Angular momentum and angular impulse

A spinning body carries angular momentum, the rotational version of linear momentum. For a particle it is $\vec{L}=\vec{r}\times\vec{p}$; for a rigid body about a fixed axis it is $L=I\omega$, the moment of inertia times the angular speed. It is a vector, with a direction set by the right-hand rule.

A torque changes that angular momentum. Over a time interval it delivers an angular impulse, $\int\tau\,dt$, which equals the change $\Delta L$. This is the rotational twin of the linear impulse $\int F\,dt=\Delta p$, and the time partner of Topic 6.2: there a torque acting over an angle did work; here a torque acting over time changes angular momentum.

Five habits carry the topic: integrate when $\tau$ varies in time, tell angular impulse apart from work, pair the moment of inertia with the angular speed, keep the $\sin\theta$ in the cross product, and find the direction with the right-hand rule.

Angular momentum comes in two equivalent forms. For a single particle, $\vec{L}=\vec{r}\times\vec{p}$, where $\vec{r}$ is the position from the reference point and $\vec{p}=m\vec{v}$ is the linear momentum. Its magnitude is $L=rp\sin\theta=rmv\sin\theta$, with $\theta$ the angle between $\vec{r}$ and $\vec{p}$.

For a rigid body spinning about a fixed axis, the particle pieces add up to $L=I\omega$: the moment of inertia times the angular speed. This is the rotational partner of $p=mv$, with $I$ standing in for mass and $\omega$ for velocity.

• $\vec{L}=\vec{r}\times\vec{p}$, magnitude $rmv\sin\theta$, for a particle.
• $L=I\omega$, for a rigid body about a fixed axis.

The partners must match. Using $mv$ alone gives linear momentum, not angular; using $M$ with $\omega$ or $I$ with a rim speed $v$ mixes a linear quantity with a rotational one. Match $I$ with $\omega$, and keep the lever arm $r$ for a particle.

Angular momentum is $\vec{L}=\vec{r}\times\vec{p}$ (magnitude $rmv\sin\theta$) for a particle, and $L=I\omega$ for a rigid body. Match $I$ with $\omega$; $mv$ on its own is linear momentum.

A torque acting over time changes angular momentum. For a constant torque the change is $\Delta L=\tau\,\Delta t$: torque times the time, in $\mathrm{kg\cdot m^2/s}$. It is the rotational version of the linear impulse $\Delta p=F\,\Delta t$.

When the torque varies with time, cut the interval into tiny steps $dt$. Over each step the torque adds $dL=\tau\,dt$. Adding the steps gives the angular impulse, the area under the torque-versus-time curve:

• $\Delta L=\displaystyle\int_{t_1}^{t_2}\tau\,dt$, the area under the $\tau$-versus-$t$ curve between the start and end times.

For example, $\tau(t)=6t$ N$\cdot$m from $t=0$ to $t=2$ s delivers $\Delta L=\int_0^2 6t\,dt=3t^2\big|_0^2=12\ \mathrm{kg\cdot m^2/s}$: the triangle under the line, not $\tau\,\Delta t$.

Angular impulse is the area under the $\tau$-versus-$t$ curve, $\Delta L=\int\tau\,dt$. For a constant torque that area is the rectangle $\tau\,\Delta t$; for a varying torque, integrate.

The same torque can be integrated two ways, giving two different quantities. Integrate over time and you get the angular impulse, $\int\tau\,dt=\Delta L$, the change in angular momentum, in $\mathrm{kg\cdot m^2/s}$. Integrate over angle and you get the work, $\int\tau\,d\theta=\Delta K$, the change in rotational kinetic energy, in joules.

Mixing them up is the common slip: giving $\tau\,\Delta\theta$ (work) when the impulse is wanted, or $\tau\,\Delta t$ (impulse) when the work is wanted. The power $P=\tau\omega$, a per-second rate in watts, is a third quantity. For a constant $\tau=8$ N$\cdot$m over $\Delta t=2$ s while the wheel turns $\Delta\theta=10$ rad, the impulse is $\tau\,\Delta t=16\ \mathrm{kg\cdot m^2/s}$ but the work is $\tau\,\Delta\theta=80$ J: different numbers, different units, different physics.

Time gives impulse: $\int\tau\,dt=\Delta L$ ($\mathrm{kg\cdot m^2/s}$). Angle gives work: $\int\tau\,d\theta=\Delta K$ (joules). Check which variable you integrate over.

For a particle, angular momentum is a cross product, so it has a magnitude and a direction. The magnitude is $L=rp\sin\theta=rmv\sin\theta$, where $r\sin\theta$ is the perpendicular lever arm, the shortest distance from the reference point to the line of motion. The $\sin\theta$ is not optional: dropping it (writing $rp$) assumes the motion is already perpendicular, and using $\cos\theta$ takes the wrong component.

The direction comes from the right-hand rule for $\vec{r}\times\vec{p}$: point the fingers along $\vec{r}$, curl them toward $\vec{p}$, and the thumb gives $\vec{L}$, perpendicular to both. For motion in a plane, $\vec{L}$ points out of the page or into it. It is not along $\vec{r}$ and not along $\vec{p}$, and reversing the order ($\vec{p}\times\vec{r}$) flips the sign. Once you pick the original spin direction as positive, a braking torque opposes the spin, so it delivers a negative $\Delta L$ that you subtract.

For $\vec{r}=3\hat{\imath}$ m and $\vec{p}=4\hat{\jmath}$ kg$\cdot$m/s, $\vec{L}=(3)(4)(\hat{\imath}\times\hat{\jmath})=12\hat{k}\ \mathrm{kg\cdot m^2/s}$: magnitude $12$, pointing out of the page.

Magnitude $L=rmv\sin\theta$ uses the perpendicular lever arm; direction comes from the right-hand rule for $\vec{r}\times\vec{p}$, perpendicular to both. Keep the $\sin\theta$ and keep the sign.

Angular-momentum problems tend to break at the same joints: mixing up angular impulse with rotational work, pairing the wrong factors, or dropping the $\sin\theta$ and the direction.

• Constant-torque shortcutUsing $\tau\,\Delta t$ when $\tau$ varies in time. Fix: integrate, $\Delta L=\int\tau\,dt$.
• Impulse vs workIntegrating torque over angle (work) when the impulse is wanted, or over time when work is wanted. Fix: time gives $\Delta L$, angle gives work.
• Mismatched partnersUsing $mv$, $M\omega$, or $Iv$ for angular momentum. Fix: $L=I\omega$ for a body, $rmv\sin\theta$ for a particle.
• Missing sineWriting $rp$ or $rp\cos\theta$ for the cross-product magnitude. Fix: $L=rp\sin\theta$, the perpendicular component.
• Direction and signTreating $\vec{L}$ as a scalar, or dropping the sign of a braking $\Delta L$. Fix: right-hand rule for direction, keep the sign.

Integrate when $\tau$ varies, tell impulse from work, match $I$ with $\omega$, keep the $\sin\theta$, and find the direction with the right-hand rule. The whole topic comes down to those five habits.