Is angular momentum conserved when an ice skater pulls in their arms?

Angular momentum is approximately conserved if the resultant external torque about the skater's rotation axis is negligible. Pulling in the arms involves internal muscular forces, which can redistribute mass but do not change the total angular momentum of the complete skater system. Friction with the ice and air resistance provide small external torques, so conservation is an idealization rather than an exact description over an unlimited time. During the short pulling-in motion, these external torques are usually small enough to ignore in an IB calculation. The appropriate equation is therefore $Ii\omegai=If\omegaf$.

Does the skater's angular momentum increase when they spin faster?

No, not in the standard idealized example. The skater's angular velocity increases, but their moment of inertia decreases at the same time. These changes compensate so that the product $L=I\omega$ remains constant. Confusing angular velocity with angular momentum is a common exam mistake because both are associated with faster rotation. Angular momentum would change only if a resultant external torque acted over time and produced an angular impulse.

Where does the skater's extra rotational kinetic energy come from?

The additional rotational kinetic energy comes from work done by the skater's muscles while pulling the arms inward. At constant angular momentum, rotational kinetic energy can be written as $Ek=L^2/(2I)$. Reducing $I$ therefore increases $Ek$, even though $L$ remains unchanged. Chemical energy stored in the skater's body is transferred into kinetic energy during the movement. Consequently, conservation of angular momentum does not imply that rotational kinetic energy must also remain constant.

What happens when a spinning skater extends their arms again?

Extending the arms moves mass farther from the rotation axis and increases the skater's moment of inertia. If the resultant external torque remains negligible, angular momentum is still conserved. The angular velocity must therefore decrease according to $Ii\omegai=If\omegaf$. In an ideal reversible model, returning to the original mass distribution returns the skater to the original angular speed. In reality, friction, air resistance, muscular control, and internal energy transfers cause some mechanical energy to be dissipated.

Why is the ice skater example part of IB Physics HL rather than SL?

Under the IB Physics course first assessed in 2025, A.4 Rigid body mechanics contains no Standard Level content and is studied at Higher Level. This area includes moment of inertia, rotational dynamics, angular momentum, angular impulse, and rotational kinetic energy. The skater example applies the HL relationship $L=I\omega$ to a system whose mass distribution changes. SL students may encounter rotational ideas informally, but they are not required to study A.4 as assessed syllabus content. HL students should be able to explain the effect conceptually and calculate changes in angular velocity or rotational energy.

Can a skater start spinning from rest simply by pulling in their arms?

No, pulling the arms inward cannot create net angular momentum when the skater initially has zero total angular momentum and no external torque acts. The movement can rearrange body parts, but the total angular momentum of the complete system must remain zero. To begin a sustained spin, the skater must obtain angular momentum through an external torque, typically by pushing against the ice. Once rotating, they can change their angular velocity by changing their moment of inertia. This distinction separates creating angular momentum from redistributing mass while conserving existing angular momentum.

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Why does a spinning ice skater speed up when they pull their arms in

• 29 Jul 2026•14

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Why does a spinning ice skater speed up when they pull their arms in

• 29 Jul 2026•14

A spinning ice skater speeds up when they pull their arms inward because doing so reduces their moment of inertia. If the resultant external torque on the skater is negligible, their angular momentum remains constant, so the decrease in moment of inertia must be accompanied by an increase in angular velocity.

This is the classic example used to explain conservation of angular momentum. For IB Physics students, it connects three quantities that must be kept distinct: angular momentum, moment of inertia, and angular velocity.

The short physics explanation

For a body rotating about a fixed axis, angular momentum is given by:

$L=I\omega$

where:

$L$ is angular momentum, measured in kg m² s⁻¹
$I$ is moment of inertia, measured in kg m²
$\omega$ is angular velocity, measured in rad s⁻¹

If the resultant external torque is zero or negligible, angular momentum is conserved:

$L_i=L_f$

Therefore:

$I_i\omega_i=I_f\omega_f$

Pulling the arms inward moves mass closer to the skater's axis of rotation. This reduces $I$ . Because the product $I\omega$ must remain constant, $\omega$ increases and the skater completes each rotation more quickly.

The relationship can be written as:

$\omega_f=\frac{I_i}{I_f}\omega_i$

If the final moment of inertia is half the initial value, the final angular velocity is twice the initial value.

Why moving the arms changes moment of inertia

Moment of inertia measures how strongly a body's mass distribution resists a change in rotational motion about a specified axis. It is the rotational analogue of mass, but unlike ordinary mass, it depends on where the mass is located.

For a collection of point masses:

$I=\sum mr^2$

Here, $r$ is the perpendicular distance from each mass to the axis of rotation. The squared distance is important: moving part of the body's mass closer to the axis can substantially reduce the moment of inertia.

Skater's position	Mass distribution	Moment of inertia	Angular velocity
Arms extended	More mass far from the axis	Larger	Smaller
Arms pulled inward	More mass close to the axis	Smaller	Larger

The skater's total mass does not change. What changes is the distribution of that mass relative to the rotation axis. RevisionDojo's explanation of why mass distribution affects rotation develops this distinction further.

A useful simplified model treats each hand as a point mass. If a hand of mass $m$ moves from distance $r$ to distance $r/2$ , its contribution to the moment of inertia changes from:

$I_i=mr^2$

to:

$I_f=m\left(\frac{r}{2}\right)^2=\frac{1}{4}mr^2$

The hand's contribution becomes one quarter of its previous value. A real skater is not a set of point masses, but the same principle applies to their arms, free leg, torso, and equipment.

Conservation of angular momentum explained

The rotational form of Newton's second law connects resultant external torque to the rate of change of angular momentum:

$\sum\tau_{\text{external}}=\frac{\Delta L}{\Delta t}$

Equivalently, over a time interval:

$\Delta L=\tau\Delta t$

If the resultant external torque is zero:

$\Delta L=0$

The total angular momentum of the chosen system therefore remains constant. This is the conservation of angular momentum condition that must be stated in a complete IB explanation.

The skater is not literally free from all external forces. Gravity acts downward, and the ice exerts an upward contact force. However, these forces produce little or no resultant torque about the approximately vertical rotation axis. Friction and air resistance do produce some opposing torque, but they can often be treated as negligible during the short interval in which the skater pulls in their arms.

The OpenStax treatment of conservation of angular momentum uses the same skater model and explicitly connects the reduction in moment of inertia to the increase in angular velocity.

Why the skater can accelerate without an external torque

Students sometimes argue that angular acceleration requires a resultant external torque, so the skater should not be able to speed up. The problem is that the familiar equation

$\tau=I\alpha$

is simplest when the moment of inertia is constant. A skater pulling in their arms is a changing-shape system, so $I$ is changing.

The more general relationship is:

$\tau_{\text{external}}=\frac{dL}{dt}=\frac{d(I\omega)}{dt}$

Using the product rule gives:

$\tau_{\text{external}}=I\frac{d\omega}{dt}+\omega\frac{dI}{dt}$

When external torque is negligible, the left-hand side is approximately zero:

$I\frac{d\omega}{dt}=-\omega\frac{dI}{dt}$

As the skater pulls inward, $dI/dt$ is negative. The right-hand side is therefore positive, so $d\omega/dt$ is positive. The angular speed can increase even though the skater's total angular momentum remains unchanged.

The forces exerted by the skater's muscles are internal to the system consisting of the complete skater. Internal forces can redistribute mass and energy, but they cannot change the total angular momentum of an isolated system. This is similar to people moving around inside a stationary boat: internal motion can rearrange the system without creating total momentum from nothing.

What happens to rotational kinetic energy

Angular momentum is conserved in the idealized skater example, but rotational kinetic energy is not constant. Rotational kinetic energy is:

$E_{k,\text{rot}}=\frac{1}{2}I\omega^2$

Using $L=I\omega$ , this can also be written as:

$E_{k,\text{rot}}=\frac{L^2}{2I}$

If $L$ remains constant while $I$ decreases, rotational kinetic energy increases. This does not violate conservation of energy because the skater does work while pulling their arms inward. Chemical energy in their muscles is transferred into rotational kinetic energy.

Suppose the moment of inertia is reduced to half its initial value:

$I_f=\frac{1}{2}I_i$

Conservation of angular momentum gives:

$\omega_f=2\omega_i$

The energy ratio is then:

$\frac{E_f}{E_i}=\frac{\frac{1}{2}I_f\omega_f^2}{\frac{1}{2}I_i\omega_i^2}$

$\frac{E_f}{E_i}=\frac{\left(\frac{1}{2}I_i\right)(2\omega_i)^2}{I_i\omega_i^2}=2$

The rotational kinetic energy doubles. The additional energy comes from the work done by the skater, not from an increase in angular momentum. OpenStax's University Physics discussion of angular momentum also distinguishes conservation of angular momentum from conservation of rotational kinetic energy.

Worked IB Physics example

A skater rotates at $2.4\ \text{rad s}^{-1}$ with a moment of inertia of $4.0\ \text{kg m}^2$ . They pull in their arms, reducing their moment of inertia to $1.6\ \text{kg m}^2$ . Assume the resultant external torque is negligible.

Step 1: State the governing principle

Because the resultant external torque is negligible, angular momentum is conserved:

$I_i\omega_i=I_f\omega_f$

Step 2: Substitute the values

$4.0\times2.4=1.6\times\omega_f$

Step 3: Calculate the final angular velocity

$\omega_f=\frac{4.0\times2.4}{1.6}=6.0\ \text{rad s}^{-1}$

The skater's angular velocity increases from 2.4 rad s⁻¹ to 6.0 rad s⁻¹.

Step 4: Check the direction of the result

The moment of inertia decreased by a factor of:

$\frac{4.0}{1.6}=2.5$

The angular velocity should therefore increase by the same factor:

$2.4\times2.5=6.0\ \text{rad s}^{-1}$

This proportional check is useful in an exam because it quickly exposes inverted ratios.

Step 5: Compare rotational kinetic energies

Initially:

$E_i=\frac{1}{2}(4.0)(2.4)^2=11.52\ \text{J}$

Finally:

$E_f=\frac{1}{2}(1.6)(6.0)^2=28.8\ \text{J}$

The rotational kinetic energy increases by:

$\Delta E_k=28.8-11.52=17.28\ \text{J}$

In the idealized model, this increase is supplied by work done by the skater's muscles.

How this fits the current IB Physics course

In the current IB Physics course, first assessed in 2025, rotational mechanics appears in A.4 Rigid body mechanics, which is HL-only content. The relevant syllabus ideas include $L=I\omega$ , conservation of angular momentum, angular impulse, and rotational kinetic energy.

The official IB Physics subject brief confirms the current assessment structure: Paper 1 contains multiple-choice and data-based questions, while Paper 2 contains short-answer and extended-response questions. An ice-skater problem can therefore appear as a conceptual comparison, a calculation, or part of a longer explanation.

For broader coverage, use IB Physics rigid body mechanics explained as the topic-wide, exam-focused overview. This article focuses only on the skater mechanism rather than repeating the full treatment of torque, equilibrium, rolling motion, and angular impulse.

Students who need the syllabus sequence can review the A.4.3 conservation of angular momentum notes alongside the A.4.2 moment of inertia notes.

How to write a strong exam explanation

For a question asking why the skater speeds up, a complete response should contain a linked chain of reasoning:

Pulling the arms inward moves mass closer to the rotation axis.
This decreases the skater's moment of inertia.
The resultant external torque is negligible, so angular momentum is conserved.
Since $L=I\omega$ , a decrease in $I$ requires an increase in $\omega$ .
Therefore, the skater spins faster.

A concise model answer would be:

Pulling the arms inward reduces the skater's moment of inertia because more of their mass is brought closer to the rotation axis. The resultant external torque is negligible, so angular momentum is conserved. Since $I_i\omega_i=I_f\omega_f$ , the decrease in moment of inertia causes angular velocity to increase.

Notice that this answer states both the conservation condition and the relevant equation. Saying only that the skater has “less rotational inertia” identifies part of the mechanism but does not fully explain why angular speed must increase.

Common misconceptions and mistakes

Mistake 1: Saying angular momentum increases

In the idealized model, angular momentum remains constant. It is angular velocity and rotational kinetic energy that increase.

Mistake 2: Assuming all energy is conserved as rotational kinetic energy

Total energy is conserved, but the skater transfers chemical energy into rotational kinetic energy by doing work. Rotational kinetic energy alone is therefore not constant.

Mistake 3: Saying there are no external forces

External forces do act on the skater. The relevant condition is that the resultant external torque about the rotation axis is negligible, not that every external force is absent.

Mistake 4: Reversing the moment-of-inertia ratio

From

$I_i\omega_i=I_f\omega_f$

it follows that

$\omega_f=\frac{I_i}{I_f}\omega_i$

If $I_f<I_i$ , the ratio $I_i/I_f$ must be greater than one. A final speed smaller than the initial speed should immediately signal an algebraic error.

Mistake 5: Treating moment of inertia as a property of mass alone

Moment of inertia depends on mass distribution and the chosen axis. The same skater can have different moments of inertia without changing total mass.

Practical revision strategy

Begin by memorizing the condition rather than only the formula: angular momentum is conserved when the resultant external torque is zero. Then practise identifying whether a question is about conservation of angular momentum, rotational kinetic energy, or angular impulse.

Use the RevisionDojo A.4 rigid body mechanics Questionbank to practise ratio problems and written explanations. The IB Physics flashcards can reinforce definitions and units, while Jojo AI can help diagnose why a calculation used the wrong ratio or conservation law.

Conclusion

A spinning ice skater speeds up when they pull their arms inward because their mass moves closer to the rotation axis, reducing their moment of inertia. With negligible resultant external torque, angular momentum remains constant, so angular velocity must increase according to $I_i\omega_i=I_f\omega_f$ .

The skater's rotational kinetic energy also increases because they do work while pulling inward. For IB exams, state the torque condition, identify the change in moment of inertia, apply conservation of angular momentum, and check that the direction of the answer is physically sensible. RevisionDojo's rigid body mechanics notes, Questionbank, Flashcards, and Jojo AI are useful for turning this explanation into reliable exam technique.

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