Topic 6.3 Angular Momentum and Angular Impulse: define angular momentum for rigid bodies and particles, relate net torque to its rate of change, and use the angular impulse-momentum theorem.

What this topic is asking

The College Board (Topic 6.3) wants you to define angular momentum for both rigid bodies ($L = I\omega$) and single particles ($\vec{L} = \vec{r}\times\vec{p}$), to relate the net torque to its rate of change, and to use the angular impulse-momentum theorem. Angular momentum is the rotational analogue of linear momentum, and the same impulse-momentum structure carries over, setting up the powerful conservation law of the next topic.

Angular momentum of a rigid body

For a spinning rigid body, $L = I\omega$ measures the "quantity of rotation", how much rotational motion the body carries. A flywheel with large rotational inertia spinning fast has large angular momentum and is hard to stop turning. The direction of $\vec{L}$ lies along the rotation axis, given by the right-hand rule (curl the fingers with the spin, the thumb points along $\vec{L}$). This is the rigid-body form you use for wheels, disks and turntables.

Angular momentum of a particle

A single particle also has angular momentum about a point, even moving in a straight line. The full definition is the cross product

where $\vec{r}$ is the position vector from the reference point to the particle and $r_\perp = r\sin\theta$ is the perpendicular distance from the point to the particle's line of motion. The striking consequence is that a particle moving in a straight line at constant velocity has constant, nonzero angular momentum about any point off that line, because $r_\perp$ is fixed. This particle form is essential for problems where an object flies in and strikes a pivoted body.

Torque as the rate of change of angular momentum

The deepest statement of rotational dynamics parallels the momentum form of Newton's second law:

The net torque equals the time rate of change of angular momentum. When the rotational inertia is constant this becomes $\tau = I\dfrac{d\omega}{dt} = I\alpha$, recovering Topic 5.6. But the angular-momentum form is more general: it handles a body whose rotational inertia changes (a spinning skater pulling in her arms), where $\tau = I\alpha$ alone would be wrong. It also leads straight to angular momentum conservation when the net torque is zero.

The angular impulse-momentum theorem

Integrating $\tau = dL/dt$ over time gives the angular impulse-momentum theorem: the angular impulse (the time integral of torque) equals the change in angular momentum.

This is the rotational counterpart of the linear impulse-momentum theorem. A torque acting over a time changes the angular momentum by the angular impulse; a constant torque over time $\Delta t$ gives $\Delta L = \tau\Delta t$. It is the natural tool when a torque acts for a known duration, such as friction slowing a wheel or a brief twist spinning one up.

Try this

Q1. A disk with $I = 0.50$ kg m squared spins at $8.0$ rad/s. Calculate its angular momentum. [2 points]

• Cue. $L = I\omega = (0.50)(8.0) = 4.0$ kg m squared per s.

Q2. A $0.20$ kg ball moves at $5.0$ m/s in a straight line whose perpendicular distance from a point is $0.40$ m. Calculate its angular momentum about that point. [2 points]

• Cue. $L = mvr_\perp = (0.20)(5.0)(0.40) = 0.40$ kg m squared per s.