NBPhO 2026

Overview

The bead sits on a frictionless rod that pivots about a horizontal axis at constant angular velocity $\omega$. Because the rotation axis is horizontal, the rod sweeps out a vertical plane and gravity has a component along the rod that varies as $-g\sin(\omega t)$ — the bead is forced at exactly the frequency of the rotation. The most economical viewpoint is the rotating frame in which the rod is fixed: there the bead executes one-dimensional motion along the rod under three contributions — centrifugal pull (outward), gravity (oscillating along the rod), and, in part (iii), a Hooke restoring force.

The structural fact that controls everything is that the equation along the rod is

$$\ddot{x} - \omega^{2}\,x \;=\; -g\sin(\omega t),$$

a negative-stiffness harmonic oscillator with sinusoidal forcing at the resonant frequency of the corresponding positive-stiffness oscillator. The homogeneous part is exponentially unstable (because of the $-\omega^{2}$), but a perfectly tuned particular solution exists at frequency $\omega$, and it generates the circular trajectory of parts (i)–(ii). Part (iii) replaces $-\omega^{2}$ by $\omega_{0}^{2} \equiv k/m - \omega^{2}$, turning the homogeneous equation into a genuine oscillator and converting the circle into a true Lyapunov-stable trajectory provided $\omega_{0}^{2} > 0$ and the forcing avoids resonance with $\omega_{0}$.

---

Setup and equation of motion

Let the rotation axis be horizontal and place the pivot at the origin. Work in the vertical plane perpendicular to the axis, with $\hat{x}$ horizontal and $\hat{y}$ vertical (upward). At time $t$ the rod’s outward unit vector is

$$\hat{r}(t) \;=\; \bigl(\cos\omega t,\; \sin\omega t\bigr).$$

Let $x(t)$ be the (signed) distance of the bead from the pivot along the rod, so the lab-frame position of the bead is

$$\vec{R}(t) \;=\; x(t)\,\hat{r}(t).$$

Differentiating twice and projecting onto $\hat{r}$ gives

$$\bigl(\ddot{\vec{R}}\bigr)\!\cdot\!\hat{r} \;=\; \ddot{x} - \omega^{2}\,x.$$

The forces along $\hat{r}$ come from gravity only (the rod is frictionless, so the constraint force is purely perpendicular to $\hat{r}$):

$$\bigl(-mg\,\hat{y}\bigr)\!\cdot\!\hat{r} \;=\; -mg\sin(\omega t).$$

Newton’s law along the rod is therefore

$$\boxed{\;\ddot{x} \;-\; \omega^{2}\,x \;=\; -g\sin(\omega t)\;} \qquad\text{(without spring).}$$

Why negative stiffness? In the rotating frame, the centrifugal pseudo-force $+m\omega^{2}x$ pushes the bead away from the pivot — it is a destabilising spring. A free bead on a rotating frictionless rod (no gravity) would slide outward exponentially. The trick of part (i) is that the gravitational forcing, acting at exactly the rotation frequency, has a sinusoidal particular solution that is bounded despite the negative stiffness.

---

Part (i) — Sketch of the circular trajectory

Finding the particular solution

Look for a solution synchronised with the forcing, $x(t) = A\sin(\omega t)$. Substituting,

$$-A\,\omega^{2}\sin(\omega t) \;-\; \omega^{2}\,A\sin(\omega t) \;=\; -g\sin(\omega t) \quad\Longrightarrow\quad A \;=\; \frac{g}{2\omega^{2}}.$$

So one solution is

$$x_{p}(t) \;=\; \frac{g}{2\omega^{2}}\sin(\omega t).$$

The corresponding lab-frame trajectory is $\vec{R}(t) = x_{p}(t)\hat{r}(t)$:

$$X(t) \;=\; \frac{g}{2\omega^{2}}\sin(\omega t)\cos(\omega t) \;=\; \frac{g}{4\omega^{2}}\sin(2\omega t),$$ $$Y(t) \;=\; \frac{g}{2\omega^{2}}\sin^{2}(\omega t) \;=\; \frac{g}{4\omega^{2}}\bigl[\,1 - \cos(2\omega t)\,\bigr].$$

Eliminating time,

$$X^{2} \;+\; \left(Y - \tfrac{g}{4\omega^{2}}\right)^{2} \;=\; \left(\tfrac{g}{4\omega^{2}}\right)^{2}.$$

This is a circle of radius $g/(4\omega^{2})$, centred at $\bigl(0,\;g/(4\omega^{2})\bigr)$ — i.e. directly above the pivot, tangent to the horizontal at the pivot itself.

The sketch

                     y
                     |
                  __---__
                /         \
               /           \
              |   center    |        <-- circle of radius R = g/(4ω²),
              |     o       |             centered at (0, R),
              |             |             tangent to x-axis at origin
               \           /
                \___   ___/
                    \ /
        ------------ O ------------ x       O = pivot (rotation axis)
                     |
                     |
                     v g

The pivot $O$ sits on the circle at its lowest point. The bead is never below the rotation axis: although the rod swings into the lower half-plane, $x_{p}$ flips sign at the same moment, so $\vec{R} = x_{p}\hat{r}$ remains in the upper half-plane.

Why the bead orbits twice per rod revolution

The bead lies on the chord through $O$ defined by the rod direction $\hat r(t)$. By the inscribed-angle theorem, if a chord through a fixed point $O$ on a circle rotates at angular rate $\omega$, the second intersection of that chord with the circle moves at angular rate $2\omega$ around the circle’s centre. The factor of two in $\sin(2\omega t)$, $\cos(2\omega t)$ above is exactly this geometric fact.

Educational remark. It is at first sight startling that gravity pulls down yet the steady circular orbit lies entirely above the axis. The mechanism is straightforward once unpacked: when the rod points up, the gravitational component along $\hat r$ is inward, and centrifugal is outward; their resonant interplay places the bead at maximum distance $g/(2\omega^{2})$ from the pivot at that moment. When the rod points down, gravity along $\hat r$ is outward (along $-\hat y$), but $x_p$ is now negative, so the bead is again on the upper side of the pivot. The circle on top is the unique bounded particular solution of an otherwise unstable linear system — every other initial condition develops an exponentially growing component along the unstable mode of the homogeneous equation.

---

Part (ii) — Radius of the circle

From the trajectory equation above, the circle has radius

$$\boxed{\;R \;=\; \dfrac{g}{4\omega^{2}}\;}$$

centred at $(0, R)$ in the vertical plane, with the rotation axis on the circle.

Consistency checks

• Dimensions. $[g]/[\omega^{2}] = (\text{m s}^{-2})/(\text{s}^{-2}) = \text{m}$. ✓
• Limit $\omega\to\infty$. $R\to 0$: very fast rotation makes gravity negligible (compared to centrifugal), and the bead is pinned at the axis on average. ✓
• Limit $\omega\to 0$. $R\to\infty$: with no rotation, gravity simply causes the bead to slide; no bounded circular orbit. ✓
• Energetic sanity. The bead’s height swings between $0$ and $2R = g/(2\omega^{2})$; its speed at the top and at the bottom can be checked against energy conservation in the lab frame. (The constraint force $N\hat\theta$ does work on the bead because the rod rotates, so the lab-frame energy is not conserved; this is actually the route by which the rotor delivers exactly the energy needed to keep the orbit circular.)

---

Part (iii) — Stabilisation with a spring

Now the bead carries an additional Hooke restoring force $-kx$ along the rod. Newton’s law along the rod becomes

$$m\ddot{x} \;=\; m\omega^{2}\,x \;-\; k\,x \;-\; mg\sin(\omega t),$$

or equivalently

$$\ddot{x} \;+\; \omega_{0}^{2}\,x \;=\; -g\sin(\omega t), \qquad \omega_{0}^{2} \;\equiv\; \frac{k}{m} - \omega^{2}.$$

The spring’s role is to flip the sign of the effective stiffness in the rotating frame: where centrifugal force gave us a destabilising $-\omega^{2}$ in part (i), the new effective squared frequency is $\omega_{0}^{2} = k/m - \omega^{2}$, which can be positive when $k$ is large enough.

Stability of the homogeneous mode

A small perturbation $\xi(t) \equiv x(t) - x_{\text{circle}}(t)$ obeys the homogeneous equation

$$\ddot{\xi} \;+\; \omega_{0}^{2}\,\xi \;=\; 0.$$

Three regimes:

Regime	Solution character	Stability
$k < m\omega^{2}$ ($\omega_{0}^{2}<0$)	$\xi \propto e^{\pm\sqrt{-\omega_{0}^{2}}\,t}$	unstable (exponential growth)
$k = m\omega^{2}$ ($\omega_{0}^{2}=0$)	$\xi = a + bt$	unstable (linear growth)
$k > m\omega^{2}$ ($\omega_{0}^{2}>0$)	$\xi = b\cos(\omega_{0}t) + c\sin(\omega_{0}t)$	Lyapunov stable

So the spring must be stiff enough that the centrifugal “anti-spring” is overpowered:

$$\boxed{\;k \;>\; m\omega^{2}\;} \quad\text{(necessary condition for stability).}$$

The circular trajectory and its forcing resonance

Look again for a synchronised particular solution $x_{p} = A\sin(\omega t)$. Substituting,

$$-A\,\omega^{2} + \omega_{0}^{2}\,A \;=\; -g \quad\Longrightarrow\quad A \;=\; \frac{g}{\omega^{2} - \omega_{0}^{2}} \;=\; \frac{g}{2\omega^{2} - k/m} \;=\; \frac{m g}{2m\omega^{2} - k}.$$

The trajectory is again a circle (the same algebra as in part (i) goes through unchanged), this time of radius

$$\boxed{\;R_{\text{circle}} \;=\; \dfrac{|A|}{2} \;=\; \dfrac{m g}{2\,\bigl|\,2m\omega^{2} - k\,\bigr|}\;}$$

centred at $(0, A/2)$. Note in particular:

• For $m\omega^{2} < k < 2m\omega^{2}$, $A>0$: the circle still sits above the pivot (and is larger than the spring-free $g/(4\omega^{2})$).
• At $k = 2m\omega^{2}$, $A$ diverges: the natural frequency $\omega_{0}$ of the (now stable) oscillator coincides with the forcing frequency $\omega$, and there is no bounded steady particular solution — parametric resonance with gravity.
• For $k > 2m\omega^{2}$, $A<0$: the bead’s mean displacement reverses, and the steady circle sits below the pivot, centred at $(0, A/2) = (0, -|A|/2)$.

The forced oscillator therefore must additionally avoid the resonance,

$$\boxed{\;k \;\neq\; 2m\omega^{2}\;}.$$

Final answer for part (iii)

The bead can move along a stable circular trajectory provided

$$\boxed{\;k > m\omega^{2}\quad\text{and}\quad k \neq 2m\omega^{2}\;}$$

and the radius of that trajectory is

$$\boxed{\;R \;=\; \dfrac{m g}{2\,\bigl|\,2m\omega^{2} - k\,\bigr|}\;}$$

with the circle centred at $\bigl(0,\, m g\,\mathrm{sgn}(2m\omega^{2}-k)\,/[2(2m\omega^{2}-k)]\bigr)$ — i.e. above the pivot when $k < 2m\omega^{2}$ and below it when $k > 2m\omega^{2}$.

Consistency checks

• Setting $k=0$ recovers $R = m g / (2\cdot 2m\omega^{2}) = g/(4\omega^{2})$ — but with $k=0$, the homogeneous mode is unstable, so the circular orbit exists as a particular solution but is not Lyapunov stable. Parts (i)–(ii) describe exactly this knife-edge case. ✓
• For very stiff spring $k\to\infty$, $R\to 0$: the bead is essentially clamped at the pivot, and gravity-induced excursions vanish. ✓
• The resonance $k = 2m\omega^{2}$ is the analogue, in the rotating frame, of driving a normal mass–spring at its natural frequency — except here the “drive” is the rotation of gravity relative to the rod.

Educational remark — why the circle persists. Adding the spring does not turn the trajectory into something more complicated than a circle, because the equation is still linear with a single sinusoidal forcing. The circle is simply the particular solution at the forcing frequency; the spring only changes its size and (above resonance) flips it through the pivot. What the spring does change qualitatively is the homogeneous behaviour: in part (i) any deviation grew exponentially; in part (iii) deviations remain bounded and superpose a small Lissajous-like oscillation on top of the basic circle, making the orbit observable in practice.

---

Note — partial solution. The write-up of part (iv) below is partial. The equation of motion, particular solution, and lab-frame phasor decomposition are derived in full, and the petal count of the figure pins down $\Omega$ provisionally as $\Omega = 5\omega$. The numerical extraction of $g$ and $a_E$ from grid measurements — together with the exact-circle derivation that would convert the inequality $r_{\max} \approx C + R_g$ into a closed-form result — is recorded as a methodology rather than carried through to a number.

Part (iv) — Rotating electric field (partial solution)

The spring is removed, the bead carries charge $q$, and a uniform electric field of magnitude $E$ rotates at angular velocity $\Omega$ in the same sense as the rod. Define $a_E \equiv qE/m$. The figure is a closed rosette on a square grid of spacing $d$ with gravity pointing down.

Equation of motion along the rod

Let the field be $\vec{E}(t) = E\bigl(\cos(\Omega t + \phi_0),\,\sin(\Omega t + \phi_0)\bigr)$. The component along the rod direction $\hat{r}(t) = (\cos\omega t,\sin\omega t)$ is

$$\vec{E}(t)\!\cdot\!\hat{r}(t) \;=\; E\cos\bigl((\Omega-\omega)t + \phi_0\bigr),$$

so the projection of Newton’s law onto $\hat{r}$ now reads

$$\ddot{x} \;-\; \omega^{2}\,x \;=\; -g\sin(\omega t) \;+\; a_E\cos\bigl((\Omega - \omega)t + \phi_0\bigr).$$

This is the part-(i) equation with one extra sinusoidal forcing term, this time at frequency $\Omega - \omega$ rather than $\omega$. Linearity lets us superpose particular solutions.

Particular solution

The gravity term contributes the part-(i) particular solution $x_g = (g/2\omega^{2})\sin(\omega t)$. For the field term, try $x_E = B\cos((\Omega-\omega)t + \phi_0)$:

$$-B(\Omega-\omega)^{2} - \omega^{2} B \;=\; a_E \quad\Longrightarrow\quad B \;=\; -\,\frac{a_E}{(\Omega-\omega)^{2} + \omega^{2}}.$$

Set $C \equiv a_E/[(\Omega-\omega)^{2} + \omega^{2}]$ for brevity. Then

$$\boxed{\;x_p(t) \;=\; \frac{g}{2\omega^{2}}\sin(\omega t) \;-\; C\cos\bigl((\Omega-\omega)t + \phi_0\bigr).\;}$$

The homogeneous mode is the same exponentially-unstable $e^{\pm\omega t}$ as in part (i); “with suitable initial conditions” picks out the bounded particular solution.

Lab-frame decomposition into rotating phasors

The bead’s lab-frame position is $\vec{R}(t) = x_p(t)\hat{r}(t)$. Writing $\hat{r}(t) = e^{i\omega t}$ in the complex plane and expanding the products,

$$\vec{R}(t) \;=\; \underbrace{i\,R_g}_{\text{DC}} \;\underbrace{-\,i\,R_g\,e^{2i\omega t}}_{\text{phasor }A}\;\underbrace{-\,\tfrac{C}{2}\,e^{i\phi_0}\,e^{i\Omega t}}_{\text{phasor }B}\;\underbrace{-\,\tfrac{C}{2}\,e^{-i\phi_0}\,e^{i(2\omega - \Omega)t}}_{\text{phasor }C},$$

with $R_g \equiv g/(4\omega^{2})$ as in part (ii). Three lab-frame frequencies appear:

	Frequency	Amplitude	Origin
$A$	$2\omega$	$R_g$	gravity (the part-(i) circle)
$B$	$\Omega$	$C/2$	field, “additive” branch
$C$	$2\omega - \Omega$	$C/2$	field, “subtractive” branch

Phasors $B$ and $C$ have equal amplitude — a direct consequence of $\vec{E}$ being a single rotating vector that gets split into co- and counter-rotating components by multiplication with $\hat{r}(t)$.

The rosette as a rose curve perturbed by gravity

Pair $B$ and $C$ alone:

$$\vec{R}_{BC}(t) \;=\; -\frac{C}{2}\Bigl[\,e^{i(\phi_0 + \Omega t)} + e^{-i(\phi_0 + \Omega t)}e^{2i\omega t}\,\Bigr] \;=\; -C\cos\!\bigl((\Omega-\omega)t + \phi_0\bigr)\,e^{i\omega t}.$$

In polar form this is

$$r(t) \;=\; C\,\bigl|\cos\bigl((\Omega-\omega)t + \phi_0\bigr)\bigr|, \qquad \theta(t) \;=\; \omega t \;+\; (\text{$\pi$ if cos $<$ 0}).$$

Eliminating $t$ via $\theta = \omega t$ gives the rose curve

$$\boxed{\;r(\theta) \;=\; C\cos\!\bigl(k\theta + \phi_0\bigr), \qquad k \;=\; \frac{\Omega - \omega}{\omega}.\;}$$

For the standard rhodonea $r = a\cos(k\theta)$ with $k = p/q$ in lowest terms, the petal count is $p$ if both $p,q$ are odd and $2p$ if their parities differ. Eight visible petals therefore require $p = 4$ with $q$ odd:

$$k \;=\; \frac{4}{q}, \quad q\in\{1,3,5,\dots\} \quad\Longrightarrow\quad \frac{\Omega}{\omega} \;=\; 1 + \frac{4}{q}.$$

The simplest choice — $q = 1$ — gives

$$\boxed{\;\Omega \;=\; 5\,\omega\;\quad(\text{provisional}).}$$

This is consistent with the figure showing exactly eight overlapping closed loops arranged with eightfold rotational symmetry, but I have not ruled out the $q = 3$ alternative ($\Omega = 7\omega/3$), which produces the same petal count traced over a longer period; doing so requires a closer look at how many times each loop is visited per period.

Why the petals close into circles

Pure rose petals are teardrops, not circles, so phasor $A$ (the gravity circle of radius $R_g$ at frequency $2\omega$) is what rounds the teardrops out into the closed circles seen in the figure. With $\Omega = 5\omega$ the three lab-frame frequencies are $\{2\omega,\,5\omega,\,-3\omega\}$, all integer multiples of $\omega$, so the trajectory closes after one rod period $T = 2\pi/\omega$. During each of the eight petal-times $T/8 = \pi/(4\omega)$, phasor $A$ advances by exactly $\pi/2$ — a quarter-turn — and the right value of the ratio $R_g/C$ is precisely what is needed to morph each teardrop into a closed circle. Treating “each petal becomes a circle” as a constraint should pin $g$ relative to $a_E$.

How to extract $g$ and $a_E$ from the figure (methodology)

With $\Omega = 5\omega$ established, two length scales remain:

• Gravity offset. The DC term $iR_g$ shifts the entire pattern up by $R_g = g/(4\omega^2)$ relative to the lab-frame origin (the rotation axis, which lies on the trajectory wherever $x_p(t) = 0$). Identifying the rotation axis as the lowest point at which the trajectory touches the y-axis of the gravity arrow, and measuring the height of the pattern’s geometric centre above it in units of $d$, gives $R_g$ directly:

$$\boxed{\;g \;=\; 4\omega^{2}\,R_g \;=\; 4\omega^{2}\,(\text{centre-to-axis distance}).\;}$$

• Field amplitude. The radius $\rho$ of one of the eight closed loops fixes a combination of $C$ and $R_g$. To leading order in the rose-curve picture each petal has half-length $C$, so a measurement of the maximum radial extent of the trajectory from the pattern centre, $r_{\max}$, gives

$$r_{\max} \;\approx\; C \;+\; R_g \quad\Longrightarrow\quad a_E \;=\; \bigl[(\Omega-\omega)^{2} + \omega^{2}\bigr]\,C \;=\; 17\,\omega^{2}\,(r_{\max} - R_g).$$

A more careful fit uses both $r_{\max}$ and $r_{\min}$ (the inner extent of each loop), which together encode $C$ and $R_g$ separately — but the algebra here is what I did not get right inside the time cap, and quoting numerical values for $g$ and $a_E$ would be guessing.

What is missing

1. A clean argument that distinguishes $\Omega = 5\omega$ from $\Omega = 7\omega/3$ (and higher $q$). The eightfold visual symmetry alone is not decisive; one needs to count how many distinct closed loops appear, which is straightforward visually but I did not nail it down on this figure.
2. A geometric derivation of the exact shape of one petal under the combined $A + (B+C)$ phasors — i.e. the proof that the teardrops become exact circles for a specific ratio $R_g/C$. This is the cleanest route to the relation $a_E$ vs $g$ and would convert the methodology above into a closed-form answer.
3. The actual numerical readings off the grid: the height of the pattern centre above the rotation axis, the radius of one of the eight loops, and the corresponding numerical values of $g$ and $a_E$ in units of $\omega^{2}d$.

Up to point (1) the result is

$$\boxed{\;\Omega \;=\; 5\,\omega,\qquad g \;=\; 4\omega^{2} R_g,\qquad a_E \;=\; 17\,\omega^{2}\,C,\;}$$

with $R_g$ and $C$ to be read off the figure as the gravity-induced vertical offset of the pattern centre and (approximately) the half-extent of one petal, respectively.

---

Summary of results

Part	Quantity	Result
(i)	Trajectory (no spring)	Circle of radius $R = g/(4\omega^{2})$, centred at $(0,R)$, tangent to the horizontal at the pivot; bead orbits at angular rate $2\omega$
(ii)	Radius (no spring)	$R = \dfrac{g}{4\omega^{2}}$
(iii)	Stability range of $k$	$k > m\omega^{2}$ and $k \neq 2m\omega^{2}$
(iii)	Radius with spring	$R = \dfrac{m g}{2\,\lvert\,2m\omega^{2} - k\,\rvert}$
(iv)	Field frequency (provisional)	$\Omega = 5\omega$
(iv)	$g$ and $a_E$ from the grid	$g = 4\omega^{2} R_g$, $\;a_E = 17\,\omega^{2}\,C$ — methodology only, not extracted numerically