Exercises — Generalized momenta and generalized forces

What we do: differentiate $L$ with respect to the velocity to get momentum, and use the potential shortcut for the force. $p_x=\frac{\partial L}{\partial \dot{x}}=m\dot{x}.$ The velocity $\dot{x}$ appears only in $\frac{1}{2}m{\dot{x}}^2$; its derivative is $m\dot{x}$. The $-\frac{1}{2}k{x}^2$ term has no $\dot{x}$, so it dies. $Q_x=-\frac{\partial V}{\partial x}=-\frac{\partial }{\partial x}\left(\frac{1}{2}k{x}^2\right)=-kx.$ Sanity: $p_x=m\dot{x}$ is ordinary momentum, $Q_x=-kx$ is Hooke's restoring force. Both correct.

$p_{\theta }=\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta },\text{units }\mathrm{kg}{\mathrm{m}}^2/s.$ That is angular momentum, not ordinary momentum — because the coordinate is an angle. $Q_{\theta }=\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta ,\text{units }\mathrm{N}\mathrm{m}.$ That is a torque. Recall $V=-mg\ell \cos \theta$, so $-\partial V/\partial \theta =-mg\ell \sin \theta$. ✓

With $V=0$, $L=T$. $p_r=\frac{\partial L}{\partial \dot{r}}=m\dot{r}.$ $p_{\theta }=\frac{\partial L}{\partial \dot{\theta }}=m{r}^2\dot{\theta }.$ Why $r^2$ appears: inside $T$ the $\dot{\theta }$ term is $\frac{1}{2}m{r}^2{\dot{\theta }}^2$; differentiating in $\dot{\theta }$ gives $m{r}^2\dot{\theta }$. This is exactly angular momentum $L_{\text{ang}}=mr{v}_{\perp }$ with $v_{\perp }=r\dot{\theta }$. See Angular momentum.

Way 1 — direct projection. We need $\partial \vec{r}/\partial s$: differentiate each component in $s$: $\frac{\partial \vec{r}}{\partial s}=(\cos \alpha ,-\sin \alpha ).$ Then $Q_s=\vec{F}\cdot \frac{\partial \vec{r}}{\partial s}=(0)(\cos \alpha )+(-mg)(-\sin \alpha )=mg\sin \alpha .$ Way 2 — potential. Height is $y=-s\sin \alpha$, so $V=mgy=-mgs\sin \alpha$, and $Q_s=-\frac{\partial V}{\partial s}=mg\sin \alpha .$ Both agree. This is the familiar "component of gravity along the slope," recovered automatically. ✓ ($s$ is a length, so $Q_s$ is a plain force in newtons.)

Cyclic coordinate: $\theta$ does not appear in $L$ (only $\dot{\theta }$ does), so $\partial L/\partial \theta =0$. By Euler–Lagrange, ${\dot{p}}_{\theta }=0$: $p_{\theta }=m{r}^2\dot{\theta }$ is conserved (angular momentum). See Noether's Theorem for the deep "symmetry ⇒ conservation" reason. Numbers. Conservation gives $r_1^2{\dot{\theta }}_1=r_2^2{\dot{\theta }}_2$ (the $m$ cancels): ${\dot{\theta }}_2={\dot{\theta }}_1\frac{r_1^2}{r_2^2}=3\cdot \frac{2^2}{1^2}=12rad/s.$ The particle spins four times faster when pulled to half the radius — the figure-skater effect.

$L=T=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\omega }^2$. Apply Euler–Lagrange $\frac{d}{dt}\frac{\partial L}{\partial \dot{r}}-\frac{\partial L}{\partial r}=0$: $p_r=\frac{\partial L}{\partial \dot{r}}=m\dot{r},\frac{d}{dt}{p}_r=m\ddot{r}.$ $\frac{\partial L}{\partial r}=mr{\omega }^2.$ So $m\ddot{r}-mr{\omega }^2=0\Rightarrow \boxed{m\ddot{r}=mr{\omega }^2}.$ Interpretation: the right side is an effective generalized force $mr{\omega }^2$ — the centrifugal push — even though $V=0$ and no real force acts along $r$. It came from the $r$-dependence of the kinetic energy. This is why the true general form is $Q_j$ hiding inside $\partial L/\partial q_j$, not merely $-\partial V/\partial q_j$.

Coordinate & velocities. One DOF: $x$. Both masses move at speed $\dot{x}$, so $T=\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{x}}^2=\frac{1}{2}(m_1+m_2){\dot{x}}^2.$ Potential. As $m_1$ drops by $x$, its height falls by $x$: contributes $-m_{1}gx$. As $m_2$ climbs the slope by $x$, its height rises by $x\sin \alpha$: 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 ${\dot{p}}_x=Q_x$: $(m_1+m_2)\ddot{x}=g(m_1-m_2\sin \alpha )\Rightarrow \boxed{\ddot{x}=\frac{g(m_1-m_2\sin \alpha )}{m_1+m_2}}.$ Check limits: $\alpha =90^{\circ }$ gives $\ddot{x}=g(m_1-m_2)/(m_1+m_2)$ — the standard Atwood machine. $\alpha =0$ gives $\ddot{x}=g{m}_1/(m_1+m_2)$ — $m_2$ on a flat floor, only $m_1$ pulls. ✓

$\ddot{x}=\frac{9.8(3-2\sin 30^{\circ })}{3+2}=\frac{9.8(3-2\cdot 0.5)}{5}=\frac{9.8\cdot 2}{5}=3.92m/{\mathrm{s}}^2.$ Positive, so $m_1$ genuinely descends — reasonable since $m_1=3>m_2\sin 30^{\circ }=1$.

$L=\frac{1}{2}m{\ell }^2({\dot{\vartheta }}^2+{\sin }^2\vartheta {\dot{\phi }}^2)+mg\ell \cos \vartheta .$ 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. $\phi$ does not appear in $L$ (only $\dot{\phi }$ does), so $\phi$ is cyclic ⇒ $p_{\phi }=m{\ell }^2{\sin }^2\vartheta \dot{\phi }$ is conserved. Physically it is the angular momentum about the vertical axis. $\vartheta$ does appear (via ${\sin }^2\vartheta$ and $\cos \vartheta$), so $p_{\vartheta }$ is not conserved. Equations of motion. For $\vartheta$: $\frac{d}{dt}{p}_{\vartheta }-\frac{\partial L}{\partial \vartheta }=0$, with $\frac{\partial L}{\partial \vartheta }=m{\ell }^2\sin \vartheta \cos \vartheta {\dot{\phi }}^2-mg\ell \sin \vartheta$: $m{\ell }^2\ddot{\vartheta }=m{\ell }^2\sin \vartheta \cos \vartheta {\dot{\phi }}^2-mg\ell \sin \vartheta .$ For $\phi$: since cyclic, $\frac{d}{dt}{p}_{\phi }=0$, i.e. $m{\ell }^2{\sin }^2\vartheta \dot{\phi }=\text{const}.$ Check: set $\dot{\phi }=0$ (planar swing) → $\ddot{\vartheta }=-\frac{g}{\ell }\sin \vartheta$, the ordinary pendulum. ✓

Set $\ddot{\vartheta }=0$ in the $\vartheta$-equation and cancel $m{\ell }^2\sin {\vartheta }_0$ (nonzero for $0<{\vartheta }_0<\pi$): $0=\cos {\vartheta }_0{\Omega }^2-\frac{g}{\ell }\Rightarrow \boxed{{\Omega }^2=\frac{g}{\ell \cos {\vartheta }_0}}.$ Degenerate checks:

• ${\vartheta }_0\to 0$ (nearly straight down): $\cos {\vartheta }_0\to 1$, ${\Omega }^2\to g/\ell$ — the small-oscillation frequency of a pendulum. ✓
• ${\vartheta }_0\to 90^{\circ }$: $\cos {\vartheta }_0\to 0$, ${\Omega }^2\to \infty$ — you cannot hold the rod horizontal by spinning at any finite rate. Physically true. Numeric: $\ell =1\mathrm{m}$, ${\vartheta }_0=60^{\circ }$, $g=9.8$: ${\Omega }^2=9.8/(1\cdot 0.5)=19.6$, so $\Omega =\sqrt{19.6}\approx 4.427rad/s$.

For any coordinate $q_j$: momentum is $p_j=\partial L/\partial {\dot{q}}_j$ ::: force is $Q_j=-\partial V/\partial q_j$ for conservative systems (or $\partial L/\partial q_j$ in general); if $q_j$ is absent from $L$, then $p_j$ is conserved.