The Tennis Racket Theorem

Set the three principal moments of inertia, the axis to spin about, how fast, and how far off that axis the spin starts. The page integrates Euler's equations as you watch, draws the body turning while its angular momentum stays where it was, and solves the time between flips exactly.

Worked example — The Tennis Racket Theorem

Spin rate	6.283 rad/s
Growth rate of a wobble about the middle axis	3.628 s⁻¹
A small wobble about axis 2	3.628 s⁻¹
Time added by each factor of ten less wobble	1.27 s
Time between flips, exact	4.38 s
Energy at the end over energy at the start	1.000000
\|L\| at the end over \|L\| at the start	1.000000
Most that L moved in space, as a fraction of \|L\|	5.9×10⁻¹⁰

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — The Tennis Racket Theorem

The tennis racket theorem, or intermediate axis theorem: a rigid body whose three principal moments of inertia differ, I₁ < I₂ < I₃, spins stably about the axes of the smallest and the largest moment and unstably about the middle one. For a box those are the axes along its longest side, its shortest side and the side in between. With no torque, a small wobble about the middle axis grows as e^λt with λ = ω√((I₂ − I₁)(I₃ − I₂)/(I₁I₃)) until the body turns over, while its angular momentum stays fixed in space. A racket thrown so that its handle makes one full turn comes back with its faces exchanged; in orbit the same motion is called the Dzhanibekov effect, after the cosmonaut who watched a spinning wing nut keep flipping in weightlessness in 1985.

What each symbol means

I₁: the smallest principal moment of inertia; for a box, about the axis along its longest side
I₂: the middle moment: the axis that will not hold a spin
I₃: the largest moment; for a thin flat body it is about the axis through the face, and I₃ = I₁ + I₂
λ: growth rate of a small wobble about the middle axis: the wobble multiplies by e every 1/λ
ε: the starting wobble, the spin about each of the other two axes as a fraction of the main spin

Assumes: A rigid body with no torque on it: no air, and no gravity gradient. Thrown on Earth, the centre of mass falls but the rotation about it is the same. The energy leak, when it is on, is a model with exactly two properties, energy out and L unchanged, at a rate this page chose; real damping depends on how the body flexes.

Spun about its middle axis, it turns over every 4.38 s and nothing pushes it 🖖

Press Middle axis, ε 10⁻³. The body has moments of inertia in the ratio 1 : 2 : 3 and spins once a second about axis 2, with a thousandth of that spin about each of the other two. At 2.37 s it is on its side and a moment later upside down; 4.38 s after that it does it again, and it goes on doing it for as long as the run lasts. The red arrow is its angular momentum L, and it does not move. After 60 s the energy and |L| are 1.000000 of their starting values, and L has wandered in space by 5.9×10⁻¹⁰ of its length, which is the integrator's own error. No torque acts, and turning over costs no energy. The second panel shows why axis 2 and no other. With E and |L| both fixed, the direction of L seen from the body can only travel along one curve of the family drawn there. Round ±1 and ±3 they are small closed loops, so a small wobble stays small. At ±2 they cross, a saddle, and the curve through a point near +2 runs all the way to −2. Press Largest moment, ε 10⁻³: the same wobble about axis 3 is a flat line in the third panel, and no flip comes.

A thousand times more care buys 3.81 seconds 🖖

Press Middle axis, ε 10⁻⁶, the same throw made a thousand times cleaner. The flips now come every 8.19 s instead of 4.38 s. The table under the results carries the pattern on: below 10⁻², every factor of ten less wobble adds the same 1.27 s, all the way down to 10⁻¹². The third panel shows where that comes from. A wobble about the middle axis grows as e^λt, a straight line on a log scale. After each flip it shrinks back to about ε, upside down, and has to grow out again before the next. The fall to ε takes about ln(1/ε)/λ and the climb back the same, so each factor of ten in ε costs 2 ln 10/λ, which is 1.27 s at λ = 3.628 s⁻¹. Change the moments and the table follows: below a wobble of 10⁻² the step stays within half a percent of 2 ln 10/λ. This is why a phone turns over when you toss it. Press A tossed phone: Michael Wheatland and colleagues worked out an iPhone 8 Plus's moments from its mass and size as 1.04, 4.23 and 5.25 ×10⁻⁴ kg·m², and at 17.002 rad/s its λ is 13.125 s⁻¹. Each factor of ten of care buys it 0.35 s, and a wobble of 0.2 turns it over at 0.32 s. Their phone's gyroscope recorded the same reversal during a real toss.

Let energy leak out and only the axis of largest moment is stable 🖖

Press Energy leaking, axis 1. Axis 1, the smallest moment, was stable a moment ago. Now the spin about it slowly opens out, reverses at 40.60 s, and settles about axis 3 with 0.333712 of the energy it started with and every bit of its angular momentum. Internal friction can take energy out of a spinning body, but it cannot change the body's total L, and at a fixed L the energy L²/2I is least where I is largest. A spin about axis 1 carries the most energy that L allows, so a leak walks L across the sphere in the second panel, over the saddle, to ±3. Explorer 1 met this in 1958. It was a long, thin body meant to spin about its long axis, the axis of least inertia, it carried four flexible antennas, and it would not keep that spin; Michael Efroimsky's 2001 paper on wobbling asteroids opens with it. The leak on this page is a model with exactly those two properties, energy out and L untouched, at a rate of the page's choosing; it is not a model of the antennas. Without a leak the flips in the first two blocks go on for ever. With one, as J. A. de la Torre and Pep Español put it, Dzhanibekov's wing nut, watched for long enough, would have been found spinning about its major axis.

Problem solved in full

How clean a throw does a sheet of paper need? 5 steps
A sheet of A4, 210 mm by 297 mm, is spun at 1 rev/s about the axis that runs along its short side, through its centre. Find the ratio of its three moments of inertia, how fast a wobble grows, and how small the starting wobble must be for ten seconds between flips, and for a minute. Press Middle axis, ε 10⁻³ to check the second step.

Open this problem in the tool
1. All of a thin sheet's mass lies in one plane. About the axis along its long side the mass is spread across the 210 mm width, so I₁ = m b²/12 with b = 210 mm; about the axis along the short side, I₂ = m a²/12 with a = 297 mm; and about the axis through the face, the perpendicular axis theorem gives I₃ = I₁ + I₂:
  
  $I_{1} : I_{2} : I_{3} = 21 0^{2} : 29 7^{2} : (21 0^{2} + 29 7^{2}) = 44, 100 : 88, 209 : 132, 309 = 1 : 2.0002 : 3.0002$
2. A-series paper is cut so that a sheet halved has the same shape, which needs a/b = √2, and then the ratio is exactly 1 : 2 : 3; 297/210 is √2 rounded to the millimetre. With ω = 2π rad/s, as in the page's first preset:
  
  $\frac{a}{b} = 2 \Rightarrow 1 : 2 : 3, λ = 2 π \frac{( 2 - 1 ) ( 3 - 2 )}{1 \times 3} = \frac{2 π}{3} = 3.628 s^{- 1}$
3. Between flips the wobble shrinks to about ε and grows back to about 1, so the interval goes as (2/λ) ln(1/ε) plus a constant. The page prints 4.38 s at ε = 10⁻³ and 1.27 s for each further factor of ten. Ten seconds needs
  
  $\frac{10 - 4 . 38}{1 . 27} = 4.43 \Rightarrow ε \approx 1 0^{- 3 - 4.43} = 3.7 \times 1 0^{- 8}$
4. A minute needs 55.62 s more than the page's preset gives:
  
  $\frac{60 - 4 . 38}{1 . 27} = \frac{55 . 62}{1 . 27} = 43.8 \Rightarrow ε \approx 1 0^{- 46.8} = 1.6 \times 1 0^{- 47}$
5. For scale, a sheet 0.3 m long whose axis was set true to the width of one atom, 10⁻¹⁰ m, would start with a wobble of about one part in 3×10⁹:
  
  $ε \approx \frac{1 0 ^{- 10} m}{0 . 3 m} = 3.3 \times 1 0^{- 10} \Rightarrow t_{flip} = 12.61 s$
Answer

The moments are 1 : 2.0002 : 3.0002 and a wobble grows at λ = 3.628 s⁻¹. Ten seconds between flips needs a starting wobble of 3.75×10⁻⁸, and a minute needs 1.54×10⁻⁴⁷, solving the exact interval rather than the decade rule. Setting the axis true to an atom's width buys about 12.6 s. No throw keeps a sheet spinning about its middle axis for a minute.

$1 : 2.0002 : 3.0002, λ = 3.628 s^{- 1}, ε_{10 s} = 3.75 \times 1 0^{- 8}, ε_{60 s} = 1.54 \times 1 0^{- 47}$

Learning path

Rotation and angular momentum

References (3)

A phone tossed in the air as a free-rotation laboratory: Euler's equations, the instability of the middle axis, and the moments of an iPhone 8 Plus, 1.04, 4.23 and 5.25 ×10⁻⁴ kg·m², computed from its measured mass and dimensions, behind the page's phone preset: M. S. Wheatland, T. Murphy, D. Naoumenko, D. van Schijndel and G. Katsifis, "The mobile phone as a free-rotation laboratory", American Journal of Physics 89, 342–348 (2021). doi:10.1119/10.0003380
Explorer 1, in 1958, was a very prolate body with four deformable antennas, meant to spin about its least-inertia axis, and refused; a body that loses energy at fixed angular momentum must end in the minimum-energy state, a spin about the maximal-inertia axis: M. Efroimsky, "Relaxation of wobbling asteroids and comets: theoretical problems, perspectives of experimental observation", Planetary and Space Science 49(9), 937–955 (2001). doi:10.1016/S0032-0633(01)00051-4
Dzhanibekov's wing nut, observed in 1985, flips for ever in Euler's equations; real bodies dissipate, and a torque-free body ends spinning about its major axis, so 'had Dzhanibekov observed his wing nut for a sufficiently long period, he would have found it spinning around the major axis': J. A. de la Torre and P. Español, "Internal dissipation in the Dzhanibekov effect", European Journal of Mechanics A/Solids 106, 105298 (2024). doi:10.1016/j.euromechsol.2024.105298

By the 42Math team · Reviewed against published sources. Corrections welcome.