Topic 6.3 Angular Momentum and Angular Impulse: define angular momentum and relate the angular impulse from a torque to the change in angular momentum.

What this topic is asking

The College Board (Topic 6.3) wants you to define angular momentum as the rotational analogue of linear momentum, and to relate the angular impulse delivered by a torque to the resulting change in angular momentum. The analogy is exact: momentum $p = mv$ becomes angular momentum $L = I\omega$, and linear impulse $F\,\Delta t$ becomes angular impulse $\tau\,\Delta t$.

What angular momentum is

Just as linear momentum $p = mv$ measures "quantity of translational motion", angular momentum $L = I\omega$ measures "quantity of rotational motion". A massive wheel spinning fast has large angular momentum; the same wheel at rest has none. For a particle moving in a straight line, angular momentum about a chosen point is still defined ($L = mvr_\perp$) and is generally nonzero, which matters when a moving object strikes and starts spinning a system.

Angular impulse changes angular momentum

This theorem is the time-based partner to the work-energy theorem of Topic 6.2. The two answer different questions: angular impulse (torque times time) changes angular momentum, while rotational work (torque times angle) changes rotational kinetic energy. Keeping these straight, the time integral relates to momentum and the displacement integral relates to energy, is the same discipline you used to separate linear impulse from linear work in Units 3 and 4.

Why the perpendicular distance matters

For a point particle, the $r_\perp$ in $L = mvr_\perp$ is the lever arm of the momentum, the perpendicular distance from the axis to the line along which the particle moves. A puck sliding past a pivot has angular momentum about that pivot even though it travels in a straight line, and that angular momentum can be transferred to a rod it strikes, setting the rod spinning. This is the bridge between the linear momentum of Unit 4 and the rotational world: when an object collides with something it can rotate about a pivot, you conserve angular momentum, not linear momentum, because the pivot exerts an external force but no torque about itself. The strategic insight is that $L = I\omega$ and $\tau\,\Delta t = \Delta L$ complete the rotational analogues of the momentum toolkit, and they set up the conservation law (Topic 6.4) that explains spinning skaters, collapsing stars, and a diver tucking into a somersault.

Try this

Q1. A disc of rotational inertia $0.80$ kg$\cdot$m squared spins at $6.0$ rad/s. Calculate its angular momentum. [2 points]

• Cue. $L = I\omega = (0.80)(6.0) = 4.8$ kg$\cdot$m squared/s.

Q2. A torque of $1.5$ N$\cdot$m acts on a wheel for $4.0$ s. Calculate the angular impulse delivered. [1 point]

• Cue. Angular impulse $= \tau\,\Delta t = (1.5)(4.0) = 6.0$ kg$\cdot$m squared/s.