Problem statement

A horizontal cylinder closed at one end is rotated with constant angular velocity ω about a vertical axis through its open end. The outside air pressure is p₀, the temperature is T, and the molar mass of air is M. Find the air pressure as a function of the distance r from the axis of rotation. Take M to be independent of r.

We assume T = 300 K, M = 29 g/mol and p₀ = 1 atm. The problem gives no numbers, so these make the drawing concrete. All are sliders.

We assume ω = 300 rad/s (about 48 turns per second) and a tube length L = 1.0 m. This means the pressure rises to about 1.7 p₀ at the closed end, which is easy to see. Both are sliders.

We assume the air has time to settle, so it rotates with the tube and the temperature is the same everywhere in it. This means the air is at rest relative to the tube.

Symbols: r is the distance from the axis, p is pressure, ρ is density, R = 8.314 J/(mol K).

What the constant Mω²/2RT means

The factor k = Mω²/(2RT) has units of 1 per m². It says how quickly the pressure rises as r grows. The pressure is multiplied by e at the distance r_e = 1/√k = √(2RT/Mω²), here m. A faster spin (larger ω), a heavier gas (larger M) or colder air (smaller T) makes r_e shorter, so the pressure climbs faster.

This is the same idea as the scale height in the Earth's atmosphere. There gravity pulls air down; here the spin pushes air outward, so the "gravity" is ω²r and grows with r. That is why the exponent has r², not r.

The gas law alone, p = ρRT/M, cannot give p(r). It only links p, ρ and T at one point. We also need Newton's second law for a slice of gas, which supplies the link between the pressure change and the spinning.

The big idea in plain words

Every ring of gas in the tube moves in a circle. A circle needs a force that points to the centre. The only forces along the tube are the pressures on the two faces of the ring. So the pressure on the outer face must be a little larger than on the inner face. The extra pressure supplies the inward push. The faster the spin and the farther from the axis, the bigger the push needed, so the pressure rises steeply toward the closed end. The denser the gas is there, the more force is needed, which makes the rise even steeper. That feedback gives an exponential in r².

The solution in each frame

Lab frame (non-rotating)

Each ring moves in a circle, so it needs a net inward force dm ω² r. This is the solution given below, and the free-body diagram on the page shows it for one ring.

Rotating frame

In the frame that spins with the tube, the gas is at rest. A slice feels an outward centrifugal force dm ω² r, balanced by the pressure difference. This gives the same equation, dp = ρ ω² r dr, and is just like the balance of a layer of the atmosphere under gravity.

The solution step by step

Take one thin ring of gas (a slice across the tube) at distance r, with thickness dr and cross-section S. This is the cyan ring on the spinning tube. Its mass is dm = ρ S dr. The free-body diagram on the page shows the forces on it along the tube. The gas on the inner side pushes it outward with force pS. The gas on the outer side pushes it inward with force (p + dp) S. The weight and the push of the wall are vertical and cancel. The ring moves in a circle of radius r, so the net inward force must equal dm ω² r:

The gas law gives ρ = pM/RT. Put this in and divide by p:

T, M and ω are constant. Integrate from the open end (r = 0, p = p₀) to r:

Numbers: Mω²/2RT = 0.029 × 300² / (2 × 8.314 × 300) = 0.523 per m². At r = 1 m, p = e^0.523 p₀ = 1.69 p₀. For the values now on screen the constant is per m², the pressure is p₀ ( atm) at r = m, and p₀ ( atm) at the closed end.

Five methods

1. Force method

This is the first step above: Newton's second law for one slice, with the net force coming from the pressure difference.

2. Energy method

In the rotating frame the centrifugal potential energy per molecule is −mω²r²/2. The Boltzmann law n ∝ e^−U/kT gives the same exponent Mω²r²/2RT. This is quick, but it takes thermal equilibrium for granted.

3. Graph method

Plot ln(p/p₀) against r². It is a straight line through the origin with slope Mω²/2RT. The page shows this line.

4. Calculus method

This is the integration above: dp/p = (Mω²/RT) r dr.

5. Check method

Units: kg/mol × (rad/s)² × m² divided by J/mol is a pure number. Limits: at r = 0 the formula gives p₀. For ω = 0 it gives p₀ everywhere. For a small exponent, p ≈ p₀(1 + Mω²r²/2RT), a parabola.

Fastest: the energy method. Most reliable: the force method with the step-by-step sum.

What each slider changes

Control	Changes the answer?	What it changes
Angular speed ω	Yes	The constant, the curve, the end pressure, the shading and the graphs
Tube length L	Not p(r)	The drawing, the end pressure and the graphs
Temperature T	Yes	The constant, the curve and the answers
Molar mass M	Yes	The constant, the curve and the answers
Outside pressure p0	Yes	All pressures in proportion; the graphs use p/p0
Ring distance r	No	Where the ring sits on the tube and in the free-body diagram, and its forces
Tube area S	Not p(r)	Every force in newtons and the ring mass
Ring thickness Δr	Not p(r)	The ring drawn, the pressure difference across it and its mass
Given-answer mode	Sets ω	The page finds the angular speed that gives the chosen end pressure

Parameters

Quantity	Symbol	Value used
Angular speed	ω	rad/s ( rpm)
Tube length	L	m
Temperature	T	K
Molar mass	M	g/mol
Outside pressure	p0	atm
Constant	Mω²/2RT	per m²
Length for a factor e	re	m
Acceleration at the closed end	ω²L	g

Exam traps

• Writing r instead of r² in the exponent. The inward force needed grows with r, so integrating r dr gives r²/2.
• Forgetting the factor 2 in the denominator: the exponent is Mω²r²/(2RT), not Mω²r²/(RT).
• Using M in grams per mole. The formula needs kilograms per mole.
• Using ω in rpm or turns per second. The formula needs radians per second.
• Expecting the pressure to fall outward. The pressure must rise toward the closed end to supply the inward force.