Classical Mechanics — the Track

Resolve forces along the incline. Gravity contributes $mg\sin \theta$ down-slope; the normal force is $N=mg\cos \theta$, so kinetic friction is $\mu N=\mu mg\cos \theta$ up-slope. Newton's second law along the slope:

$$ma=mg\sin \theta -\mu mg\cos \theta ⟹a=g(\sin \theta -\mu \cos \theta ).$$

It slides only if that acceleration is positive, i.e. $\tan \theta >\mu$; otherwise static friction holds it and $a=0$.

They share a speed; take down-positive for $m_1$ and up-positive for $m_2$. Newton's second law on each is $m_{1}g-T=m_{1}a$ and $T-m_{2}g=m_{2}a$. Add to cancel the tension:

$$a=\frac{m_1-m_2}{m_1+m_2}g,T=\frac{2m_1{m}_2}{m_1+m_2}g.$$

Checks: equal masses give $a=0$, $T=mg$; letting $m_2\to 0$ gives free fall, $a\to g$.

In a short interval the rocket throws back mass $-dm>0$ at relative speed $u$; momentum conservation gives $mdv=-udm$, i.e. $dv=-udm/m$. Integrate from $m_0$ to $m$:

$$v=u\ln \frac{m_0}{m}.$$

The Tsiolkovsky equation: final speed grows only with the logarithm of the mass ratio, which is why reaching orbit is mostly fuel.

Momentum and kinetic energy are both conserved; the clean shortcut is that in an elastic collision the relative velocity reverses, $v_1^{\prime }-v_2^{\prime }=-(v_1-v_2)$. Together with momentum:

$$v_1^{\prime }=\frac{m_1-m_2}{m_1+m_2}v,v_2^{\prime }=\frac{2m_1}{m_1+m_2}v.$$

Equal masses exchange velocities ($v_1^{\prime }=0,v_2^{\prime }=v$); a light ball off a heavy wall ($m_2\gg m_1$) bounces straight back.

Two stages, two different conserved quantities. The collision is inelastic, so use momentum: $mv=(m+M)V$ gives $V=\frac{mv}{m+M}$. After the collision energy is conserved, $\frac{1}{2}(m+M)V^2=(m+M)gh$:

$$h=\frac{V^2}{2g}=\frac{1}{2g}{\left(\frac{mv}{m+M}\right)}^2.$$

You can't run energy through the collision — most of the bullet's kinetic energy goes to heat and deformation; only momentum survives it.

The tension is radial, so it exerts no torque about the hole and angular momentum $L=mvr$ is conserved. Halving the radius doubles the speed:

$$v^{\prime }=2v,\frac{1}{2}m{v}^{\prime 2}=4\times \frac{1}{2}m{v}^2.$$

The kinetic energy quadruples — that energy is exactly the work you did pulling the string inward against the centripetal pull.

Equilibrium balances spring and weight: $k{x}_0=mg$, so $x_0=mg/k$. Measure the displacement $y=x-x_0$ from there; the net force is $-k(x_0+y)+mg=-ky$ — the constant terms cancel — so

$$m\ddot{y}=-ky⟹\omega =\sqrt{k/m}.$$

Gravity only shifts the equilibrium point. The restoring curvature is still $k$, so the frequency is the same as with no gravity at all.

Try $x=e^{rt}$: the characteristic equation $m{r}^2+br+k=0$ gives $r=-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$. When $\gamma <{\omega }_0$ the roots are complex, $r=-\gamma \pm i{\omega }^{\prime }$, a decaying oscillation:

$$x(t)=e^{-\gamma t}\cos ({\omega }^{\prime }t+\phi ),{\omega }^{\prime }=\sqrt{{\omega }_0^2-{\gamma }^2}.$$

Critical damping is the boundary $\gamma ={\omega }_0$, i.e. $b=2\sqrt{mk}$ — the fastest return to rest without overshooting.

The length is $S=\int \sqrt{1+y^{\prime 2}}dx$, so the integrand $F=\sqrt{1+y^{\prime 2}}$ depends on $y^{\prime }$ but not on $y$. The Euler–Lagrange equation $\frac{d}{dx}\frac{\partial F}{\partial y^{\prime }}=\frac{\partial F}{\partial y}=0$ then makes $\partial F/\partial y^{\prime }$ constant:

$$\frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}=\text{const}⟹y^{\prime }=\text{const}⟹y=mx+c.$$

A constant slope is a straight line. The same machinery with a different integrand gives the brachistochrone and the hanging catenary.

Use the angle $\theta$ from vertical. The bob's speed is $\ell \dot{\theta }$, so the kinetic energy is $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$, and taking the pivot as the reference, $U=-mg\ell \cos \theta$. The Lagrangian:

$$L=T-U=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+mg\ell \cos \theta .$$

The Euler–Lagrange equation $\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta }=0$ gives $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0$, i.e.

$$\ddot{\theta }=-\frac{g}{\ell }\sin \theta \stackrel{\theta \ll 1}{\to }\ddot{\theta }=-\frac{g}{\ell }\theta ,\omega =\sqrt{g/\ell }.$$

The bead's speed has two pieces: $R\dot{\theta }$ along the hoop and $R\sin \theta \omega$ around the axis. So $T=\frac{1}{2}m{R}^2({\dot{\theta }}^2+{\omega }^2{\sin }^2\theta )$ and $U=-mgR\cos \theta$. The Euler–Lagrange equation gives

$$m{R}^2\ddot{\theta }=mR\sin \theta (R{\omega }^2\cos \theta -g).$$

Equilibria: $\theta =0$ (the bottom) and $\cos \theta =g/(R{\omega }^2)$. The tilted one exists only once ${\omega }^2>g/R$; past that spin rate it becomes stable and the bottom goes unstable — a pitchfork bifurcation.

The radial coordinate sees the effective potential $U_{\mathrm{eff}}(r)=-\frac{k}{r}+\frac{L^2}{2m{r}^2}$, the second term being the centrifugal barrier. A circular orbit sits at its minimum, $d{U}_{\mathrm{eff}}/dr=0$:

$$\frac{k}{r^2}-\frac{L^2}{m{r}^3}=0⟹r=\frac{L^2}{mk}.$$

Since $U_{\mathrm{eff}}$ has a true minimum there, the orbit is stable — small radial perturbations oscillate, and for a $1/r^2$ force the orbit closes into an ellipse.

In the rotating frame the Coriolis acceleration is $-2\mathbf{\omega }\times \mathbf{v}$. For a body falling at speed $v=gt$ the eastward component is $2\omega gt\cos \lambda$; integrating twice from rest gives $x=\frac{1}{3}\omega g\cos \lambda t^3$, and the fall time is $t=\sqrt{2h/g}$:

$$x=\frac{1}{3}\omega \cos \lambda \sqrt{\frac{8h^3}{g}}.$$

Eastward in either hemisphere, largest at the equator — about $2\text{cm}$ for a 100 m drop.

In the rotating frame the bob feels a centrifugal acceleration ${\omega }^{2}R\cos \lambda$ directed away from the spin axis. The component perpendicular to the local vertical (the part that tilts the string) is ${\omega }^{2}R\cos \lambda \sin \lambda$, so

$$\alpha \approx \frac{{\omega }^{2}R\cos \lambda \sin \lambda }{g}=\frac{{\omega }^{2}R}{2g}\sin 2\lambda .$$

Zero at the equator and poles, largest at $45^{\circ }$ where it is about $0.1^{\circ }$ — the reason "down" isn't exactly toward the center.

Only the vertical component of Earth's rotation, $\omega \sin \lambda$, twists the local horizontal plane through the Coriolis force, so the swing plane precesses at

$$\Omega =\omega \sin \lambda ,T_{\mathrm{precess}}=\frac{2\pi }{\omega \sin \lambda }=\frac{\text{1 sidereal day}}{\sin \lambda }.$$

One day at the pole, never at the equator, about 32 hours in Paris ($\lambda \approx 49^{\circ }$) — Foucault's 1851 public proof that the Earth turns.

About the end the rod's moment of inertia is $I=\frac{1}{3}m{L}^2$. Gravity acts at the center, a distance $L/2$ away, giving a restoring torque $-mg\frac{L}{2}\sin \theta \approx -mg\frac{L}{2}\theta$. Then $I\ddot{\theta }=-mg\frac{L}{2}\theta$, so

$$\omega =\sqrt{\frac{mgL/2}{\frac{1}{3}m{L}^2}}=\sqrt{\frac{3g}{2L}},T=2\pi \sqrt{\frac{2L}{3g}}.$$

Shorter period than a simple pendulum of length $L$ (which has $T=2\pi \sqrt{L/g}$): the rod behaves like a simple pendulum of length $2L/3$, its center of oscillation.

Let $x_1,x_2$ be the displacements. The equations of motion are $m{\ddot{x}}_1=-(k+k^{\prime })x_1+k^{\prime }{x}_2$ and $m{\ddot{x}}_2=k^{\prime }{x}_1-(k+k^{\prime })x_2$, i.e. $m\ddot{\mathbf{x}}=-K\mathbf{x}$ with the stiffness matrix

$$K=\left(\begin{array}{cc}k+k^{\prime }& -k^{\prime }\\ -k^{\prime }& k+k^{\prime }\end{array}\right).$$

The normal modes are the eigenvectors of $K$, with ${\omega }^2=\lambda /m$. Its eigenvalues are $k$ (eigenvector $(1,1)$) and $k+2k^{\prime }$ (eigenvector $(1,-1)$), so

$${\omega }_1=\sqrt{k/m}\text{(masses move together, coupling spring relaxed)},{\omega }_2=\sqrt{(k+2k^{\prime })/m}\text{(opposed, coupling spring doubly stretched)}.$$

This is exactly expanding in the eigenbasis: the normal coordinates are the eigenvectors of $K$, and in them the coupled system decouples into two independent oscillators.

With momentum $p=m\dot{x}$, the Hamiltonian is the total energy written in the variables $(x,p)$:

$$H=\frac{p^2}{2m}+\frac{1}{2}k{x}^2.$$

Hamilton's equations $\dot{x}=\partial H/\partial p$ and $\dot{p}=-\partial H/\partial x$ give $\dot{x}=p/m$ and $\dot{p}=-kx$. Eliminating $p$ returns $\ddot{x}=-{\omega }^{2}x$ with $\omega =\sqrt{k/m}$ — and the phase-space orbits are ellipses of constant $H$.