Irodov 1.110: Two Equilibria on a Spinning Hoop – Amit & Deepali Sinha Foundation

Problem Irodov 1.110

A sleeve $A$ slides without friction on a rod bent into a half circle of radius $R$ ; the rod’s two ends lie on a common vertical line $OO'$ , and that same line is the axis about which the whole assembly rotates at a constant angular velocity $\omega$ . Find the angle $\theta$ , measured at the circle’s center between the downward direction along the axis and the radius to the sleeve, at which the sleeve rests in its steady position as the system turns.

The half-circular rod, its ends fixed to the vertical axis OO', which turns in bearings at O and O'. Sleeve A sits on the curve at angle θ, measured at the center C from the downward vertical CO'. — The half-circular rod, its ends fixed to the vertical axis $OO'$ , which turns in bearings at $O$ and $O'$ . Sleeve $A$ sits on the curve at angle $\theta$ , measured at the center $C$ from the downward vertical $CO'$ .

Given

$R$
$\omega$

Find

The angle

\theta

at which the sleeve stays put as the assembly turns

Solution

Step 1Working in the frame that rotates with the rod

The sleeve stays at a fixed point on the rod once the assembly is turning steadily, so in the frame that rotates with it the sleeve is in static equilibrium under three forces.

Free body of the sleeve in the rotating frame: the weight m g, the smooth rod's reaction N (radial, along CA) and the centrifugal force mω^2x (horizontal, outward), with x=Rθ the distance from the axis. — Free body of the sleeve in the rotating frame: the weight $m\vect g$ , the smooth rod’s reaction $\vect N$ (radial, along $CA$ ) and the centrifugal force $m\omega^2x$ (horizontal, outward), with $x=R\sin\theta$ the distance from the axis.

\begin{aligned} &m\vect g && \text{(weight, downward)} \\ &\vect N && \text{(rod's reaction, radial along } CA\text{, smooth)} \\ &m\omega^2 x && \text{(centrifugal, horizontal, outward)} \end{aligned}

where $x=R\sin\theta$ is the sleeve’s distance from $OO'$ . I take $\theta$ as the angle between $CA$ and the downward vertical $CO'$ , matching the direction gravity itself points from the center; this is the natural reference for a sleeve released under gravity alone.

Step 2Resolving along the rod

Since the rod is smooth, $\vect N$ has no component along the rod, so equilibrium requires the tangential components of gravity and the centrifugal force to cancel. Projecting both onto the tangent at $A$ (perpendicular to $CA$ ),

\underbrace{-mg\sin\theta}_{\text{gravity}} + \underbrace{m\omega^2 R\sin\theta\cos\theta}_{\text{centrifugal}} = 0 \tag{1}

Factoring out $\sin\theta$ gives the equilibrium condition

\sin\theta\left(\omega^2R\cos\theta - g\right) = 0

(2)

Step 3The two branches

Equation (2) is satisfied in two distinct ways. Either

\sin\theta = 0 \quad\Longrightarrow\quad \theta_1 = 0,

the sleeve sitting at the bottom of the loop, on the axis itself, which is an equilibrium at any $\omega$ ; or

\cos\theta = \frac{g}{\omega^2R} \quad\Longrightarrow\quad \theta_2 = \arccos\frac{g}{\omega^2R},

which has a solution only when $\omega^2R > g$ , since a cosine cannot exceed 1. The sleeve resting on the axis itself, $\theta_1=0$ , is always a solution; whenever $\omega^2R>g$ a second, off-axis position also satisfies (2):

\theta_2 = \arccos\dfrac{g}{\omega^2R}

(3)

Step 4Which position is steady

Collect the two tangential terms of (1) into the net tangential force on the sleeve at a general angle, counted positive in the direction of increasing $\theta$ , and rewrite the product of sine and cosine as a double angle ready for differentiating.

\begin{align*}F_t(\theta) &= m\sin\theta\left(\omega^2R\cos\theta-g\right) \\ &= m\left(\frac{\omega^2R}{2}\sin2\theta - g\sin\theta\right)\end{align*}

Near the bottom of the loop $\theta$ is small, so $\sin\theta\approx\theta$ and $\cos\theta\approx1$ in the first form,

F_t \approx m\theta\left(\omega^2R-g\right)

A small displacement $\theta>0$ therefore feels a force back toward the axis while $\omega^2R<g$ , and a force away from it once $\omega^2R>g$ : the on-axis position $\theta_1=0$ is stable below the threshold and unstable above it, where the sleeve slides out to $\theta_2$ instead.

For the off-axis branch, differentiate the same tangential force and evaluate at $\theta_2$ , using $\cos\theta_2=g/\omega^2R$ together with $\cos2\theta_2=2\cos^2\theta_2-1$ .

\begin{align*}\frac{dF_t}{d\theta} &= m\left(\omega^2R\cos2\theta - g\cos\theta\right) \\ &= m\left[\omega^2R\left(\frac{2g^2}{\omega^4R^2}-1\right)-\frac{g^2}{\omega^2R}\right] \\ &= m\left(\frac{2g^2}{\omega^2R}-\omega^2R-\frac{g^2}{\omega^2R}\right) \\ &= -\frac{m}{\omega^2R}\left(\omega^4R^2-g^2\right)\end{align*}

The branch $\theta_2$ exists only where $\omega^2R>g$ , so $\omega^4R^2>g^2$ there and this derivative is negative: displaced either way, the sleeve feels a tangential force back toward $\theta_2$ . The off-axis position is therefore the stable one wherever it exists, and it is the steady position the problem asks for; below the threshold, where $\theta_2$ does not exist, $\theta_1=0$ is the only resting point and is stable, exactly as it would be with no rotation at all.

Check

As $\omega^2R\to g^+$ , $\theta_2\to0$ : the new equilibrium is born continuously out of $\theta_1=0$ exactly at the threshold, rather than appearing at a finite angle out of nowhere, which is what a smooth loss of stability should look like. As $\omega\to\infty$ , $\theta_2\to90^\circ$ : the sleeve is flung out to the rod’s widest point, the farthest it can get from the axis.

Remark

The transition at $\omega^2R=g$ is the same exchange of stability seen in a textbook pitchfork bifurcation, though on this half-circle rod’s restricted domain $\theta\in[0,90^\circ]$ only one arm of the full pitchfork is available: the sleeve is confined to a single planar rod, one degree of freedom, so it can only reach the single angle $\theta_2=\arccos(g/\omega^2R)$ found above. The full symmetric pitchfork, with its mirror pair $\theta=\pm\arccos(g/\omega^2R)$ , lives on a complete rotating hoop instead: a bead threaded on a full circular hoop (still one degree of freedom, but now ranging over the whole circle) is free to settle at either sign of $\theta$ , the two mirror images of the same equilibrium found here, both stable once $\omega^2R>g$ . A continuous ring of equilibria is a different problem again, a bead on a rotating sphere or bowl, where the extra azimuthal freedom around the vertical axis turns a discrete pair into a whole circle of them.