Angular Momentum

Stand on a turntable that turns without friction, hold a dumbbell in each hand, spin, and pull them in. Set the spin, the dumbbells, your body's own moment of inertia and how far out and in your hands go. The page integrates the motion live, prints what stays fixed and what does not, and adds up the work your arms do.

Worked example — Angular Momentum

Moment of inertia, hands out	3.600 kg·m²
Moment of inertia, hands in	1.200 kg·m²
Angular momentum, the same throughout	22.62 kg·m²/s
Spin with the hands in	3.000 rev/s
Spin if the dumbbells reached the axis	3.462 rev/s
Kinetic energy, hands out	71.06 J
Kinetic energy, hands in	213.18 J
Spin and kinetic energy both rise by	3.000

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 — Angular Momentum

The angular momentum of a body turning about a fixed axis is L = Iω, its moment of inertia about that axis times its angular velocity. The moment of inertia adds up each piece of mass times the square of its distance from the axis, so a dumbbell whose centre is at r carries m r² + I_own, the parallel-axis theorem; for two plates d either side of the hand, I_own = m d². With no external torque about the axis L cannot change, so I_out ω_out = I_in ω_in, while the kinetic energy ½ I ω² = L²/2I changes by the work the arms do.

What each symbol means

I: moment of inertia about the turntable's axis, kg·m²
ω: angular velocity in rad/s; the page prints f = ω/2π in rev/s
L: angular momentum about the axis, Iω, in kg·m²/s
I_b: the body and turntable's own moment of inertia, which pulling in does not change
d: half the length of a dumbbell's bar, 0.1 m: its plates sit d either side of the hand

Assumes: A turntable with no friction in its bearing and nothing else touching the body, so there is no torque about the axis. The arms have no mass; real arms, a few kilograms each, would add to the moment of inertia with the hands out and to the gain. Each dumbbell is two point plates on a light bar, held level and across the arm, and the body's own moment of inertia stays fixed whatever the arms do. The work counted is that of the pull mω²r which keeps the dumbbells on their circles; the small energy of their in-and-out motion is zero at both ends and is left out.

Pull the dumbbells in and the kinetic energy triples 🖖

Press Pull in, 2 kg and watch the two arrows on the axis. L keeps its length while the spin grows threefold, from 1 to 3.000 rev/s: nothing outside exerts a torque about the axis, so L = Iω has to survive the moment of inertia falling from 3.600 to 1.200 kg·m². The kinetic energy, ½Iω² = L²/2I, has no such protection, and it rises by the same factor, from 71.06 J to 213.18 J. The extra energy has a source. To keep the two dumbbells going round at radius r the hands must pull them inwards with 2mω²r, and drawing them in along that pull is work: the shaded area under the curve in the second panel. The results table adds it up over a thousand strips and gets 142.12 J, the gain to the last digit printed. Push the hands back out and the same pull does work on the arms instead, and the spin falls back to 1 rev/s.

Hands four times closer, and only three times the spin 🖖

If the dumbbells were all there was, as points in the hands, L = 2mr²ω would make bringing them from 0.8 m to 0.2 m spin you sixteen times faster. The page says 3.000. Most of what turns does not move when you pull. Your body and the turntable keep their 1 kg·m², and each dumbbell keeps the part of its moment that belongs to its own size, md² by the parallel-axis theorem, however close you bring it. So there is a ceiling: with both dumbbells on the axis the row under the spin gives 3.462 rev/s, and no amount of pulling gets past it. Heavier dumbbells are a bigger share of what turns. Heavier, 10 kg multiplies the spin by 7.000, but starting from only a quarter turn a second the hands already have to hold 483.6 N at the end.

With no angular momentum at all, you can still turn round 🖖

Press From rest, arm stroke. L is zero from start to finish, and each stroke still leaves the body 25.00° round. Sweep the arms 45° with the hands out and the body turns 32.50° the other way, since the dumbbells are 2.600 of the 3.600 kg·m² that turns. Bring them in and sweep back, and the body comes back only 7.50°, because now they are 0.200 of 1.200. The arms end where they began, relative to the body; the body does not. Zero angular momentum forbids a spin that carries on: between sweeps the body stands still, and Stroke, hands out both ways brings it back to 0.00°. A cat dropped upside down from rest has no angular momentum either, and lands on its feet. Thomas Kane and M. P. Scher published a dynamical explanation of that in 1969, and in 1970 a study of how a person can turn themselves round by moving their limbs.

Problem solved in full

Where along the way is the pull hardest? 4 steps
The 2 kg dumbbells come in from 0.8 m to 0.2 m with L = 22.62 kg·m²/s. The pull the two hands hold is 126.3 N at the start and 284.2 N at the end. Is it hardest at the end? If not, where, and how hard? Press Pull in, 2 kg to check the first step.

Open this problem in the tool
1. The hands keep the dumbbells on their circle with 2mω²r, and ω = L/I(r), so the pull depends on r alone:
  
  $F (r) = 2 m (\frac{L}{I ( r )})^{2} r = \frac{2 \times 2 \times 22 . 6 2 ^{2} r}{( 1 . 04 + 4 r ^{2} ) ^{2}} F (0.8) = 126.3 N, F (0.2) = 284.2 N$
2. The table's 0.30 m row, 313.2 N, is above both ends, so the largest pull is somewhere inside. Set the derivative of r/I² to zero, with I′ = 4mr:
  
  $\frac{d}{d r} \frac{r}{I ^{2}} = \frac{I - 2 r I ^{'}}{I ^{3}} = \frac{I - 8 m r ^{2}}{I ^{3}} = 0 \Rightarrow I = 8 m r^{2}$
3. Write I out and solve for r:
  
  $I_{b} + 2 m d^{2} + 2 m r^{2} = 8 m r^{2} \Rightarrow r^{* 2} = \frac{I _{b} + 2 m d ^{2}}{6 m} = \frac{1 . 04}{12} = 0.08667 \Rightarrow r^{*} = 0.2944 m$
4. Put r* back into I and into the pull:
  
  $I (r^{*}) = 8 \times 2 \times 0.08667 = 1.3867 kg m^{2} F (r^{*}) = \frac{4 \times 22 . 6 2 ^{2} \times 0 . 2944}{1 . 386 7 ^{2}} = 313.3 N$
Answer

The pull is hardest with the hands 0.2944 m from the axis: 313.3 N on the two hands, 156.7 N each, about the weight of a 16 kg mass. Closer in the spin still rises, but r falls faster than ω² grows, and the last 9 cm need less pull.

$r^{*} = 0.2944 m : F_{m a x} = 313.3 N = 2 \times 156.7 N (\approx 16 kg a hand)$

Learning path

Rotation and angular momentum

References (3)

Conservation of angular momentum when the moment of inertia changes, and the skater's kinetic energy rising by I/I′ because of the work she does pulling her arms in, the page's first two claims: W. Moebs, S. J. Ling and J. Sanny, University Physics Volume 1, OpenStax, Houston, 2016. Section 11.3, "Conservation of Angular Momentum".
The falling cat, which turns over with no angular momentum, given a dynamical explanation: T. R. Kane and M. P. Scher, "A dynamical explanation of the falling cat phenomenon", International Journal of Solids and Structures 5(7), 663–670, 1969.
A person turning themselves round by moving their limbs, the page's third claim in a human body: T. R. Kane and M. P. Scher, "Human self-rotation by means of limb movements", Journal of Biomechanics 3(1), 39–49, 1970.

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