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6.3 - Angular Momentum and Angular Impulse

Angular Momentum of Rotating Systems

Angular momentum is the rotational equivalent of linear momentum. Just as linear momentum is p=mvp = mv, angular momentum for a rotating system is:
L=IωL = I\omega
where LL is angular momentum in kg·m²/s, II is moment of inertia in kg·m², and ω\omega is angular velocity in rad/s.
Linear
Rotational
p=mvp = mv
L=IωL = I\omega
mm
II
vv
ω\omega
Example: a 2.0 kg solid disk of radius 0.30 m spins at 12 rad/s. Its rotational inertia is I=12MR2=12(2.0)(0.30)2=0.090I = \frac{1}{2}MR^2 = \frac{1}{2}(2.0)(0.30)^2 = 0.090 kg·m², so its angular momentum is L=Iω=(0.090)(12)=1.08≈1.1L = I\omega = (0.090)(12) = 1.08 \approx 1.1 kg·m²/s.

Angular Momentum of a Single Object

For a single object (not a rigid rotating system) moving about some reference point, angular momentum can be calculated as:
L=mvrsin⁡θL = mvr\sin\theta
where mm is the object’s mass in kg, vv is its linear velocity in m/s, rr is the distance from the reference point to the object in meters, and θ\theta is the angle between the velocity vector and the position vector.
Example: a planet orbiting a star with mass m=6.0×1024m = 6.0 \times 10^{24} kg, orbital velocity v=3.0×104v = 3.0 \times 10^4 m/s, and orbital radius r=1.5×1011r = 1.5 \times 10^{11} m, moving perpendicular to the radius (sin⁡θ=1\sin\theta = 1):
L=mvrsin⁡(90°)=(6.0×1024)(3.0×104)(1.5×1011)(1)≈2.7×1040kg⋅m2/sL = mvr\sin(90°) = (6.0 \times 10^{24})(3.0 \times 10^4)(1.5 \times 10^{11})(1) \approx 2.7 \times 10^{40} \mathrm{ kg \cdot m}^2\text{/s}
This formula is especially useful for analyzing linear motion that creates rotational effects, like someone running tangentially and pushing a merry-go-round.

Angular Momentum of an Object Moving in a Straight Line

An object doesn’t have to spin, or even curve, to have angular momentum. A puck sliding in a straight line has angular momentum about any point that isn’t on its line of motion. The formula sheet writes the equation as L=rmvsin⁡θL = rmv\sin\theta, the same thing as mvrsin⁡θmvr\sin\theta above.
6.3 - Angular Momentum and Angular Impulse
Look at the rsin⁡θr\sin\theta part. In the right triangle in the figure, rsin⁡θr\sin\theta is the side opposite θ\theta, which is the perpendicular distance dd from the reference point to the line of motion (the same lever-arm idea as r⊥r_\perp in torque). So the angular momentum depends on four things:
  • Mass mm: twice the mass, twice the angular momentum.
  • Speed vv: twice the speed, twice the angular momentum.
  • Distance rr from the reference point to the object.
  • Angle θ\theta between rr and the velocity. Together, rr and θ\theta set the perpendicular distance d=rsin⁡θd = r\sin\theta.
As the puck slides past, rr and θ\theta both change, but dd doesn’t, because the line of motion stays the same distance from PP. So an object moving in a straight line at constant speed has constant angular momentum about any fixed point.
The reference point matters: the value of LL depends on which point or axis you measure it about. About QQ, a point on the puck’s line of motion, d=0d = 0 and L=0L = 0. About a point farther from the line, LL is larger. The same goes for rigid objects: a rod has a different II about its center than about its end, so its L=IωL = I\omega is different too. Always state the axis, and use the same axis for every part of a problem.

Worked Example: Angular Momentum of a Sliding Puck

A 0.16 kg hockey puck slides in a straight line at a constant 12 m/s. At its closest approach it passes 1.5 m from point PP. Find its angular momentum about PP (a) at the closest point and (b) later, when it’s 2.5 m from PP. (c) What is its angular momentum about point QQ on its path?
(a) At the closest point, the velocity is perpendicular to rr, so θ=90∘\theta = 90^\circ:
L=rmvsin⁡θ=(1.5)(0.16)(12)sin⁡90∘=2.88≈2.9 kg⋅m2/sL = rmv\sin\theta = (1.5)(0.16)(12)\sin 90^\circ = 2.88 \approx 2.9\ \mathrm{kg\cdot m^2/s}
(b) Now r=2.5r = 2.5 m, and from the triangle, sin⁡θ=dr=1.52.5=0.60\sin\theta = \frac{d}{r} = \frac{1.5}{2.5} = 0.60 (so θ=37∘\theta = 37^\circ):
L=(2.5)(0.16)(12)(0.60)=2.88≈2.9 kg⋅m2/sL = (2.5)(0.16)(12)(0.60) = 2.88 \approx 2.9\ \mathrm{kg\cdot m^2/s}
Same as (a). No net force acts on the puck, so there’s no torque about PP, and its angular momentum about PP can’t change.
(c) QQ is on the line of motion, so θ=0∘\theta = 0^\circ (or 180∘180^\circ) and L=0L = 0 about QQ. Same puck, same motion, different answer, because the reference point changed.

Angular Impulse-Momentum Theorem

Just as linear impulse changes linear momentum, angular impulse changes angular momentum.
Angular impulse is the torque exerted on an object multiplied by the time interval the torque acts. Its units are N·m·s, which is the same as kg·m²/s, the unit of angular momentum.
angular impulse=τΔt\text{angular impulse} = \tau\Delta t
Direction (sign): angular impulse has the same direction as the torque that delivers it. With counterclockwise as positive, a counterclockwise torque gives a positive angular impulse and a clockwise torque gives a negative one. A torque in the same direction as the rotation speeds the object up, and a torque against the rotation slows it down.
Linear: FΔt=ΔpF\Delta t = \Delta p. Angular: τΔt=ΔL\tau\Delta t = \Delta L
where τ\tau is net torque in N·m, Δt\Delta t is the time interval in seconds, and ΔL\Delta L is the change in angular momentum in kg·m²/s. Expanded:
τΔt=Lf−Li=Iωf−Iωi\tau\Delta t = L_f - L_i = I\omega_f - I\omega_i
This is an extension of Newton’s Second Law for rotation, τ=dLdt\tau = \frac{dL}{dt}.
Where it comes from: start with Newton’s second law in rotational form, τnet=Iα\tau_{\text{net}} = I\alpha, and write the angular acceleration as α=ΔωΔt\alpha = \frac{\Delta\omega}{\Delta t}. If the rotational inertia stays constant:
τnet=Iα=IΔωΔt=Iωf−IωiΔt=ΔLΔt\tau_{\text{net}} = I\alpha = I\frac{\Delta\omega}{\Delta t} = \frac{I\omega_f - I\omega_i}{\Delta t} = \frac{\Delta L}{\Delta t}
Multiply both sides by Δt\Delta t and you get τnetΔt=ΔL\tau_{\text{net}}\Delta t = \Delta L, written ΔL=τΔt\Delta L = \tau\Delta t on the formula sheet. It’s the rotational version of FnetΔt=ΔpF_{\text{net}}\Delta t = \Delta p from Unit 4.

Graphical Analysis: Angular Momentum vs. Time

Finding torque from an L vs. t graph: the net torque equals the slope of an angular momentum vs. time graph:
τ=ΔLΔt\tau = \frac{\Delta L}{\Delta t}
6.3 - Angular Momentum and Angular Impulse
For a curved graph, the instantaneous torque at any point is the slope of the tangent line there.
Finding angular impulse from a τ vs. t graph: the area under a torque vs. time graph equals the angular impulse, or change in angular momentum:
ΔL=∫τ dt=Area under τ vs. t graph\Delta L = \int \tau \, dt = \text{Area under } \tau \text{ vs. } t \text{ graph}
6.3 - Angular Momentum and Angular Impulse
Example with the graph above: suppose the torque rises from 0 at t=1.0t = 1.0 s to 4.0 N·m at t=3.0t = 3.0 s, holds at 4.0 N·m until 7.0 s, then falls back to 0 at 9.0 s (the shape shown). The area is two triangles and a rectangle: 12(2.0)(4.0)+(4.0)(4.0)+12(2.0)(4.0)=24\frac{1}{2}(2.0)(4.0) + (4.0)(4.0) + \frac{1}{2}(2.0)(4.0) = 24 N·m·s. So the angular impulse is 24 N·m·s, and the object’s angular momentum goes up by 24 kg·m²/s. If it started at rest with I=3.0I = 3.0 kg·m², it ends at ω=243.0=8.0\omega = \frac{24}{3.0} = 8.0 rad/s.

Worked Example: Braking a Bicycle Wheel

A bicycle wheel with I=0.12I = 0.12 kg·m² spins counterclockwise at 8.0 rad/s. A brake pad exerts a constant friction torque of 0.30 N·m, clockwise, for 2.0 s. Find the angular impulse and the wheel’s final angular velocity.
Signs first: counterclockwise is positive, so ωi=+8.0\omega_i = +8.0 rad/s and τ=−0.30\tau = -0.30 N·m.
Angular impulse: τΔt=(−0.30)(2.0)=−0.60\tau\Delta t = (-0.30)(2.0) = -0.60 N·m·s. It’s negative, the same direction as the clockwise torque.
Initial angular momentum: Li=Iωi=(0.12)(8.0)=0.96L_i = I\omega_i = (0.12)(8.0) = 0.96 kg·m²/s.
Final angular momentum and angular velocity:
Lf=Li+τΔt=0.96+(−0.60)=0.36 kg⋅m2/sωf=LfI=0.360.12=3.0 rad/sL_f = L_i + \tau\Delta t = 0.96 + (-0.60) = 0.36\ \mathrm{kg\cdot m^2/s} \qquad \omega_f = \frac{L_f}{I} = \frac{0.36}{0.12} = 3.0\ \mathrm{rad/s}
The wheel is still turning counterclockwise, just slower. To stop it completely, the brake would need an angular impulse of −0.96-0.96 N·m·s, which takes Δt=0.960.30=3.2\Delta t = \frac{0.96}{0.30} = 3.2 s.

Worked Example: Comparing Torques from L vs. t Graphs

Two objects, A and B, each have straight-line angular momentum vs. time graphs. Object A’s line goes from (0,0) to (5, 20), and Object B’s line goes from (0,0) to (5, 10), half the slope. Additionally, IB=2IAI_B = 2I_A. Find how the average torques on the two objects compare.
Finding average torques: the average torque equals the slope of each L vs. t graph.
τA=20−05−0=4 N⋅mτB=10−05−0=2 N⋅m\tau_A = \frac{20 - 0}{5 - 0} = 4 \ \mathrm{N \cdot m} \qquad \tau_B = \frac{10 - 0}{5 - 0} = 2 \ \mathrm{N \cdot m}
Taking the ratio: τAτB=42=2\frac{\tau_A}{\tau_B} = \frac{4}{2} = 2, so τA=2τB\tau_A = 2\tau_B, or equivalently τB=τA2\tau_B = \frac{\tau_A}{2}.
Notice the rotational inertias never entered the torque calculation: net torque is the slope of the LL vs. tt graph, whatever II is. The inertias only matter for the angular accelerations, α=τI\alpha = \frac{\tau}{I}: αA=4IA\alpha_A = \frac{4}{I_A} and αB=22IA=1IA\alpha_B = \frac{2}{2I_A} = \frac{1}{I_A}, so A’s angular velocity grows 4 times as fast as B’s.

Common Mistakes

Confusing which formula to use: use L=IωL = I\omega for rigid rotating systems (disks, wheels, rotating rods), and L=mvrsin⁡θL = mvr\sin\theta for single objects moving about a point (particles, satellites, orbiting objects).
Forgetting the angle in L=mvrsin⁡θL = mvr\sin\theta: always identify the angle between velocity and position vectors. When velocity is perpendicular to the radius, sin⁡θ=1\sin\theta = 1 (the most common case); when velocity is parallel to the radius, sin⁡θ=0\sin\theta = 0, giving L=0L = 0.
Incorrectly finding slope from graphs: don’t read just one point, you need two to find slope, slope=ΔLΔt=L2−L1t2−t1\text{slope} = \frac{\Delta L}{\Delta t} = \frac{L_2 - L_1}{t_2 - t_1}.
Mixing up slope and area: the slope of an L vs. t graph gives torque, while the area under a τ\tau vs. t graph gives the change in angular momentum, don’t swap these.
Ignoring moment of inertia differences: two objects can share the same angular momentum but have different angular velocities if their moments of inertia differ, since L=IωL = I\omega means a larger II requires a smaller ω\omega for the same LL.

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