Exercises — Lagrangian L = T − V

WHAT: name the two energies. WHY: $L=T-V$ needs both. Kinetic energy of a moving mass is $T=\frac{1}{2}m{\dot{x}}^2$ — the "moving-energy". Since motion is horizontal, gravity's potential $V=mgy$ never changes, so we take $V=0$ (constant potential contributes nothing to the equation of motion). $L=T-V=\frac{1}{2}m{\dot{x}}^2-0=\boxed{\frac{1}{2}m{\dot{x}}^2}$ WHAT IT LOOKS LIKE: a bead sliding on a flat wire; nothing pushes it, so $L$ is pure kinetic.

A stretched/compressed spring stores energy: $V=\frac{1}{2}k{x}^2$. The kinetic term is the same $T=\frac{1}{2}m{\dot{x}}^2$. $L=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{x}^2.$ WHY the sign: the spring energy is subtracted because $L=T-V$; that subtraction will produce the restoring force $-kx$ with the correct sign.

A generalized coordinate is any independent number that fixes the configuration. The bob is tied to a rigid rod, so $x=\ell \sin \theta$ and $y=-\ell \cos \theta$ are not independent — they obey the constraint $x^2+y^2={\ell }^2$. That's two Cartesian numbers minus one constraint $=$ one true degree of freedom, exactly captured by $\theta$. Choosing $\theta$ makes the constraint vanish automatically and the rod's tension never appears.

$L=\frac{1}{2}m{\dot{y}}^2-mgy$. Step 1 (momentum term): $\frac{\partial L}{\partial \dot{y}}=m\dot{y}$. WHY: this is the generalized momentum $p_y$. Step 2 (its rate): $\frac{d}{dt}(m\dot{y})=m\ddot{y}$. Step 3 (position term): $\frac{\partial L}{\partial y}=-mg$. Step 4 (EL): $m\ddot{y}-(-mg)=0\Rightarrow \boxed{\ddot{y}=-g}$. WHAT IT LOOKS LIKE: acceleration points down (negative), matching free fall. ✓

$\frac{\partial L}{\partial \dot{x}}=m\dot{x}\Rightarrow \frac{d}{dt}=m\ddot{x}$. And $\frac{\partial L}{\partial x}=-kx$. EL: $m\ddot{x}-(-kx)=0\Rightarrow m\ddot{x}+kx=0$. Comparing with the standard form $\ddot{x}=-{\omega }^{2}x$ gives ${\omega }^2=k/m$, so $\boxed{\omega =\sqrt{k/m}}$. Numeric check: if $m=2\text{ kg}$, $k=8\text{ N/m}$, then $\omega =\sqrt{8/2}=2\text{ rad/s}$.

$L=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\varphi }}^2)-V(r)$. WHAT: notice $\varphi$ does not appear in $L$ (only $\dot{\varphi }$ does). Such a coordinate is called cyclic or ignorable. WHY it matters: $\frac{\partial L}{\partial \varphi }=0$, so the EL equation for $\varphi$ becomes $\frac{d}{dt}\frac{\partial L}{\partial \dot{\varphi }}=0$ — meaning $\frac{\partial L}{\partial \dot{\varphi }}$ is constant in time. Compute it: $\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }$. This is conserved — it is the angular momentum $\ell =m{r}^2\dot{\varphi }$. WHAT IT LOOKS LIKE: whenever a coordinate is missing from $L$, its momentum is a conserved constant — Noether's idea in miniature.

$\frac{\partial L^{\prime }}{\partial \dot{x}}=m\dot{x}\Rightarrow \frac{d}{dt}=m\ddot{x}$; $\frac{\partial L^{\prime }}{\partial x}=+kx$. EL: $m\ddot{x}-kx=0\Rightarrow \ddot{x}=+\frac{k}{m}x$. WHY unphysical: the acceleration points away from equilibrium (same sign as $x$). Solutions are $e^{\pm t\sqrt{k/m}}$ — exponential blow-up, not oscillation. A real spring pulls back. This is the concrete proof that only $L=T-V$ gives correct physics.

Bob speed $=\ell \dot{\theta }$, so $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$. Gravity potential $=mg\ell (1-\cos \theta )$; spring potential $=\frac{1}{2}\kappa {\theta }^2$. Total $V=mg\ell (1-\cos \theta )+\frac{1}{2}\kappa {\theta }^2$. $L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2-mg\ell (1-\cos \theta )-\frac{1}{2}\kappa {\theta }^2.$ $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }\Rightarrow \frac{d}{dt}=m{\ell }^2\ddot{\theta }$. $\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta -\kappa \theta$. EL: $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta +\kappa \theta =0$. Small angle ($\sin \theta \approx \theta$): $m{\ell }^2\ddot{\theta }+(mg\ell +\kappa )\theta =0\Rightarrow \boxed{\ddot{\theta }=-\frac{mg\ell +\kappa }{m{\ell }^2}\theta }$. WHAT IT LOOKS LIKE: SHM with $\omega =\sqrt{(mg\ell +\kappa )/(m{\ell }^2)}$ — the spring stiffens the pendulum, raising the frequency, exactly as intuition demands.

WHAT: check whether $L$ depends explicitly on time $t$. Here $\Omega$ is a fixed constant, so $L$ has no explicit $t$ — thus the Jacobi energy function (the Hamiltonian-like quantity) $h=\dot{\theta }\frac{\partial L}{\partial \dot{\theta }}-L$ is conserved. Compute: $h=\dot{\theta }(m{R}^2\dot{\theta })-L=\frac{1}{2}m{R}^2{\dot{\theta }}^2-\frac{1}{2}m{R}^2{\Omega }^2{\sin }^2\theta +mgR(1-\cos \theta )$. WHY it's NOT $T+V$: the ${\Omega }^2{\sin }^2\theta$ term enters $h$ with a minus sign, unlike in $T+V$ where it's plus. The rotation constraint is rheonomic (imposed externally), so the naive $E=T+V$ is not the conserved quantity — $h$ is. This is the subtle payoff: "$H=T+V$" holds only for time-independent, natural systems; here the driven spin breaks that identity. Takeaway: conservation follows from symmetry of $L$ (here time-translation), not from a memorized formula.

Level-1 skill? ::: Identify $T$, $V$, assemble $L=T-V$. Why does tension never appear in the pendulum/Atwood solutions? ::: Coordinates run along allowed motion, so ideal constraint forces do no work and drop out. What signals a conserved momentum? ::: A coordinate missing from $L$ (cyclic) ⇒ its $\partial L/\partial \dot{q}$ is constant. Critical spin of the rotating hoop? ::: ${\Omega }_c=\sqrt{g/R}$, where a pitchfork bifurcation lifts the bead off the bottom. When is $h=T+V$? ::: Only for natural, time-independent (scleronomic) systems; else derive $h=\sum \dot{q}\partial L/\partial \dot{q}-L$.