Angular momentum and angular impulse

Topic 6.2 read the signed area under a $\tau$-vs-$\theta$ graph and got rotational work. This topic reads the partner graph, $\tau$-vs-$t$, and gets angular impulse. Same shape on the page; two different physical quantities; two different units.

Angular momentum $L = I\omega$ is the rotational version of linear momentum $p = mv$: inertia times motion. Angular impulse $J = \tau\,\Delta t$ is the rotational version of linear impulse $J = F\,\Delta t$ from Topic 4.2. One rule ties them together: $J_{\text{net}} = \Delta L$, the angular impulse-momentum theorem.

Once you see that this is the old impulse-momentum theorem with new letters, the rest is reading the right graph for the right quantity and keeping signs honest.

Two definitions and one graph reading.

Units: $L$ in $\text{kg}\cdot\text{m}^2/\text{s}$, $J$ in $\text{N}\cdot\text{m}\cdot\text{s}$. Both are signed components about the chosen axis: pick a positive direction at the start of the problem and stick with it. A negative $L$ means the wheel spins in the negative direction; it is not a small $L$.

• $L$: angular momentum about the chosen axis (kg·m²/s), signed.
• $I$: moment of inertia about that axis (kg·m²), always positive.
• $\omega$: angular velocity (rad/s), signed.
• $\tau$: net torque about that axis (N·m), signed.
• $J$: angular impulse (N·m·s), signed. Same letter as linear impulse; context tells you which.

When the torque varies, the rule $J = \tau\,\Delta t$ stops working as written. The general reading is: angular impulse is the signed area under the $\tau$-vs-$t$ graph. Rectangle for constant $\tau$; triangle for a $\tau$ that rises linearly from zero; trapezoid for a $\tau$ that varies linearly but starts at a nonzero value.

Start from $\tau = I\alpha$, the rotational form of Newton's second law. Multiply both sides by a duration $\Delta t$ over which $\tau$ is held constant:

So $J = \Delta L$ for a constant torque. When $\tau$ varies, the same logic applies in each tiny slice of time, and the slices add up to the signed area under the $\tau$-vs-$t$ graph. Either way:

Two graphs, two physical quantities. This is the trap 6.2 set up and 6.3 closes. Both plot $\tau$ on the vertical axis. The difference is the horizontal axis.

• $\tau$-vs-$\theta$ graph: signed area is rotational work, $W = \tau\,\Delta\theta$ for constant $\tau$. Units: $\text{N}\cdot\text{m}$ (joules).
• $\tau$-vs-$t$ graph: signed area is angular impulse, $J = \tau\,\Delta t$ for constant $\tau$. Units: $\text{N}\cdot\text{m}\cdot\text{s}$.

One $\tau$ value, two different things you can do with it. Angle on the bottom: you are reading work. Time on the bottom: you are reading impulse. Glance at the axis label before you compute area.

A point particle moving in a straight line still has angular momentum. Angular momentum is defined about a chosen axis. For a particle of mass $m$ moving with speed $v$ along a line that sits a perpendicular distance $d$ from the axis,

$L$ is zero only when $d = 0$, meaning the line of motion passes through the axis. Otherwise the particle carries angular momentum about that axis, even though it never circles around. You'll use this in Topic 6.4, where one object in a collision is often a point mass passing a pivot.

Angular impulse trips students who trust the wrong graph, equate a bigger torque with a bigger effect, or assume straight-line motion carries no angular momentum. Your diagnostic checks all three.