stable unstable and neutral equilibrium

Equilibrium of a rigid body means that the conditions $\sum \vec F = 0$ and $\sum \vec M = 0$ hold, but these conditions say nothing about how the body behaves after a small displacement.

Therefore three types of equilibrium are distinguished. If a small displacement produces a torque that turns the body back, the equilibrium is stable; if the body keeps moving away from the equilibrium position, it is unstable; if the displacement produces neither return nor departure, it is neutral.

Energy criterion: in stable equilibrium the potential energy $U$ is a minimum, in unstable equilibrium it is a maximum, and in neutral equilibrium it is constant. For a small angle $\theta$ the torque is $M \approx -\left.\dfrac{d^2U}{d\theta^2}\right|_0 \theta$, so the sign of the second derivative fixes the type of equilibrium.

Practical rule: for a small displacement the center of gravity of a stable system rises, of an unstable one falls, and of a neutral one stays at the same height $z_{\text{cg}} = \text{const}$.

This is why a body on a broad support is stable, a body on a sharp tip is unstable, and a ball on a horizontal plane is in neutral equilibrium.

A uniform rod of length $L = 1{,}0$ m is mounted on a horizontal axle through its midpoint and rests horizontally. The rod is tilted through a small angle $5^\circ$ and released. Determine the type of equilibrium.

A block of mass $m = 12$ kg rests on a horizontal floor; the base width is $b = 0{,}30$ m and the height is $h = 0{,}60$ m. The block is tilted by $\theta = 10^\circ$ about its lower edge. Find the restoring torque and the toppling limit angle.

A hole for an axle is drilled in a uniform rod of length $L = 0{,}80$ m at a distance $d$ from its upper end. The rod hangs on a horizontal axle and can rotate freely in a vertical plane. Determine the type of vertical equilibrium for $d = 0{,}10$ m, $d = 0{,}40$ m and $d = 0{,}60$ m, and find the stability condition in general form.

Qualitative table-top experiment (safe). Take three objects: a flat coin, an eraser and a marble. Carry out all actions on the floor or on a low, wide table, away from the edge.

1. Place the marble on the horizontal table surface. Give it a gentle push: it neither returns nor runs away on its own — a model of neutral equilibrium ($U = \text{const}$).
2. Stand the eraser on its broad base. Tilt it through a small angle $\theta \approx 10^\circ$ and release it: it returns — stable equilibrium, the center of gravity rose.
3. Stand the eraser on its narrow edge (or balance the coin on its rim) and try to hold it vertical: the slightest deflection leads to falling — unstable equilibrium ($U$ is maximal).
4. Quantitative part. Measure the base width $b$ and the height $h$ of the eraser with a ruler and compute $\theta_c = \arctan\frac{b}{h}.$ Check that for a tilt smaller than $\theta_c$ the object returns, while for a larger tilt it topples.

If no steady surface is available for tilting experiments, use a computer equilibrium simulation (for example, a pendulum model with the axle above and below the center of mass): set the axle position $d$ and observe how the type of equilibrium changes. A simulation is preferable for any experiment involving a tall support, glass, or heavy loads.

Show that the type of equilibrium is determined unambiguously by how the height of the center of gravity (and the potential energy) changes under a small displacement.

A three-panel sketch of a ball on a surface: (a) in a concave bowl — stable equilibrium, an arrow returns the ball to the bottom; (b) on a hilltop — unstable, an arrow carries the ball away; (c) on a horizontal plane — neutral, no arrow. Beside it, a common graph of $U(\theta)$: a minimum at $\theta = 0$ for the stable case, a maximum for the unstable case, and a horizontal line $U = \text{const}$ for the neutral case.

A gymnast of mass $50\ \text{kg}$ stands on a beam $10\ \text{cm}$ wide; her centre of mass is about $1.1\ \text{m}$ above the top of the beam. Leaning by $3°$ shifts her centre of mass sideways by $1.1 \times \sin 3° \approx 5.8\ \text{cm}$ — more than half the width of her base of support ($5\ \text{cm}$). Once the vertical line through her centre of mass passes outside the beam edge, her weight $mg = 50 \times 9.8 = 490\ \text{N}$ produces about this edge a toppling torque of $490 \times 0.058 \approx 28\ \text{N·m}$, and she falls unless she pulls her body back in time.

This is an unstable equilibrium: rocking about the edge of the base lowers the centre of mass, so the potential energy $U = mgR\cos\theta$ is a maximum at $\theta = 0$. The classification rests on the sign of the torque of gravity about the edge of support: inside the $10\ \text{cm}$ base it restores the body, outside it accelerates the fall.

A double-decker bus has a track (distance between the wheels of one axle) of $2.0\ \text{m}$ and its centre of mass is $1.3\ \text{m}$ above the road. In a bend of radius $50\ \text{m}$ at $8\ \text{m/s}$ ($29\ \text{km/h}$) the sideways acceleration is $a = v^2/r = 1.28\ \text{m/s}^2 \approx 0.13g$. About the outer wheels the centrifugal effect gives a toppling torque $m a h = m \times 1.28 \times 1.3 = 1.7m\ \text{N·m}$, while gravity holds the bus with $m g (t/2) = m \times 9.8 \times 1.0 = 9.8m\ \text{N·m}$. The bus rolls over when $a/g > (t/2)/h = 1.0/1.3 = 0.77$, that is above $v = \sqrt{0.77 \times 9.8 \times 50} \approx 19\ \text{m/s} \approx 70\ \text{km/h}$.

Stability here is decided by the direction of the total torque about the line of contact of the outer wheels: gravity rights the bus, the inertial term tips it. Describing rocking about that line by the potential energy $U = mgR\cos\theta$ with $R = \sqrt{1.0^2 + 1.3^2} = 1.64\ \text{m}$, tilting outward raises the centre of mass from $1.30\ \text{m}$ to $1.64\ \text{m}$ — $U$ grows, the signature of stable equilibrium; at $a = 0.77g$ the effective gravity is tilted exactly enough for this growth to vanish at small $\theta$.

A smartphone is $150\ \text{mm}$ long, $70\ \text{mm}$ wide, $8\ \text{mm}$ thick and has a mass of $200\ \text{g}$. Stood on its long edge it rests on a strip only $8\ \text{mm}$ wide, with its centre of mass $75\ \text{mm}$ up; a tilt of just $\arctan(4/75) \approx 3°$ carries the vertical through the centre of mass past the edge and the phone falls. The same phone lying flat slides to wherever it is pushed and stays there. Tilting the flat phone about one long edge raises its centre of mass by $\sqrt{35^2 + 4^2} - 4 \approx 31\ \text{mm}$ before it can pass over that edge, an energy barrier of $mg\Delta h = 0.2 \times 9.8 \times 0.031 \approx 0.06\ \text{J}$.

The three cases are read off the shape of $U(\theta) = mgR\cos\theta$: a maximum for the phone standing on its edge (unstable), a minimum for the flat phone tilted about an edge (stable), and $U = \text{const}$ for sliding on the horizontal table (neutral). What decides the case is the height of the centre of mass $h(\theta)$, not the values of the forces: in all three positions the sum of forces and the sum of torques are zero.