Exercises — Kinetic energy in generalized coordinates

Rule yeh hai: $T_1$ aur $T_0$ tab hi aate hain jab $\partial \mathbf{r}/\partial t\ne 0$, yaani jab position map mein literally $t$ ho. (a) $x=r\cos \theta ,y=r\sin \theta$ mein koi $t$ nahi → sirf $T_2$. (b) Map hai $x=r\cos \omega t$: $t$ contain karta hai → $T_2$ aur $T_0$; $T_1=0$ yahan kyunki ek moving coordinate ka basis vector drift ke perpendicular hai (L2.2 mein dikhaya gaya). (c) Pivot position mein imposed clock term $A\cos (\Omega t)$ (driving frequency $\Omega$) hai, toh $t$ explicitly aata hai → generally teeno, $T_2,T_1,T_0$.

Parent worked example se: $M_{rr}=m$, $M_{\theta \theta }=m{r}^2$, $M_{r\theta }=M_{\theta r}=0$. $M=\left(\begin{array}{cc}m& 0\\ 0& m{r}^2\end{array}\right).$ Yeh symmetric hai ($M_{r\theta }=M_{\theta r}$) aur yeh position par depend karta hai $r^2$ ke through: angular motion ke liye "effective mass" bahar jaane par badhti hai — yeh ek lever-arm effect hai.

Map ko har coordinate ke respect mein differentiate karo (yehi poora method hai — $M_{ij}$ in ke dot products hain): $\frac{\partial \mathbf{r}}{\partial r}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta ),$ $\frac{\partial \mathbf{r}}{\partial \theta }=(r\cos \theta \cos \varphi ,r\cos \theta \sin \varphi ,-r\sin \theta ),$ $\frac{\partial \mathbf{r}}{\partial \varphi }=(-r\sin \theta \sin \varphi ,r\sin \theta \cos \varphi ,0).$ Ab dot products lo, baar baar ${\cos }^2+{\sin }^2=1$ use karte hue:

• $M_{rr}=m({\sin }^2\theta {\cos }^2\varphi +{\sin }^2\theta {\sin }^2\varphi +{\cos }^2\theta )=m({\sin }^2\theta +{\cos }^2\theta )=m.$
• $M_{\theta \theta }=m{r}^2({\cos }^2\theta {\cos }^2\varphi +{\cos }^2\theta {\sin }^2\varphi +{\sin }^2\theta )=m{r}^2.$
• $M_{\varphi \varphi }=m{r}^2{\sin }^2\theta ({\sin }^2\varphi +{\cos }^2\varphi )=m{r}^2{\sin }^2\theta .$
• Off-diagonals: jaise $M_{r\theta }=mr(\sin \theta \cos \theta {\cos }^2\varphi +\sin \theta \cos \theta {\sin }^2\varphi -\sin \theta \cos \theta )=0$. Isi tarah $M_{r\varphi }=M_{\theta \varphi }=0$ (spherical basis orthogonal hai). $\boxed{T=\frac{1}{2}m\left({\dot{r}}^2+r^2{\dot{\theta }}^2+r^2{\sin }^2\theta {\dot{\varphi }}^2\right)}$

Kyunki coordinate $r$ hai, hum map ko $r$ ke respect mein differentiate karte hain (yeh poore mein $q$ ki jagah play karta hai): $\frac{\partial \mathbf{r}}{\partial r}=(\cos \omega t,\sin \omega t),\frac{\partial \mathbf{r}}{\partial t}=(-r\omega \sin \omega t,r\omega \cos \omega t).$ $T_2$: $M_{rr}=m({\cos }^2\omega t+{\sin }^2\omega t)=m$, toh $T_2=\frac{1}{2}m{\dot{r}}^2$. $T_1$: $a_r=m[\cos \omega t(-r\omega \sin \omega t)+\sin \omega t(r\omega \cos \omega t)]=0$ — do products equal aur opposite hain. Yeh $T_1=0$ hai kyunki $\partial \mathbf{r}/\partial r$ wire ke saath point karta hai jabki $\partial \mathbf{r}/\partial t$ us ke across point karta hai (perpendicular); unka dot product zero hota hai. $T_0=\frac{1}{2}m(r\omega {)}^2$. $\boxed{T=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\omega }^2}$

Neeche diya figure yeh draw karta hai: kisi instant par straight wire (white), bead (yellow dot), yellow arrow $\partial \mathbf{r}/\partial r$ wire ke saath bahar ki taraf point karta hua (tumhari free direction), aur blue arrow $\partial \mathbf{r}/\partial t$ sideways point karta hua (bead ko spin drag kar raha hai). Pink right-angle mark dikhata hai ki woh $90^{\circ }$ par milte hain — woh perpendicularity exactly woh reason hai kyun unka dot product, aur hence $T_1$, zero hai.

$\frac{\partial \mathbf{r}}{\partial q}=(1,kq).$ $M_{qq}=m(1\cdot 1+kq\cdot kq)=m(1+k^2{q}^2).$ Map mein koi $t$ nahi, toh $T=T_2$: $\boxed{T=\frac{1}{2}m(1+k^2{q}^2){\dot{q}}^2}$ Notice karo ki effective mass $m(1+k^2{q}^2)$ bottom se door badhti hai: parabola ke steeper part par wahi $\dot{q}$ zyada real speed carry karta hai.

$T_1$ overlap $\partial \mathbf{r}/\partial q\cdot \partial \mathbf{r}/\partial t$ measure karta hai: "kya woh direction jisme tum freely move kar sakte ho constraint ke drag direction ke saath koi component share karta hai?" Ek wire jo sirf rotate kare, usme tumhari free direction (radially, wire ke saath) exactly drag direction ke perpendicular hoti hai (tangential, spin se). Perpendicular ⇒ dot product $=0$ ⇒ $T_1=0$. Yeh exactly L2.2 figure mein draw ki gayi perpendicular arrows ki pair hai (yellow wire ke saath, blue us ke across).

$T_1\ne 0$ banane ke liye, constraint ko partly tumhari free direction ke saath drag karne do. Example: ek wire jo apni khud ki length ke saath translate ho rahi ho, $x=q+vt$ (coordinate $q$, wire $x$-direction mein speed $v$ se slide karti hai). Tab $\partial \mathbf{r}/\partial q=(1,0)$ aur $\partial \mathbf{r}/\partial t=(v,0)$ parallel hain, giving $a_q=mv\ne 0$ aur $T_1=mv\dot{q}$ (yeh parent Example 3 hai).

$T=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\dot{\theta }}^2$ ke saath: $\frac{\partial T}{\partial \dot{r}}=m\dot{r},\frac{\partial T}{\partial \dot{\theta }}=m{r}^2\dot{\theta }.$ $\dot{r}(m\dot{r})+\dot{\theta }(m{r}^2\dot{\theta })=m{\dot{r}}^2+m{r}^2{\dot{\theta }}^2=2\left(\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\dot{\theta }}^2\right)=2T.✓$ Kyun matter karta hai: yeh factor of $2$ exactly woh reason hai kyun scleronomic systems ke liye energy function ${\sum }_i{p}_i{\dot{q}}_i-L$ equals $2T_2-(T_2-V)=T+V$, giving $H=T+V$. Dekho Energy function and Hamiltonian aur Generalized momentum and conserved quantities.

Pehle $T$ build karo. $\dot{x}=\dot{q}+v$ aur $\dot{y}=\dot{w}$ ke saath, $T=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2)=\frac{1}{2}m[(\dot{q}+v{)}^2+{\dot{w}}^2]=\frac{1}{2}m({\dot{q}}^2+{\dot{w}}^2)+mv\dot{q}+\frac{1}{2}m{v}^2.$ Toh $T_2=\frac{1}{2}m({\dot{q}}^2+{\dot{w}}^2)$, $T_1=mv\dot{q}$, $T_0=\frac{1}{2}m{v}^2$, coordinates $(q,w)$. $\frac{\partial T}{\partial \dot{q}}=m\dot{q}+mv,\frac{\partial T}{\partial \dot{w}}=m\dot{w}.$ $\dot{q}(m\dot{q}+mv)+\dot{w}(m\dot{w})=m{\dot{q}}^2+mv\dot{q}+m{\dot{w}}^2.$ Compare karo $2T=m{\dot{q}}^2+m{\dot{w}}^2+2mv\dot{q}+m{v}^2$ se: $\left({\sum }_i{\dot{q}}_i\frac{\partial T}{\partial {\dot{q}}_i}\right)-2T=-mv\dot{q}-m{v}^2=-(T_1+2T_0).$ Kyun: Euler's theorem degree × term deta hai: $T_2\to 2T_2$, $T_1\to 1\cdot T_1$, $T_0\to 0$. Toh $\sum {\dot{q}}_i{\partial }_{{\dot{q}}_i}T=2T_2+T_1$, jo $2T=2T_2+2T_1+2T_0$ se exactly $-(T_1+2T_0)$ differ karta hai. Leftover time-dependent ($T_1,T_0$) pieces se produce hota hai — exactly isliye rheonomic systems ke liye $H\ne T+V$.

Velocities: ${\dot{x}}_1={\ell }_1\cos {\theta }_1{\dot{\theta }}_1,{\dot{y}}_1={\ell }_1\sin {\theta }_1{\dot{\theta }}_1\Rightarrow {\dot{x}}_1^2+{\dot{y}}_1^2={\ell }_1^2{\dot{\theta }}_1^2.$ ${\dot{x}}_2={\ell }_1\cos {\theta }_1{\dot{\theta }}_1+{\ell }_2\cos {\theta }_2{\dot{\theta }}_2,{\dot{y}}_2={\ell }_1\sin {\theta }_1{\dot{\theta }}_1+{\ell }_2\sin {\theta }_2{\dot{\theta }}_2.$ Square karo aur add karo, $\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2=\cos ({\theta }_1-{\theta }_2)$ use karte hue: ${\dot{x}}_2^2+{\dot{y}}_2^2={\ell }_1^2{\dot{\theta }}_1^2+{\ell }_2^2{\dot{\theta }}_2^2+2{\ell }_1{\ell }_2\cos ({\theta }_1-{\theta }_2){\dot{\theta }}_1{\dot{\theta }}_2.$ $T=\frac{1}{2}{m}_1({\dot{x}}_1^2+{\dot{y}}_1^2)+\frac{1}{2}{m}_2({\dot{x}}_2^2+{\dot{y}}_2^2)$ assemble karo: $\boxed{T=\frac{1}{2}(m_1+m_2){\ell }_1^2{\dot{\theta }}_1^2+\frac{1}{2}{m}_2{\ell }_2^2{\dot{\theta }}_2^2+m_2{\ell }_1{\ell }_2\cos ({\theta }_1-{\theta }_2){\dot{\theta }}_1{\dot{\theta }}_2.}$ Metric read off karo (yaad raho $\frac{1}{2}{\sum }_{i,j}$ pair $\{1,2\}$ ko do baar count karta hai, toh ${\dot{\theta }}_1{\dot{\theta }}_2$ ka coefficient $M_{12}$ hai, $2M_{12}$ nahi): $M_{11}=(m_1+m_2){\ell }_1^2,M_{22}=m_2{\ell }_2^2,M_{12}=M_{21}=m_2{\ell }_1{\ell }_2\cos ({\theta }_1-{\theta }_2).$ Yeh scleronomic hai (koi $t$ nahi) toh $T=T_2$. Coupling poori tarah position-dependent $M_{12}$ mein hai.

Figure geometry dikhata hai: upar fixed pivot, pehli rod (white, length $L_1$) bob $m_1$ tak angle ${\theta }_1$ par vertical dashed line se, aur doosri rod (blue, length $L_2$) $m_1$ se latakkar bob $m_2$ tak angle ${\theta }_2$ par apne vertical se. Do rods ke beech ka angle difference ${\theta }_1-{\theta }_2$ woh hai jo $M_{12}$ ke cosine ke andar appear karta hai — jab rods align hoti hain, woh cosine $1$ hota hai aur coupling strongest hoti hai.

${\theta }_1={\theta }_2$ par: $\cos ({\theta }_1-{\theta }_2)=1$. Equal masses/lengths ke saath: $M=\left(\begin{array}{cc}2m{\ell }^2& m{\ell }^2\\ m{\ell }^2& m{\ell }^2\end{array}\right).$ Positive-definite test (Sylvester): top-left entry $2m{\ell }^2>0$; determinant $\det M=(2m{\ell }^2)(m{\ell }^2)-(m{\ell }^2{)}^2=2m^2{\ell }^4-m^2{\ell }^4=m^2{\ell }^4>0.$ Dono leading minors positive ⇒ positive-definite ✓. Yeh guarantee karta hai ki $T_2>0$ kisi bhi nonzero velocity ke liye, jaisa physics demand karta hai (motion ki energy negative nahi ho sakti). Dekho Configuration space and the metric tensor.

Coordinate hai $q=\alpha$; spin $\omega t$ imposed hai (rheonomic). Do vectors compute karo. $\frac{\partial \mathbf{r}}{\partial \alpha }=(R\cos \alpha \cos \omega t,R\cos \alpha \sin \omega t,R\sin \alpha ),$ $\frac{\partial \mathbf{r}}{\partial t}=(-R\sin \alpha \omega \sin \omega t,R\sin \alpha \omega \cos \omega t,0).$ $T_2$: $M_{\alpha \alpha }=m(R^2{\cos }^2\alpha {\cos }^2\omega t+R^2{\cos }^2\alpha {\sin }^2\omega t+R^2{\sin }^2\alpha )=m{R}^2({\cos }^2\alpha +{\sin }^2\alpha )=m{R}^2.$ Toh $T_2=\frac{1}{2}m{R}^2{\dot{\alpha }}^2.$ $T_1$: $a_{\alpha }=m{\partial }_{\alpha }\mathbf{r}\cdot {\partial }_t\mathbf{r}=m[R\cos \alpha \cos \omega t(-R\sin \alpha \omega \sin \omega t)+R\cos \alpha \sin \omega t(R\sin \alpha \omega \cos \omega t)+0]$. Pehle do terms $\mp m{R}^2\omega \cos \alpha \sin \alpha \cos \omega t\sin \omega t$ hain — equal magnitude, opposite sign — toh $a_{\alpha }=0\Rightarrow T_1=0.$ (Phir se: hoop ke saath free direction spin-drag direction ke perpendicular hai.) $T_0$: $\frac{1}{2}m[(R\sin \alpha \omega \sin \omega t{)}^2+(R\sin \alpha \omega \cos \omega t{)}^2]=\frac{1}{2}m{R}^2{\omega }^2{\sin }^2\alpha .$ $\boxed{T=\frac{1}{2}m{R}^2{\dot{\alpha }}^2+\frac{1}{2}m{R}^2{\omega }^2{\sin }^2\alpha .}$ Physically, $T_0=\frac{1}{2}m{R}^2{\omega }^2{\sin }^2\alpha$ sirf position par depend karta hai, toh Lagrangian mein yeh negative potential ki tarah behave karta hai — yeh centrifugal push drive karta hai jo bead ko bahar flingta hai (bada $\sin \alpha$), classic mechanism jo bead ko ek critical $\omega$ se upar bottom se utha sakta hai.

Figure vertical circular hoop dikhata hai, spin axis (vertical diameter) dashed line ke roop mein draw kiya hua, bead bottom se angle $\alpha$ par, aur horizontal radius $R\sin \alpha$ jo bead hoop ke ghoomne par trace karta hai — yahi horizontal radius, squared, $T_0$ ke andar ${\sin }^2\alpha$ produce karta hai.

Effective potential $V_{\text{eff}}(\alpha )=V-T_0=-mgR\cos \alpha -\frac{1}{2}m{R}^2{\omega }^2{\sin }^2\alpha .$ Stationary points: $V_{\text{eff}}^{\prime }(\alpha )=mgR\sin \alpha -m{R}^2{\omega }^2\sin \alpha \cos \alpha =mR\sin \alpha (g-R{\omega }^2\cos \alpha )$. $\alpha =0$ par, $\sin \alpha =0$ ⇒ yeh hamesha ek equilibrium hai. Uski stability: $\alpha =0$ par second derivative expand karo, $V_{\text{eff}}^{\prime \prime }(0)=mgR\cos 0-m{R}^2{\omega }^2({\cos }^{2}0-{\sin }^{2}0)=mgR-m{R}^2{\omega }^2=mR(g-R{\omega }^2).$ Stable (minimum) ke liye $V_{\text{eff}}^{\prime \prime }(0)>0\Rightarrow g>R{\omega }^2$ chahiye. Yeh stability tab kho deta hai jab $\boxed{{\omega }_c=\sqrt{\frac{g}{R}}.}$ ${\omega }_c$ se upar bottom ek maximum ban jaata hai aur ek naya stable angle ${\alpha }_{\star }$ jisme $\cos {\alpha }_{\star }=g/(R{\omega }^2)$ hota hai appear karta hai — bead spinning hoop par upar ride karta hai. Yeh poora analysis $T_0$ ko sahi se alag karne par powered hai; equation of motion ke liye dekho Euler-Lagrange equations jo isko feed karta hai.

Kaun sa coordinate quantity decide karta hai ki $T_1,T_0$ exist karte hain ya nahi? ::: Kya position map $\mathbf{r}(q,t)$ mein ek explicit, imposed $t$ hai (rheonomic) — agar nahi, toh $T=T_2$ sirf. Double pendulum ke liye $M_{12}$ kya hai? ::: $m_2{\ell }_1{\ell }_2\cos ({\theta }_1-{\theta }_2)$. Critical spin for rotating hoop bead? ::: ${\omega }_c=\sqrt{g/R}$.