Topic 6.4 Conservation of Angular Momentum: state that angular momentum is conserved when the net external torque is zero, and apply it to changing rotational inertia and rotational collisions.

What this topic is asking

The College Board (Topic 6.4) wants you to state that angular momentum is conserved when the net external torque is zero, and to apply this to systems with changing rotational inertia (the spinning skater) and to rotational collisions (a particle striking a pivoted body). It is the rotational analogue of linear momentum conservation and one of the most elegant tools in mechanics.

The condition for conservation

The justification is $\vec{\tau}_{ext} = \dfrac{d\vec{L}}{dt}$ from the previous topic: if the net external torque is zero, the total angular momentum does not change. Internal torques come in equal-and-opposite pairs (Newton's third law for rotation) and cancel when summed over the system, so they cannot alter the total. This mirrors linear momentum conservation exactly, with torque in place of force and angular momentum in place of linear momentum.

Changing rotational inertia: the skater effect

The most vivid consequence appears when a body's rotational inertia changes while no external torque acts. Then $I\omega$ is constant, so

A figure skater spinning with arms out has a large $I$ and slow $\omega$; pulling her arms in decreases $I$, so $\omega$ increases to keep $L$ fixed, and she spins faster. The same principle speeds up a collapsing star and lets a diver control rotation by tucking. Note that the rotational kinetic energy $\tfrac{1}{2}I\omega^2 = \tfrac{1}{2}L\omega$ rises when $\omega$ rises, the extra energy comes from the work the skater does pulling her arms inward against the outward pull, so energy is not conserved even though angular momentum is.

Rotational collisions

When a moving object strikes a pivoted body and the two then rotate together, angular momentum about the pivot is conserved (the impact forces are internal, and the pivot exerts no torque about itself). Treat the incoming object as a particle with angular momentum $L = mvr_\perp$ about the pivot, and set the total before equal to the total after:

This is the rotational analogue of a perfectly inelastic collision, and like that case, kinetic energy is lost to the impact. The method, particle angular momentum in, combined rotational inertia out, handles a bullet embedding in a rod, a child jumping on a merry-go-round, or a ball of clay striking a turntable.

Try this

Q1. A turntable with $I = 0.40$ kg m squared spins at $6.0$ rad/s. A lump of clay is dropped on, raising the inertia to $0.60$ kg m squared. Calculate the new angular velocity. [2 points]

• Cue. $I_i\omega_i = I_f\omega_f \Rightarrow \omega_f = (0.40)(6.0)/0.60 = 4.0$ rad/s.

Q2. Explain why a skater spins faster when she pulls her arms in, in terms of conserved quantities. [2 points]

• Cue. No external torque, so $L = I\omega$ is conserved; reducing $I$ increases $\omega$.