Exercises — Applying E-L equations to various systems

Step 1 — DOF: ek seedhi line ⇒ 1 DOF, coordinate $q=x$. Step 2 — $T$: speed hai $\dot{x}$, isliye $T=\frac{1}{2}m{\dot{x}}^2$. Step 3 — $V$: kuch push nahi kar raha ⇒ $V=0$. Step 4 — $L$: $L=\frac{1}{2}m{\dot{x}}^2$. Derivatives: $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$ (yeh ordinary momentum hai!), aur $\frac{\partial L}{\partial x}=0$ ($L$ mein koi $x$ nahi hai). Step 5 — E–L: $\frac{d}{dt}(m\dot{x})-0=0\Rightarrow \boxed{m\ddot{x}=0}$. Matlab kya hai: koi force nahi ⇒ koi acceleration nahi ⇒ constant velocity. Yeh Newton's first law hai jo recipe se khud nikal aayi.

Coordinate/sign convention: $y$ ko neeche ki taraf natural length se measure kiya gaya hai, isliye $y$ badhne ka matlab hai mass neeche gaya. $\frac{1}{2}m{\dot{y}}^2$ mein $\dot{y}$ hai (ek velocity) ⇒ yeh $T$ hai. $\frac{1}{2}k{y}^2-mgy$ sirf position $y$ par depend karta hai ⇒ yeh $V$ hai. Iske do pieces:

• spring stored energy $\frac{1}{2}k{y}^2$ (hamesha positive, stretch ke saath badhti hai);
• gravitational potential $V_{\text{grav}}=-mgy$, kyunki $y$ neeche ki taraf measure hai aur $V_{\text{grav}}=mg(\text{height})$ jahan height $=-y$ hai. Isliye neeche jaana (bada $y$) $V_{\text{grav}}$ ghataata hai, jo minus sign se match karta hai. $\boxed{L=\frac{1}{2}m{\dot{y}}^2-\frac{1}{2}k{y}^2+mgy}.$

Is page par full E–L derivation (kuch bhi defer nahi kiya): $\theta$ ko neeche ki vertical se measure karo.

• Step 2 — $T$: bob ek circle par $\ell$ radius se move karta hai, isliye arc-speed hai $\ell \dot{\theta }$ aur $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$.
• Step 3 — $V$: lowest point se height hai $\ell (1-\cos \theta )$, isliye $V=mg\ell (1-\cos \theta )$.
• Step 4 — $L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2-mg\ell (1-\cos \theta )$, jisme $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }$ aur $\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta$.
• Step 5 — E–L: $\frac{d}{dt}(m{\ell }^2\dot{\theta })+mg\ell \sin \theta =0\Rightarrow \ddot{\theta }=-\frac{g}{\ell }\sin \theta$.

Chhote $\theta$ ke liye, $\sin \theta \approx \theta$, jisse milta hai $\ddot{\theta }=-\frac{g}{\ell }\theta$ — SHM form $\ddot{\theta }=-{\omega }^2\theta$. Read off: $\omega =\sqrt{g/\ell }=\sqrt{9.8/0.25}=\sqrt{39.2}\approx 6.261\text{rad/s}$. Period $T_{\text{period}}=\frac{2\pi }{\omega }\approx \frac{6.2832}{6.261}\approx 1.0035\text{s}$.

Recipe skip mat karo — pehle EOM banao.

• Step 1 — DOF: block slope surface par stuck hai ⇒ 1 DOF, coordinate $q=s$ = incline ke saath saath travel ki gayi distance.
• Step 2 — $T$: block ki slope ke saath speed hai $\dot{s}$, isliye $T=\frac{1}{2}m{\dot{s}}^2$.
• Step 3 — $V$: $s$ distance slope se neeche slide karne par height $s\sin \alpha$ ghatti hai, isliye $V=-mgs\sin \alpha$.
• Step 4 — $L=\frac{1}{2}m{\dot{s}}^2+mgs\sin \alpha$, jisme $\frac{\partial L}{\partial \dot{s}}=m\dot{s}$ aur $\frac{\partial L}{\partial s}=mg\sin \alpha$.
• Step 5 — E–L: $\frac{d}{dt}(m\dot{s})-mg\sin \alpha =0\Rightarrow \ddot{s}=g\sin \alpha$. (Normal force kabhi appear nahi hua — $s$ ko slope ke saath choose karne se constraint absorb ho gaya.)

Ab numbers daalo: $a=\ddot{s}=g\sin 30^{\circ }=9.8\times 0.5=4.9{\text{m/s}}^2$ (constant). Kyunki $a$ constant hai aur rest se start karta hai, $v^2=2as$ se $v=\sqrt{2\times 4.9\times 2}=\sqrt{19.6}\approx 4.427\text{m/s}$. Sanity check: height dropped $h=s\sin \alpha =1\text{m}$; energy conservation $mgh=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{2gh}=\sqrt{19.6}$. Same number ✔.

Recipe se EOM banao (koi memorized formula nahi).

• Step 1 — DOF: inextensible string dono masses ko tie karta hai: agar $m_1$ $x$ neeche jaata hai, toh $m_2$ $x$ upar aata hai. Isliye ek coordinate $x$ = $m_1$ ka descent sab kuch fix kar deta hai ⇒ 1 DOF. String tension is choice se absorb ho jaata hai.
• Step 2 — $T$: dono masses $|\dot{x}|$ speed se move karte hain, isliye $T=\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{x}}^2=\frac{1}{2}(m_1+m_2){\dot{x}}^2$.
• Step 3 — $V$: $m_1$ $x$ neeche girta hai (PE lose karta hai: $-m_{1}gx$); $m_2$ $x$ upar jaata hai (PE gain karta hai: $+m_{2}gx$). Isliye $V=-m_{1}gx+m_{2}gx=-(m_1-m_2)gx$.
• Step 4 — $L=\frac{1}{2}(m_1+m_2){\dot{x}}^2+(m_1-m_2)gx$, jisme $\frac{\partial L}{\partial \dot{x}}=(m_1+m_2)\dot{x}$ aur $\frac{\partial L}{\partial x}=(m_1-m_2)g$.
• Step 5 — E–L: $(m_1+m_2)\ddot{x}-(m_1-m_2)g=0\Rightarrow \ddot{x}=\frac{(m_1-m_2)g}{m_1+m_2}$.

Numbers daalo: $\ddot{x}=\frac{(3-2)(9.8)}{3+2}=\frac{9.8}{5}=1.96{\text{m/s}}^2.$ Positive ⇒ bhaari $m_1$ neeche jaata hai, jaisa expected tha.

Coordinate/sign convention: horizontal plane mein kaam karo. Origin ko rotation axis par rakho; wire ek rotating radial direction mein lie karta hai. $r\ge 0$ axis se distance hai (positive outward along wire, red arrow) aur $\theta$ plane mein wire ka angle hai. Step 1 — DOF: wire force karta hai ki angle $\theta =\Omega t$ ho (free nahi!), isliye sirf $r$ ek free coordinate hai ⇒ 1 DOF, $q=r$. Step 2 — $T$: plane polar coordinates mein $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$ — velocity ke do pieces hain red arrow ($\dot{r}$, radial) aur blue arrow ($r\dot{\theta }$, tangential). Lekin $\dot{\theta }=\Omega$ constraint se fixed hai, isliye $T=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\Omega }^2.$ Step 3 — $V$: horizontal, frictionless ⇒ $V=0$. Step 4 — $L=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\Omega }^2$. $\frac{\partial L}{\partial \dot{r}}=m\dot{r}$, $\frac{\partial L}{\partial r}=mr{\Omega }^2$. Step 5 — E–L: $\frac{d}{dt}(m\dot{r})-mr{\Omega }^2=0\Rightarrow \boxed{\ddot{r}={\Omega }^{2}r}$. Matlab kya hai: plus sign note karo. Yeh oscillation nahi hai ($\ddot{r}=-{\omega }^{2}r$ hota) balki runaway growth hai: $r(t)=A{e}^{\Omega t}+B{e}^{-\Omega t}$. Bead exponentially bahar fling hota hai. $mr{\Omega }^2$ term exactly centrifugal effect hai, jo automatically appear hota hai kyunki humne constraint ko $T$ mein plug kiya.

Is page par conserved quantity derive karte hain (kuch defer nahi):

• Step 2 — $T$: polar coordinates mein velocity ka ek radial part $\dot{r}$ aur ek tangential part $r\dot{\theta }$ hota hai, isliye $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$.
• Step 3 — $V=V(r)$, $\theta$ se independent.
• Step 4 — $L=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)-V(r)$. Dhyan do ki $\theta$ $L$ mein appear nahi karta — sirf $\dot{\theta }$ karta hai. Aisa coordinate cyclic (ignorable) kehlaata hai (dekho 2.1.07-Cyclic-coordinates-and-conservation-laws).
• $\theta$ E–L: $\frac{\partial L}{\partial \theta }=0$, isliye $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta }}\right)=0$, matlab $\frac{\partial L}{\partial \dot{\theta }}$ time mein constant hai. Calculate karo: $p_{\theta }\equiv \frac{\partial L}{\partial \dot{\theta }}=m{r}^2\dot{\theta }=\text{const}.$ Yeh conserved $p_{\theta }$ centre ke baare mein particle ka angular momentum hai: mass $\times$ radius $\times$ tangential speed $=m\cdot r\cdot (r\dot{\theta })=m{r}^2\dot{\theta }$. (General principle "ek symmetry ek conserved quantity deti hai" Noethers-theorem hai: yahan $V(r)$ ki rotational symmetry angular momentum conserve karti hai.)

Ab numbers. Conservation ka matlab hai $m{r}_1^2{\dot{\theta }}_1=m{r}_2^2{\dot{\theta }}_2$, isliye ${\dot{\theta }}_2={\dot{\theta }}_1\frac{r_1^2}{r_2^2}=2\cdot \frac{1^2}{2^2}=0.5\text{rad/yr}.$ Matlab kya hai: radius-squared chaar guna ⇒ sweep chaar guna slow. Yeh Kepler ka "equal areas in equal times" cyclic $\theta$ coordinate ke andar chhupa hua hai.

Coordinate/sign convention: fixed lab axes jinka origin support ke starting point par hai; $x$ right point karta hai, $y$ up point karta hai (jaisa figure s02 mein hai). Support $(X(t),0)$ par hai aur $\theta$ downward vertical se measure kiya gaya hai (positive = bob right swing kiya). $X(t)$ ke baare mein ek baat: kyunki $X(t)$ ek prescribed function of time hai, Lagrangian time par explicitly depend karta hai ($\dot{X}$ aur $\ddot{X}$ ke through). Yeh ek rheonomic (time-dependent) situation hai, lekin Euler–Lagrange equation $\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta }=0$ ek free coordinate $\theta$ ke liye bilkul unchanged rehti hai — Hamilton's principle ne kabhi require nahi kiya ki $L$ time-independent ho. Step 1 — DOF: support motion $X(t)$ given hai (free nahi), isliye sirf $\theta$ free hai ⇒ 1 DOF. Step 2 — $T$: bob ki position (figure s02 se read karo): $x=X(t)+\ell \sin \theta$, $y=-\ell \cos \theta$. Velocities: $\dot{x}=\dot{X}+\ell \dot{\theta }\cos \theta$, $\dot{y}=\ell \dot{\theta }\sin \theta$. Speed squared: $v^2={\dot{x}}^2+{\dot{y}}^2={\dot{X}}^2+2\ell \dot{X}\dot{\theta }\cos \theta +{\ell }^2{\dot{\theta }}^2.$ (Cross terms mein ${\cos }^2+{\sin }^2=1$ se ${\ell }^2{\dot{\theta }}^2$ pieces collapse ho jaate hain.) Isliye $T=\frac{1}{2}m{v}^2$. Step 3 — $V$: height hai $y=-\ell \cos \theta$ ⇒ $V=-mg\ell \cos \theta$. Step 4 — $L=\frac{1}{2}m({\dot{X}}^2+2\ell \dot{X}\dot{\theta }\cos \theta +{\ell }^2{\dot{\theta }}^2)+mg\ell \cos \theta .$ Derivatives ($\dot{X},\dot{\theta }$ ko independent variables ki tarah treat karo): $\frac{\partial L}{\partial \dot{\theta }}=m\ell \dot{X}\cos \theta +m{\ell }^2\dot{\theta }.$ $\frac{\partial L}{\partial \theta }=-m\ell \dot{X}\dot{\theta }\sin \theta -mg\ell \sin \theta .$ Momentum ki time-derivative: $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta }}\right)=m\ell \ddot{X}\cos \theta -m\ell \dot{X}\dot{\theta }\sin \theta +m{\ell }^2\ddot{\theta }.$ Step 5 — E–L ($\partial L/\partial \theta$ subtract karo): $-m\ell \dot{X}\dot{\theta }\sin \theta$ terms cancel ho jaate hain, bacha: $m{\ell }^2\ddot{\theta }+m\ell \ddot{X}\cos \theta +mg\ell \sin \theta =0.$ $m{\ell }^2$ se divide karo: $\boxed{\ddot{\theta }=-\frac{g}{\ell }\sin \theta -\frac{\ddot{X}}{\ell }\cos \theta }.$ Matlab kya hai: agar support still hai ($\ddot{X}=0$) toh ordinary pendulum milta hai. Extra $-\frac{\ddot{X}}{\ell }\cos \theta$ accelerating support se ek pseudo-forcing hai — support ko sideways accelerate karna ek extra horizontal gravity jaisa feel hota hai.

Coordinate/sign convention: fixed lab axes jinka origin cart ke starting point par hai; $x$ right point karta hai, $y$ up point karta hai (figure s03). Cart ka top-attachment point $(X,0)$ par hai (cart table par rehta hai, height fixed) aur $\theta$ downward vertical se measure kiya gaya hai. Problem 4.1 se unlike, yahan $X$ ek free coordinate hai, prescribed function nahi. Step 1 — DOF: cart free ($X$) + pendulum free ($\theta$) ⇒ 2 DOF, coordinates $\{X,\theta \}$. Step 2 — $T$: cart: $T_M=\frac{1}{2}M{\dot{X}}^2$. Bob position: $x=X+\ell \sin \theta$, $y=-\ell \cos \theta$; velocities se milta hai (Problem 4.1 jaisi binomial expansion): $v_m^2={\dot{X}}^2+2\ell \dot{X}\dot{\theta }\cos \theta +{\ell }^2{\dot{\theta }}^2.$ Isliye $T=\frac{1}{2}M{\dot{X}}^2+\frac{1}{2}m({\dot{X}}^2+2\ell \dot{X}\dot{\theta }\cos \theta +{\ell }^2{\dot{\theta }}^2).$ Step 3 — $V$: $V=-mg\ell \cos \theta$ (sirf bob ki height hai; cart level rehta hai). Step 4 — $L=\frac{1}{2}(M+m){\dot{X}}^2+m\ell \dot{X}\dot{\theta }\cos \theta +\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+mg\ell \cos \theta .$

$X$-equation — note karo ki $X$ $L$ mein appear nahi karta (sirf $\dot{X}$ karta hai), isliye $X$ ek cyclic coordinate hai: $\frac{\partial L}{\partial \dot{X}}=(M+m)\dot{X}+m\ell \dot{\theta }\cos \theta ,\frac{\partial L}{\partial X}=0.$ E–L ⇒ yeh generalized momentum conserved hai: $\boxed{(M+m)\dot{X}+m\ell \dot{\theta }\cos \theta =\text{const}}$ — total horizontal momentum (cart + bob) conserved hai, kyunki kuch bhi system ko sideways push nahi kar raha. Yeh 2.1.07-Cyclic-coordinates-and-conservation-laws action mein hai. Is conserved quantity ko time mein ek baar differentiate karne par cart ka EOM milta hai: $\boxed{(M+m)\ddot{X}+m\ell \ddot{\theta }\cos \theta -m\ell {\dot{\theta }}^2\sin \theta =0}.$

$\theta$-equation — ab $\theta$ par full E–L karo: $\frac{\partial L}{\partial \dot{\theta }}=m\ell \dot{X}\cos \theta +m{\ell }^2\dot{\theta }.$ $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta }}\right)=m\ell \ddot{X}\cos \theta -m\ell \dot{X}\dot{\theta }\sin \theta +m{\ell }^2\ddot{\theta }.$ $\frac{\partial L}{\partial \theta }=-m\ell \dot{X}\dot{\theta }\sin \theta -mg\ell \sin \theta .$ $\frac{\partial L}{\partial \theta }$ ko time-derivative se subtract karo; dono $-m\ell \dot{X}\dot{\theta }\sin \theta$ terms cancel ho jaate hain: $m\ell \ddot{X}\cos \theta +m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0.$ $m\ell$ se divide karo: $\boxed{\ell \ddot{\theta }+\ddot{X}\cos \theta +g\sin \theta =0}.$ Consistency check: agar $M\to \infty$ toh cart move nahi kar sakta, $\ddot{X}\to 0$, aur $\theta$-equation ban jaata hai $\ell \ddot{\theta }+g\sin \theta =0$, yaani $\ddot{\theta }=-\frac{g}{\ell }\sin \theta$ — fixed pendulum. ✔ Dono boxed equations (cart + bob) is 2-DOF system ke complete pair of equations of motion hain.

Problem 5.1 mein mila conserved horizontal momentum use karo. Woh start mein $0$ tha (system at rest) aur $0$ rehta hai: $(M+m)\dot{X}+m\ell \dot{\theta }\cos \theta =0.$ $\theta =0$ par, $\cos \theta =1$: $\dot{X}=-\frac{m\ell \dot{\theta }}{M+m}=-\frac{(1)(2)(1)}{3+1}=-0.5\text{m/s}.$ Minus sign dikhata hai ki cart bob ke swing ke opposite recoil karta hai — Newton's third law, bina koi force likhe mila.

• Cyclic coordinate kaise pehchante ho? ::: woh $L$ mein sirf $\dot{q}$ ke roop mein appear karta hai, $q$ ke roop mein kabhi nahi.
• Sign test: $\ddot{q}=+cq$ ka matlab? ::: exponential runaway, oscillation nahi.
• Cross term $2\ell \dot{X}\dot{\theta }\cos \theta$ kyun matter karta hai? ::: yeh drive aur swing ko couple karta hai; ise drop karne se interaction khatam ho jaata hai.
• Movable-support pendulum EOM? ::: $\ell \ddot{\theta }+\ddot{X}\cos \theta +g\sin \theta =0$.
• Rotating-wire bead EOM? ::: $\ddot{r}={\Omega }^{2}r$ (exponential fling-out).