial L}{\partial X}=-\frac{1}{2}k\cdot 2(x-X)\cdot (-1)=k(x-X).$ $\Rightarrow M\ddot{X}-k(x-X)=0\Rightarrow \boxed{M\ddot{X}=k(x-X)}.$ $x$ ke liye equation: $\frac{\partial L}{\partial \dot{x}}=m\dot{x},\frac{\partial L}{\partial x}=-k(x-X).$ $\Rightarrow \boxed{m\ddot{x}=-k(x-X)}.$ Newton's third law dikh raha hai: dono forces equal aur opposite hain. ✓

Bob position (pivot at height $y_p$): $x=\ell \sin \theta ,y=y_p-\ell \cos \theta .$ Velocities: $\dot{x}=\ell \dot{\theta }\cos \theta$, $\dot{y}={\dot{y}}_p+\ell \dot{\theta }\sin \theta$. ${\dot{x}}^2+{\dot{y}}^2={\ell }^2{\dot{\theta }}^2+{\dot{y}}_p^2+2\ell \dot{\theta }{\dot{y}}_p\sin \theta .$ $T=\frac{1}{2}m({\ell }^2{\dot{\theta }}^2+{\dot{y}}_p^2+2\ell \dot{\theta }{\dot{y}}_p\sin \theta ),V=mgy=mg(y_p-\ell \cos \theta ).$ ${\dot{y}}_p^2$ aur $mg{y}_p$ pieces sirf time par depend karti hain, $\theta$ ya $\dot{\theta }$ par nahi, toh ye EL derivatives se drop ho jaati hain. Relevant parts rakho: $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }+m\ell {\dot{y}}_p\sin \theta .$ $\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\ddot{\theta }+m\ell {\ddot{y}}_p\sin \theta +m\ell {\dot{y}}_p\dot{\theta }\cos \theta .$ $\frac{\partial L}{\partial \theta }=m\ell \dot{\theta }{\dot{y}}_p\cos \theta -mg\ell \sin \theta .$ Subtract karo; $m\ell {\dot{y}}_p\dot{\theta }\cos \theta$ terms cancel ho jaate hain: $m{\ell }^2\ddot{\theta }+m\ell {\ddot{y}}_p\sin \theta +mg\ell \sin \theta =0.$ ${\ddot{y}}_p=-a{\Omega }^2\cos (\Omega t)$ ke saath: $\boxed{\ddot{\theta }=-\frac{1}{\ell }(g-a{\Omega }^2\cos \Omega t)\sin \theta }.$ Drive ek time-varying effective gravity $g_{\text{eff}}(t)=g-a{\Omega }^2\cos \Omega t$ ki tarah act karta hai. (Ye famous inverted-pendulum stabilization ka seed hai.) ✓

Koi potential nahi, toh $\frac{d}{dt}\frac{\partial T}{\partial \dot{x}}-\frac{\partial T}{\partial x}=Q_x$ use karo. $T=\frac{1}{2}m{\dot{x}}^2,Q_x=\vec{F}\cdot \frac{\partial \vec{r}}{\partial x}=F(t)\cdot 1=F_0\sin (\Omega t).$ $m\ddot{x}=F_0\sin (\Omega t)\Rightarrow \ddot{x}=\frac{F_0}{m}\sin (\Omega t).$ Ek baar integrate karo ($\dot{x}(0)=0$): $\dot{x}=\frac{F_0}{m\Omega }(1-\cos \Omega t).$ Dobara integrate karo ($x(0)=0$): $\boxed{x(t)=\frac{F_0}{m\Omega }\left(t-\frac{\sin \Omega t}{\Omega }\right)}.$ Bead drift karta hai ($t$ term) saath mein wiggle karte hue — ek oscillating force se net secular motion. ✓

$\varphi$ cyclic hai matlab $L$ mein $\varphi$ khud nahi aata (sirf $\dot{\varphi }$). Indeed $\partial L/\partial \varphi =0$. $\varphi$ Euler-Lagrange equation tab ye hai: $\frac{d}{dt}\frac{\partial L}{\partial \dot{\varphi }}=0\Rightarrow \frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }=\text{const}\equiv p_{\varphi }.$ Toh angular momentum $p_{\varphi }=m{r}^2\dot{\varphi }$ conserved hai — ek symmetry (rotational invariance of $L$) se conservation law milta hai, exactly Noether's theorem miniature mein. Radial equation: $\frac{\partial L}{\partial \dot{r}}=m\dot{r}\Rightarrow \frac{d}{dt}=m\ddot{r},\frac{\partial L}{\partial r}=mr{\dot{\varphi }}^2-kr.$ $m\ddot{r}-mr{\dot{\varphi }}^2+kr=0.$ $\dot{\varphi }=p_{\varphi }/(m{r}^2)$ eliminate karo: $\boxed{m\ddot{r}=\frac{p_{\varphi }^2}{m{r}^3}-kr}.$ Pehla term centrifugal barrier hai, doosra spring pull. ✓

$T=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}I{\dot{\varphi }}^2,V=0(\text{level ground}).$ Constraint $\dot{x}-R\dot{\varphi }=0$ non-holonomic hai (ismein velocities hain); $\delta q_i$ independent nahi hain, toh hum multiplier terms append karte hain. Modified equations hain $\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_i}-\frac{\partial L}{\partial q_i}=\lambda a_i$, jahan $a_x=1,a_{\varphi }=-R$ constraint $a_x\dot{x}+a_{\varphi }\dot{\varphi }=0$ se aate hain. $x$ 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): $\ddot{x}=R\ddot{\varphi }$. $\varphi$ equation mein $\ddot{\varphi }=\ddot{x}/R$ substitute karo: $\frac{1}{2}mR\cdot \ddot{x}/R=-\lambda \Rightarrow \frac{1}{2}m\ddot{x}=-\lambda$. $m\ddot{x}=\lambda$ ke saath combine karo: $\frac{1}{2}(\lambda )=-\lambda \Rightarrow \frac{3}{2}\lambda =0\Rightarrow \boxed{\lambda =0,\ddot{x}=0}.$ Level ground par bina kisi driving ke, friction (constraint) force $\lambda$ zero hai aur hoop constant speed par roll karta hai — friction sirf tab zaroori hoti hai jab kuch rolling change karne ki koshish kare. Multiplier $\lambda$ wahi friction force hai. Dekho Lagrange multipliers for non-holonomic constraints. ✓

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