Exercises — Generalized momenta and generalized forces

Hum kya karte hain: momentum paane ke liye $L$ ko velocity ke saath differentiate karo, aur force ke liye potential shortcut use karo. $p_x=\frac{\partial L}{\partial \dot{x}}=m\dot{x}.$ Velocity $\dot{x}$ sirf $\frac{1}{2}m{\dot{x}}^2$ mein aati hai; uska derivative $m\dot{x}$ hai. $-\frac{1}{2}k{x}^2$ wale term mein koi $\dot{x}$ nahi, isliye woh zero ho jaata hai. $Q_x=-\frac{\partial V}{\partial x}=-\frac{\partial }{\partial x}\left(\frac{1}{2}k{x}^2\right)=-kx.$ Sanity check: $p_x=m\dot{x}$ ordinary momentum hai, $Q_x=-kx$ Hooke ka restoring force hai. Dono sahi hain.

$p_{\theta }=\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta },\text{units }\mathrm{kg}{\mathrm{m}}^2/s.$ Ye angular momentum hai, ordinary momentum nahi — kyunki coordinate ek angle hai. $Q_{\theta }=\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta ,\text{units }\mathrm{N}\mathrm{m}.$ Ye ek torque hai. Yaad karo $V=-mg\ell \cos \theta$, isliye $-\partial V/\partial \theta =-mg\ell \sin \theta$. ✓

$V=0$ ke saath, $L=T$. $p_r=\frac{\partial L}{\partial \dot{r}}=m\dot{r}.$ $p_{\theta }=\frac{\partial L}{\partial \dot{\theta }}=m{r}^2\dot{\theta }.$ $r^2$ kyun aata hai: $T$ ke andar $\dot{\theta }$ wala term $\frac{1}{2}m{r}^2{\dot{\theta }}^2$ hai; $\dot{\theta }$ mein differentiate karne par $m{r}^2\dot{\theta }$ milta hai. Ye bilkul angular momentum $L_{\text{ang}}=mr{v}_{\perp }$ hai, jahan $v_{\perp }=r\dot{\theta }$. Dekho Angular momentum.

Tarika 1 — direct projection. Humein $\partial \vec{r}/\partial s$ chahiye: har component ko $s$ mein differentiate karo: $\frac{\partial \vec{r}}{\partial s}=(\cos \alpha ,-\sin \alpha ).$ Phir $Q_s=\vec{F}\cdot \frac{\partial \vec{r}}{\partial s}=(0)(\cos \alpha )+(-mg)(-\sin \alpha )=mg\sin \alpha .$ Tarika 2 — potential. Height hai $y=-s\sin \alpha$, isliye $V=mgy=-mgs\sin \alpha$, aur $Q_s=-\frac{\partial V}{\partial s}=mg\sin \alpha .$ Dono agree karte hain. Ye wahi familiar "gravity ka slope ke saath component" hai, jo automatically recover ho gaya. ✓ ($s$ ek length hai, isliye $Q_s$ newtons mein ek plain force hai.)

Cyclic coordinate: $\theta$ $L$ mein appear nahi karta (sirf $\dot{\theta }$ karta hai), isliye $\partial L/\partial \theta =0$. Euler–Lagrange se, ${\dot{p}}_{\theta }=0$: $p_{\theta }=m{r}^2\dot{\theta }$ conserved hai (angular momentum). Deep "symmetry ⇒ conservation" reason ke liye dekho Noether's Theorem. Numbers. Conservation se milta hai $r_1^2{\dot{\theta }}_1=r_2^2{\dot{\theta }}_2$ ($m$ cancel ho jaata hai): ${\dot{\theta }}_2={\dot{\theta }}_1\frac{r_1^2}{r_2^2}=3\cdot \frac{2^2}{1^2}=12rad/s.$ Particle chaar guna tez spin karta hai jab aadhe radius tak kheencha jaaye — figure-skater effect.

$L=T=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\omega }^2$. Euler–Lagrange apply karo $\frac{d}{dt}\frac{\partial L}{\partial \dot{r}}-\frac{\partial L}{\partial r}=0$: $p_r=\frac{\partial L}{\partial \dot{r}}=m\dot{r},\frac{d}{dt}{p}_r=m\ddot{r}.$ $\frac{\partial L}{\partial r}=mr{\omega }^2.$ Toh $m\ddot{r}-mr{\omega }^2=0\Rightarrow \boxed{m\ddot{r}=mr{\omega }^2}.$ Interpretation: right side ek effective generalized force $mr{\omega }^2$ hai — centrifugal push — even though $V=0$ hai aur $r$ ke along koi real force nahi lagta. Ye kinetic energy ki $r$-dependence se aaya. Isliye true general form mein $Q_j$ $\partial L/\partial q_j$ ke andar chhupa hota hai, sirf $-\partial V/\partial q_j$ nahi.

Coordinate & velocities. Ek DOF: $x$. Dono masses speed $\dot{x}$ se move karte hain, isliye $T=\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{x}}^2=\frac{1}{2}(m_1+m_2){\dot{x}}^2.$ Potential. Jab $m_1$ $x$ neeche girta hai, uski height $x$ se girti hai: contributes $-m_{1}gx$. Jab $m_2$ slope par $x$ upar chadhta hai, uski height $x\sin \alpha$ se badhti hai: contributes $+m_{2}gx\sin \alpha$. $V=-m_{1}gx+m_{2}gx\sin \alpha .$ Lagrangian. $L=\frac{1}{2}(m_1+m_2){\dot{x}}^2+m_{1}gx-m_{2}gx\sin \alpha .$ Momentum & force. $p_x=\frac{\partial L}{\partial \dot{x}}=(m_1+m_2)\dot{x},Q_x=-\frac{\partial V}{\partial x}=m_{1}g-m_{2}g\sin \alpha .$ Equation of motion ${\dot{p}}_x=Q_x$: $(m_1+m_2)\ddot{x}=g(m_1-m_2\sin \alpha )\Rightarrow \boxed{\ddot{x}=\frac{g(m_1-m_2\sin \alpha )}{m_1+m_2}}.$ Limit check: $\alpha =90^{\circ }$ deta hai $\ddot{x}=g(m_1-m_2)/(m_1+m_2)$ — standard Atwood machine. $\alpha =0$ deta hai $\ddot{x}=g{m}_1/(m_1+m_2)$ — $m_2$ flat floor par, sirf $m_1$ kheenchta hai. ✓

$\ddot{x}=\frac{9.8(3-2\sin 30^{\circ })}{3+2}=\frac{9.8(3-2\cdot 0.5)}{5}=\frac{9.8\cdot 2}{5}=3.92m/{\mathrm{s}}^2.$ Positive hai, isliye $m_1$ sach mein neeche jaata hai — reasonable hai kyunki $m_1=3>m_2\sin 30^{\circ }=1$.

$L=\frac{1}{2}m{\ell }^2({\dot{\vartheta }}^2+{\sin }^2\vartheta {\dot{\phi }}^2)+mg\ell \cos \vartheta .$ Momenta. $p_{\vartheta }=\frac{\partial L}{\partial \dot{\vartheta }}=m{\ell }^2\dot{\vartheta },p_{\phi }=\frac{\partial L}{\partial \dot{\phi }}=m{\ell }^2{\sin }^2\vartheta \dot{\phi }.$ Conservation. $\phi$ $L$ mein appear nahi karta (sirf $\dot{\phi }$ karta hai), isliye $\phi$ cyclic hai ⇒ $p_{\phi }=m{\ell }^2{\sin }^2\vartheta \dot{\phi }$ conserved hai. Physically ye vertical axis ke baare mein angular momentum hai. $\vartheta$ zaroor appear karta hai (${\sin }^2\vartheta$ aur $\cos \vartheta$ ke through), isliye $p_{\vartheta }$ conserved nahi hai. Equations of motion. $\vartheta$ ke liye: $\frac{d}{dt}{p}_{\vartheta }-\frac{\partial L}{\partial \vartheta }=0$, jahan $\frac{\partial L}{\partial \vartheta }=m{\ell }^2\sin \vartheta \cos \vartheta {\dot{\phi }}^2-mg\ell \sin \vartheta$: $m{\ell }^2\ddot{\vartheta }=m{\ell }^2\sin \vartheta \cos \vartheta {\dot{\phi }}^2-mg\ell \sin \vartheta .$ $\phi$ ke liye: cyclic hone ki wajah se, $\frac{d}{dt}{p}_{\phi }=0$, yani $m{\ell }^2{\sin }^2\vartheta \dot{\phi }=\text{const}.$ Check: $\dot{\phi }=0$ set karo (planar swing) → $\ddot{\vartheta }=-\frac{g}{\ell }\sin \vartheta$, ordinary pendulum. ✓

$\vartheta$-equation mein $\ddot{\vartheta }=0$ set karo aur $m{\ell }^2\sin {\vartheta }_0$ cancel karo ($0<{\vartheta }_0<\pi$ ke liye nonzero): $0=\cos {\vartheta }_0{\Omega }^2-\frac{g}{\ell }\Rightarrow \boxed{{\Omega }^2=\frac{g}{\ell \cos {\vartheta }_0}}.$ Degenerate checks:

• ${\vartheta }_0\to 0$ (almost seedha neeche): $\cos {\vartheta }_0\to 1$, ${\Omega }^2\to g/\ell$ — pendulum ki small-oscillation frequency. ✓
• ${\vartheta }_0\to 90^{\circ }$: $\cos {\vartheta }_0\to 0$, ${\Omega }^2\to \infty$ — rod ko horizontal kisi bhi finite rate par spin karke hold nahi kar sakte. Physically sach hai. Numeric: $\ell =1\mathrm{m}$, ${\vartheta }_0=60^{\circ }$, $g=9.8$: ${\Omega }^2=9.8/(1\cdot 0.5)=19.6$, isliye $\Omega =\sqrt{19.6}\approx 4.427rad/s$.

Kisi bhi coordinate $q_j$ ke liye: momentum hai $p_j=\partial L/\partial {\dot{q}}_j$ ::: force hai $Q_j=-\partial V/\partial q_j$ conservative systems ke liye (ya general mein $\partial L/\partial q_j$); agar $q_j$ $L$ mein absent hai, toh $p_j$ conserved hai.