6.3 Angular Momentum and Angular Impulse

Learning objectives

Choose the reference first

Angular momentum describes rotational motion relative to a specified axis or point. Changing that reference can change the rotational inertia, perpendicular geometry, and therefore the measured angular momentum.

Match the model to the motion

Physical model	Angular-momentum magnitude	Geometry
Rigid system rotating about an axis	$L=I\omega$	$I$ and $\omega$ use the same axis
Object moving relative to a point	$L=rmv\sin\theta$	$\theta$ is between $\vec r$ and $\vec v$

Calculate rigid-system momentum

Rigid system: $I=1.2\,\text{kg}\,\text{m}^2$ and $\omega=4.0\,\text{rad}\,\text{s}^{-1}$ .

$L=I\omega=(1.2)(4.0)=4.8\,\text{kg}\,\text{m}^2\,\text{s}^{-1}$ .

Calculate moving-object momentum

Moving object: $m=2.0\,\text{kg}$ , $v=3.0\,\text{m}\,\text{s}^{-1}$ , $r=0.50\,\text{m}$ , and $\theta=90^\circ$ .

$L=rmv\sin\theta=(0.50)(2.0)(3.0)\sin90^\circ=3.0\,\text{kg}\,\text{m}^2\,\text{s}^{-1}$ .

Keep the perpendicular geometry

The distance $r$ alone does not determine a moving object's angular momentum. Only the velocity component perpendicular to $\vec r$ contributes: radial motion has $\theta=0^\circ$ and therefore $L=0$ about that point.

Connect torque and duration

Angular impulse measures the effect of a torque acting over a time interval. Within a one-dimensional sign convention, it has the same signed rotational sense as the torque.

$J_{\mathrm{ang}}=\tau\,\Delta t$

Calculate constant-torque impulse

Constant torque: a signed torque of $+5.0\,\text{N}\,\text{m}$ acts for $0.40\,\text{s}$ .

$J_{\mathrm{ang}}=(+5.0)(0.40)=+2.0\,\text{N}\,\text{m}\,\text{s}$ .

Read signed torque–time area

Torque–time graph feature	Angular impulse
Area above the time axis	Positive
Area below the time axis	Negative
Total signed area	Angular impulse over the interval

Calculate varying-torque impulse

Varying torque: torque rises linearly from $0$ to $8.0\,\text{N}\,\text{m}$ over $0.50\,\text{s}$ . The triangular graph area is

$J_{\mathrm{ang}}=\tfrac12(0.50)(8.0)=2.0\,\text{N}\,\text{m}\,\text{s}$ .

Use the correct graph and method

Angular impulse uses area on a torque–time graph. Area on a torque–angular-position graph represents work instead. Use $\tau\Delta t$ only when torque is constant over the interval.

Measure final minus initial momentum

$\Delta L=L_f-L_i$

Equate impulse and momentum change

$\Delta L=J_{\mathrm{ang}}=\tau_{\mathrm{net}}\Delta t$

When rotational inertia is constant, the rotational second law produces the impulse–momentum theorem:

$\tau_{\mathrm{net}}=\dfrac{\Delta L}{\Delta t}=I\dfrac{\Delta\omega}{\Delta t}=I\alpha$ .

Multiplying by $\Delta t$ gives $\tau_{\mathrm{net}}\Delta t=\Delta L$ .

Connect slope, area, torque, and change

Graph	Operation	Result
Angular momentum $L$ vs. time $t$	Slope	Net torque $\tau_{\mathrm{net}}$
Net external torque $\tau$ vs. time $t$	Signed area	Change in angular momentum $\Delta L$

Calculate a momentum change

Constant inertia: $I=0.50\,\text{kg}\,\text{m}^2$ and angular velocity changes from $2.0$ to $6.0\,\text{rad}\,\text{s}^{-1}$ .

$\Delta L=I(\omega_f-\omega_i)=(0.50)(6.0-2.0)=2.0\,\text{kg}\,\text{m}^2\,\text{s}^{-1}$ .

The delivered angular impulse is therefore $2.0\,\text{N}\,\text{m}\,\text{s}$ .

Use one-dimensional signs only

AP Physics 1 uses one-dimensional signed conventions to manipulate angular-momentum and angular-impulse magnitudes. Full vector directions for these quantities are beyond scope. Keep one sign convention consistent when subtracting $L_i$ from $L_f$ .

Next Topic

—6.4 Conservation of Angular MomentumRead