Exercises — Applying E-L equations to various systems

Step 1 — DOF: one straight line ⇒ 1 DOF, coordinate $q=x$. Step 2 — $T$: speed is $\dot{x}$, so $T=\frac{1}{2}m{\dot{x}}^2$. Step 3 — $V$: nothing pushes it ⇒ $V=0$. Step 4 — $L$: $L=\frac{1}{2}m{\dot{x}}^2$. Derivatives: $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$ (this is ordinary momentum!), and $\frac{\partial L}{\partial x}=0$ (no $x$ appears in $L$). Step 5 — E–L: $\frac{d}{dt}(m\dot{x})-0=0\Rightarrow \boxed{m\ddot{x}=0}$. What it means: no force ⇒ no acceleration ⇒ constant velocity. This is Newton's first law falling out of the recipe.

Coordinate/sign convention: $y$ is measured downward from the natural length, so increasing $y$ means the mass has descended. $\frac{1}{2}m{\dot{y}}^2$ contains $\dot{y}$ (a velocity) ⇒ it is $T$. $\frac{1}{2}k{y}^2-mgy$ depends only on position $y$ ⇒ it is $V$. Its two pieces:

• spring stored energy $\frac{1}{2}k{y}^2$ (always positive, grows with stretch);
• gravitational potential $V_{\text{grav}}=-mgy$, because $y$ is measured downward and $V_{\text{grav}}=mg(\text{height})$ with height $=-y$. So dropping (larger $y$) lowers $V_{\text{grav}}$, matching the minus sign. $\boxed{L=\frac{1}{2}m{\dot{y}}^2-\frac{1}{2}k{y}^2+mgy}.$

Full E–L derivation on this page (so nothing is deferred): measure $\theta$ from the downward vertical.

• Step 2 — $T$: the bob moves on a circle of radius $\ell$, so its arc-speed is $\ell \dot{\theta }$ and $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$.
• Step 3 — $V$: its height above the lowest point is $\ell (1-\cos \theta )$, so $V=mg\ell (1-\cos \theta )$.
• Step 4 — $L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2-mg\ell (1-\cos \theta )$, with $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }$ and $\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta$.
• Step 5 — E–L: $\frac{d}{dt}(m{\ell }^2\dot{\theta })+mg\ell \sin \theta =0\Rightarrow \ddot{\theta }=-\frac{g}{\ell }\sin \theta$.

For small $\theta$, $\sin \theta \approx \theta$, giving $\ddot{\theta }=-\frac{g}{\ell }\theta$ — the 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 $T_{\text{period}}=\frac{2\pi }{\omega }\approx \frac{6.2832}{6.261}\approx 1.0035\text{s}$.

Do not skip the recipe — build the EOM first.

• Step 1 — DOF: the block is stuck to the slope surface ⇒ 1 DOF, coordinate $q=s$ = distance travelled along the incline.
• Step 2 — $T$: the block's speed along the slope is $\dot{s}$, so $T=\frac{1}{2}m{\dot{s}}^2$.
• Step 3 — $V$: sliding a distance $s$ down the slope lowers the height by $s\sin \alpha$, so $V=-mgs\sin \alpha$.
• Step 4 — $L=\frac{1}{2}m{\dot{s}}^2+mgs\sin \alpha$, with $\frac{\partial L}{\partial \dot{s}}=m\dot{s}$ and $\frac{\partial L}{\partial s}=mg\sin \alpha$.
• Step 5 — E–L: $\frac{d}{dt}(m\dot{s})-mg\sin \alpha =0\Rightarrow \ddot{s}=g\sin \alpha$. (The normal force never appeared — choosing $s$ along the slope absorbed the constraint.)

Now plug numbers: $a=\ddot{s}=g\sin 30^{\circ }=9.8\times 0.5=4.9{\text{m/s}}^2$ (constant). Since $a$ is constant and it starts from rest, $v^2=2as$ gives $v=\sqrt{2\times 4.9\times 2}=\sqrt{19.6}\approx 4.427\text{m/s}$. 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 ✔.

Build the EOM from the recipe (no memorized formula).

• Step 1 — DOF: the inextensible string ties the two masses together: if $m_1$ descends by $x$, then $m_2$ rises by $x$. So one coordinate $x$ = descent of $m_1$ fixes everything ⇒ 1 DOF. The string tension is absorbed by this choice.
• Step 2 — $T$: both masses move at speed $|\dot{x}|$, so $T=\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{x}}^2=\frac{1}{2}(m_1+m_2){\dot{x}}^2$.
• Step 3 — $V$: $m_1$ drops by $x$ (loses PE: $-m_{1}gx$); $m_2$ rises by $x$ (gains PE: $+m_{2}gx$). So $V=-m_{1}gx+m_{2}gx=-(m_1-m_2)gx$.
• Step 4 — $L=\frac{1}{2}(m_1+m_2){\dot{x}}^2+(m_1-m_2)gx$, with $\frac{\partial L}{\partial \dot{x}}=(m_1+m_2)\dot{x}$ and $\frac{\partial L}{\partial x}=(m_1-m_2)g$.
• Step 5 — E–L: $(m_1+m_2)\ddot{x}-(m_1-m_2)g=0\Rightarrow \ddot{x}=\frac{(m_1-m_2)g}{m_1+m_2}$.

Plug numbers: $\ddot{x}=\frac{(3-2)(9.8)}{3+2}=\frac{9.8}{5}=1.96{\text{m/s}}^2.$ Positive ⇒ heavier $m_1$ descends, as expected.

Coordinate/sign convention: work in the horizontal plane. Put the origin on the rotation axis; the wire lies along a rotating radial direction. Let $r\ge 0$ be distance from the axis (positive outward along the wire, the red arrow) and $\theta$ the wire's angle in the plane. Step 1 — DOF: the wire forces the angle to be $\theta =\Omega t$ (not free!), so only $r$ is a free coordinate ⇒ 1 DOF, $q=r$. Step 2 — $T$: in plane polar coordinates $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$ — the two velocity pieces are the red arrow ($\dot{r}$, radial) and the blue arrow ($r\dot{\theta }$, tangential). But $\dot{\theta }=\Omega$ is fixed by the constraint, so $T=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\Omega }^2.$ 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: $\frac{d}{dt}(m\dot{r})-mr{\Omega }^2=0\Rightarrow \boxed{\ddot{r}={\Omega }^{2}r}$. What it means: note the plus sign. This is not oscillation ($\ddot{r}=-{\omega }^{2}r$ would be) but runaway growth: $r(t)=A{e}^{\Omega t}+B{e}^{-\Omega t}$. The bead is flung outward exponentially. The $mr{\Omega }^2$ term is exactly the centrifugal effect, appearing automatically because we plugged the constraint into $T$.

Deriving the conserved quantity on this page (nothing deferred):

• Step 2 — $T$: in polar coordinates the velocity has a radial part $\dot{r}$ and a tangential part $r\dot{\theta }$, so $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$.
• Step 3 — $V=V(r)$, independent of $\theta$.
• Step 4 — $L=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)-V(r)$. Notice $\theta$ does not appear in $L$ — only $\dot{\theta }$ does. Such a coordinate is called cyclic (ignorable) (see 2.1.07-Cyclic-coordinates-and-conservation-laws).
• $\theta$ E–L: $\frac{\partial L}{\partial \theta }=0$, so $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta }}\right)=0$, meaning $\frac{\partial L}{\partial \dot{\theta }}$ is constant in time. Computing it: $p_{\theta }\equiv \frac{\partial L}{\partial \dot{\theta }}=m{r}^2\dot{\theta }=\text{const}.$ This conserved $p_{\theta }$ is the particle's angular momentum about the centre: mass $\times$ radius $\times$ tangential speed $=m\cdot r\cdot (r\dot{\theta })=m{r}^2\dot{\theta }$. (The general principle "a symmetry gives a conserved quantity" is Noethers-theorem: here rotational symmetry of $V(r)$ conserves angular momentum.)

Now the numbers. Conservation means $m{r}_1^2{\dot{\theta }}_1=m{r}_2^2{\dot{\theta }}_2$, so ${\dot{\theta }}_2={\dot{\theta }}_1\frac{r_1^2}{r_2^2}=2\cdot \frac{1^2}{2^2}=0.5\text{rad/yr}.$ What it means: four times the radius-squared ⇒ four times slower sweep. This is Kepler's "equal areas in equal times" hiding inside the cyclic $\theta$ coordinate.

Coordinate/sign convention: fixed lab axes with origin at the support's starting point; $x$ points right, $y$ points up (as in figure s02). The support sits at $(X(t),0)$ and $\theta$ is measured from the downward vertical (positive = bob swung to the right). A note on $X(t)$ being given: because $X(t)$ is a prescribed function of time, the Lagrangian depends on time explicitly (through $\dot{X}$ and $\ddot{X}$). This is a rheonomic (time-dependent) situation, but the Euler–Lagrange equation $\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta }=0$ still holds unchanged for the one free coordinate $\theta$ — Hamilton's principle never required $L$ to be time-independent. Step 1 — DOF: the support motion $X(t)$ is given (not free), so only $\theta$ is free ⇒ 1 DOF. Step 2 — $T$: the bob's position (read off figure s02): $x=X(t)+\ell \sin \theta$, $y=-\ell \cos \theta$. Velocities: $\dot{x}=\dot{X}+\ell \dot{\theta }\cos \theta$, $\dot{y}=\ell \dot{\theta }\sin \theta$. Speed squared: $v^2={\dot{x}}^2+{\dot{y}}^2={\dot{X}}^2+2\ell \dot{X}\dot{\theta }\cos \theta +{\ell }^2{\dot{\theta }}^2.$ (The cross terms ${\cos }^2+{\sin }^2=1$ collapse the ${\ell }^2{\dot{\theta }}^2$ pieces.) So $T=\frac{1}{2}m{v}^2$. Step 3 — $V$: height is $y=-\ell \cos \theta$ ⇒ $V=-mg\ell \cos \theta$. 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 (treat $\dot{X},\dot{\theta }$ as independent variables): $\frac{\partial L}{\partial \dot{\theta }}=m\ell \dot{X}\cos \theta +m{\ell }^2\dot{\theta }.$ $\frac{\partial L}{\partial \theta }=-m\ell \dot{X}\dot{\theta }\sin \theta -mg\ell \sin \theta .$ Time-derivative of the momentum: $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta }}\right)=m\ell \ddot{X}\cos \theta -m\ell \dot{X}\dot{\theta }\sin \theta +m{\ell }^2\ddot{\theta }.$ Step 5 — E–L (subtract $\partial L/\partial \theta$): the $-m\ell \dot{X}\dot{\theta }\sin \theta$ terms cancel, leaving $m{\ell }^2\ddot{\theta }+m\ell \ddot{X}\cos \theta +mg\ell \sin \theta =0.$ Divide by $m{\ell }^2$: $\boxed{\ddot{\theta }=-\frac{g}{\ell }\sin \theta -\frac{\ddot{X}}{\ell }\cos \theta }.$ What it means: if the support is still ($\ddot{X}=0$) we recover the ordinary pendulum. The extra $-\frac{\ddot{X}}{\ell }\cos \theta$ is a pseudo-forcing from the accelerating support — driving the pendulum sideways feels like an extra horizontal gravity.

Coordinate/sign convention: fixed lab axes with origin at the cart's starting point; $x$ points right, $y$ points up (figure s03). The cart's top-attachment point is at $(X,0)$ (cart stays on the table, height fixed) and $\theta$ is measured from the downward vertical. Unlike Problem 4.1, here $X$ is a free coordinate, not a prescribed function. 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 give (same binomial expansion as Problem 4.1): $v_m^2={\dot{X}}^2+2\ell \dot{X}\dot{\theta }\cos \theta +{\ell }^2{\dot{\theta }}^2.$ So $T=\frac{1}{2}M{\dot{X}}^2+\frac{1}{2}m({\dot{X}}^2+2\ell \dot{X}\dot{\theta }\cos \theta +{\ell }^2{\dot{\theta }}^2).$ Step 3 — $V$: $V=-mg\ell \cos \theta$ (only the bob has height; cart stays level). 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 $X$ does not appear in $L$ (only $\dot{X}$), so $X$ is a cyclic coordinate: $\frac{\partial L}{\partial \dot{X}}=(M+m)\dot{X}+m\ell \dot{\theta }\cos \theta ,\frac{\partial L}{\partial X}=0.$ E–L ⇒ this generalized momentum is conserved: $\boxed{(M+m)\dot{X}+m\ell \dot{\theta }\cos \theta =\text{const}}$ — the total horizontal momentum (cart + bob) is conserved, because nothing pushes the system sideways. This is 2.1.07-Cyclic-coordinates-and-conservation-laws in action. Differentiating this conserved quantity once in time gives the cart's EOM: $\boxed{(M+m)\ddot{X}+m\ell \ddot{\theta }\cos \theta -m\ell {\dot{\theta }}^2\sin \theta =0}.$

$\theta$-equation — now do the full E–L on $\theta$: $\frac{\partial L}{\partial \dot{\theta }}=m\ell \dot{X}\cos \theta +m{\ell }^2\dot{\theta }.$ $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta }}\right)=m\ell \ddot{X}\cos \theta -m\ell \dot{X}\dot{\theta }\sin \theta +m{\ell }^2\ddot{\theta }.$ $\frac{\partial L}{\partial \theta }=-m\ell \dot{X}\dot{\theta }\sin \theta -mg\ell \sin \theta .$ Subtract $\frac{\partial L}{\partial \theta }$ from the time-derivative; the two $-m\ell \dot{X}\dot{\theta }\sin \theta$ terms cancel: $m\ell \ddot{X}\cos \theta +m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0.$ Divide by $m\ell$: $\boxed{\ell \ddot{\theta }+\ddot{X}\cos \theta +g\sin \theta =0}.$ Consistency check: if $M\to \infty$ the cart cannot move, $\ddot{X}\to 0$, and the $\theta$-equation becomes $\ell \ddot{\theta }+g\sin \theta =0$, i.e. $\ddot{\theta }=-\frac{g}{\ell }\sin \theta$ — the fixed pendulum. ✔ The two boxed equations (cart + bob) are the complete pair of equations of motion for this 2-DOF system.

Use the conserved horizontal momentum found in Problem 5.1. It was $0$ at the start (system at rest) and stays $0$: $(M+m)\dot{X}+m\ell \dot{\theta }\cos \theta =0.$ At $\theta =0$, $\cos \theta =1$: $\dot{X}=-\frac{m\ell \dot{\theta }}{M+m}=-\frac{(1)(2)(1)}{3+1}=-0.5\text{m/s}.$ The minus sign shows the cart recoils opposite to the bob's swing — Newton's third law, obtained without ever writing a force.

• How do you spot a cyclic coordinate? ::: it appears in $L$ only as $\dot{q}$, never as $q$.
• Sign test: $\ddot{q}=+cq$ means? ::: exponential runaway, not oscillation.
• Why does the cross term $2\ell \dot{X}\dot{\theta }\cos \theta$ matter? ::: it couples drive and swing; dropping it kills the interaction.
• 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).