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In the rotating-hoop bead problem, use a signed angle in the hoop plane from the downward vertical. The gravitational potential energy is $-mga\cos \theta$, the rotating-frame speed is $a\dot{\theta }$, and the distance from the rotation axis is $a\sin \theta$. The centrifugal potential givesGravity is stationary in these rotating coordinates, and the frictionless constraint does no work. Projecting the rotating equation along the hoop, or differentiating its effective potential in the one-dimensional Euler-Lagrange equation, givesIf ${\omega }^2>g/a$, the two off-axis positions areTheir unsigned angle to the downward vertical is the same; they lie on opposite sides of the hoop. The second derivative of the effective potential there is $m{\omega }^2{a}^2{\sin }^2{\theta }_0>0$. Equivalently, writing $\theta ={\theta }_0+\delta$ gives to first orderThe same holds at $-{\theta }_0$. Thus both equilibria are stable, with small-oscillation angular frequency $|\omega |\sin {\theta }_0$.

For $0<v<c$, the Lorentz transformation with coincident origins isFor two simultaneous events in $S$, $\Delta t=0$, so $\Delta t^{\prime }=-\gamma v\Delta x/c^2$. Distinct spatial positions therefore generally give different times in the boosted frame: this is relativity of simultaneity.

Choose the emission events at $(t,x)=(0,0)$ and $(0,d)$. After emission, the photon world lines are $x=ct$ and $x=d+ct$. Applying the Lorentz transformation to either givesThus the two photons move at speed $c$ in $S^{\prime }$ on parallel lines whose intercepts differ by $\gamma (1+v/c)d$. Measuring their positions at the same $t^{\prime }$ after both emissions gives the photon separation under a collinear Lorentz boost:This separation is constant. It is not the contracted distance between the stationary sources: the two emission events are not simultaneous in $S^{\prime }$, and one photon has already moved when the other is emitted.

Set $\theta =0$ initially and choose its positive sense to agree with the initial tangential motion. The inward radial component is $\dot{r}(0)=-4\cos (\pi /3)=-2$, while the tangential speed is $4\sin (\pi /3)=2\sqrt{3}$. HenceSubstitution in the Binet equation gives $12(u^{\prime \prime }+u)=3+9u$, or $u^{\prime \prime }+u/4=1/4$. This is an orbit equation for combined inverse-square and inverse-cube attraction. Solving the constant-coefficient equation and applying both initial conditions givesOn $0\le \theta \le 2\pi$, the sine is nonnegative, so $r$ stays finite and positive. At $\theta =2\pi$, $u=1$, hence $r=1$ and the particle is back at its original spatial position after one revolution. But $u^{\prime }(2\pi )=-1/\sqrt{3}$, so its radial velocity is now $+2$, rather than the original $-2$. The return is not a periodic return in position and velocity.

After that revolution, the first zero of $u$ is at $\theta =8\pi /3$, where $\sin (\theta /2)=-\sqrt{3}/2$. The physical branch therefore has $r\to \infty$ as $\theta \uparrow 8\pi /3$; it must not be continued into negative $u$. Since $dt/d\theta =1/(l{u}^2)$ and $u$ has a simple zero, reaching infinity takes infinite time. As a check, the specific potential energy is $\Phi (r)=-3/r-9/(2r^2)$ and the conserved energy per unit mass is $4^2/2+\Phi (1)=1/2$, giving escape speed $1$ at infinity, also obtained from $-l{u}^{\prime }(8\pi /3)=1$.

Differentiation gives ${\mathbf{r}}^{\prime }=e^t(\sqrt{2},-\sin t-\cos t,\cos t-\sin t)$. The squared speed is $4e^{2t}$, so the arc length isThe speed never vanishes, so this is a regular curve throughout its parameter range.

Use the arch geometry of two hinged rods sliding on a floor, with $A=O=(0,0)$ and $C$ on the other side of $B$. As long as $C$ remains in contact with the floor,The centres of mass of the two rods are $(l\cos \theta ,l\sin \theta )$ and $(3l\cos \theta ,l\sin \theta )$. Their squared speeds are respectively $l^2{\dot{\theta }}^2$ and $l^2(9{\sin }^2\theta +{\cos }^2\theta ){\dot{\theta }}^2$. Each uniform rod has moment of inertia $M{l}^2/3$ about its midpoint, and their angular velocities are $\dot{\theta }$ and $-\dot{\theta }$. Applying the kinetic energy decomposition about the center of mass to each rod givesThe gravitational potential energy is $U=2Mgl\sin \theta$. The fixed pivot does no work; neither does the frictionless floor reaction while $C$ slides horizontally. Conservation of energy from rest therefore givesOn the falling branch $\dot{\theta }$ is the negative square root. Just before the hinge reaches the floor, $\theta =0$, and its speed is $|\dot{B}|=2l|\dot{\theta }|$. Thus, under the maintained-contact interpretation,

There is a contact hypothesis behind this calculation. A floor can push upward but cannot pull downward. For $s=\sin \theta$ and $a=\sin \alpha$, the required normal reaction at $C$, found from the second rod's linear and angular equations, isFor the usual acute arch, this stays nonnegative throughout $0\le \theta \le \alpha$ exactly when $\sin \alpha \le 35/54$. Indeed, for $s>0$, its nonnegativity is $a\le 1/(18s)+11s/6+2s^3$, whose minimum is $35/54$ at $s=1/6$. For larger initial angles, the endpoint lifts before the hinge hits the floor, and the one-coordinate formula no longer governs the whole motion. The displayed impact speed is the intended floor-constrained answer; for unrestricted initial angles, maintained contact is an additional assumption. Subsequent free motion and possible reimpact would require a different model and contact/impact conditions.

The PDF gives initial position $(x^{\prime },y^{\prime },z^{\prime })=(1,0,0)$ and initial rotating-frame velocity zero. Use those values; the TeX erroneously places dots over the position coordinates.

For rotation about the third axis, the supplied acceleration relation and Newton's second law give the equations of motion in a rotating frameThus $z^{\prime }=0$. Set $\zeta =x^{\prime }+i{y}^{\prime }$ to combine the planar equations:The two characteristic roots are $i\omega$ and $-3i\omega$. From $\zeta (0)=1$ and $\dot{\zeta }(0)=0$,The triple-angle formulas therefore give the particularly simple astroid motion of a harmonic oscillator in a rotating frameThe orbit satisfies $|x^{\prime }{|}^{2/3}+|y^{\prime }{|}^{2/3}=1$; the zero speed at each cusp is consistent with the smooth time-parametrized motion.

Its rotating-frame speed isHence, for $\omega >0$, the first maximum occurs at $t=\pi /(4\omega )$, andThe maximum repeats periodically. For a signed negative angular frequency, use $\pi /(4|\omega |)$ and $3|\omega |/2$; at $\omega =0$ the particle remains at rest.