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What Is Angular Momentum and When Is It Conserved?

Angular momentum is the rotational equivalent of linear momentum. It is conserved when the resultant external torque acting on an athlete or object is zero, or is negligible during the period being considered.

The Reasoning and Mechanism

For an athlete rotating about an axis, angular momentum is calculated using:

L=IωL = I\omegaL=Iω

Here, LLL is angular momentum, III is the moment of inertia, and ω\omegaω is angular velocity. The SI unit of angular momentum is kg m2s−1\text{kg m}^2\text{s}^{-1}kg m2s−1.

Moment of inertia depends on the athlete's mass and how far that mass is distributed from the axis of rotation. When no resultant external torque acts, total angular momentum remains constant:

I1ω1=I2ω2I_1\omega_1 = I_2\omega_2I1​ω1​=I2​ω

For example, suppose a diver has I1=6 kg m2I_1 = 6\text{ kg m}^2I1​=6 kg m2 and ω1=2 rad s−1\omega_1 = 2\text{ rad s}^{-1}ω. The diver's angular momentum is . If tucking reduces the moment of inertia to , then .

SituationEffect on angular momentum
Airborne athlete with negligible external torqueLLL remains constant
Athlete tucks or draws limbs inwardIII decreases, so ω\omegaω increases
Athlete extends arms or legsIII increases, so ω\omegaω decreases
Ground reaction force produces torque at take-off changes

This reflects Newton's laws in rotational motion: torque causes a change in angular momentum. Internal movements can redistribute mass and alter angular velocity, but they cannot change the system's total angular momentum without external torque.

A common misconception is that angular momentum is conserved whenever an athlete is rotating. It is conserved only when resultant external torque is zero or negligible; ground contact can create substantial torque.

Exam Technique

For IB SEHS subtopic B.2.1, define angular momentum, state L=IωL = I\omegaL=Iω, give the conservation condition, and apply it to a sporting movement. Clearly distinguish angular momentum from angular velocity and explain the inverse relationship between III and ω\omegaω.

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B.2.1 Newton’s laws of motion

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2
​
1
​
=
2 rad s−1
L=6×2=12 kg m2s−1L = 6 \times 2 = 12\text{ kg m}^2\text{s}^{-1}L=6×2=12 kg m2s−1
3 kg m23\text{ kg m}^23 kg m2
ω2=123=4 rad s−1\omega_2 = \frac{12}{3} = 4\text{ rad s}^{-1}ω2​=312​=4 rad s−1
LL
L

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