7.4 Angular momentum and angular impulse

Achievable AP Physics 1

7. Torque and rotational mechanics

Our AP Physics 1 course is currently in development and is a work-in-progress.

Angular momentum and angular impulse

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In linear motion, momentum $p = m v$ tells you how difficult it is to change an object’s motion. In rotational motion, the analogous quantity is angular momentum, which tells you how difficult it is to change an object’s spin.

In this sub-chapter we cover:

The concept of angular momentum
The algebraic definition of angular momentum
Concept and application of the angular impulse-momentum theorem.

Angular momentum of a particle

Consider a particle of mass $m$ at position $r$ (measured from an origin $O$ ) moving with velocity $v$ . Its angular momentum about $O$ is defined as

$L = r \times (m v),$

and its magnitude is

$L = m r v sin ϕ,$

where $ϕ$ is the angle between $r$ and $v$ .

A useful special case is when the particle moves directly toward or away from the origin. Then the velocity is purely radial ( $ϕ = 0$ ), so $sin ϕ = 0$ and therefore $L = 0$ .

Angular momentum of a rigid body

For a rigid body rotating about a fixed axis with angular speed $ω$ , each mass element $m_{i}$ at distance $r_{i}$ from the axis moves with tangential speed $r_{i} ω$ . That element contributes angular momentum

$L_{i} = m_{i} (r_{i} ω) r_{i} = m_{i} r_{i}^{2} ω .$

Adding the contributions from all mass elements gives

$L = i \sum L_{i} = ω i \sum m_{i} r_{i}^{2} = I ω,$

where

$I = i \sum m_{i} r_{i}^{2}$

is the moment of inertia about that axis.

Common forms of $I$ :

$Thin hoop: Solid disk: Solid sphere: Rod about center: Rod about end: I I I I I = M R^{2}, = \frac{1}{2} M R^{2}, = \frac{2}{5} M R^{2}, = \frac{1}{12} M L^{2}, = \frac{1}{3} M L^{2} .$

Conservation of angular momentum

If a system experiences no net external torque about a chosen axis, then the system’s total angular momentum about that axis stays constant.

Equivalently,

$i \sum τ_{ext, i} = 0 ⟹ L_{initial} = L_{final} .$

Here, depending on the situation, you’ll typically use

$L = I ω (for a rigid body about a fixed axis), or L = r \times p (for a point mass).$

“No net external torque” means all torques due to forces from outside the system sum to zero.
Internally generated torques (action-reaction pairs within the system) can’t change the system’s total $L$ .
This is the rotational analogue of Newton’s first law for linear momentum. Illustration 1:

Angular momentum of a satellite orbiting the earth

A satellite moves under gravity, which always acts along the line from Earth’s center to the satellite (the radial vector $r$ ). Since torque is $τ = r \times F$ and $r ∥ F$ , we have $τ = 0$ . With no external torque, the satellite’s angular momentum $L = m r v_{⊥}$ remains constant throughout its orbit. Illustration 2:

No net external torque, hence angular momentum is conserved

This diagram shows two masses $m_{1}$ and $m_{2}$ connected by an internal “spring” (coiled line). The forces $F_{1 \to 2}$ and $F_{2 \to 1}$ act along the line joining them. Each mass is at a lever arm $r_{i}$ from the rotation axis at $O$ , producing internal torques $τ_{1}$ and $τ_{2}$ . Because these torques are equal in magnitude and opposite in direction, they cancel pairwise. That leaves no net external torque, so the total angular momentum of the system is conserved.

Angular impulse-momentum theorem

Torque plays the same role in rotation that force plays in translation.

A constant force $F$ acting for time $Δ t$ gives a linear impulse $J = F Δ t$ and changes momentum by $Δ p = J$ .
A constant torque $τ$ acting for time $Δ t$ gives an angular impulse

$J_{θ} = τ Δ t,$

and changes angular momentum by

$Δ L = L_{f} - L_{i} = J_{θ} .$

If the torque varies with time, the total impulse is found by adding up (or integrating) the contributions from small time intervals. In AP-level problems, you’ll most often work with constant-torque cases.

Angular impulse-momentum theorem illustrated

Reasoning practice

You whirl a small stone on a string in a horizontal circle. Without external torque, you pull the string in so that the stone’s radius decreases.

Questions:

What happens to the stone’s angular speed $ω$ ?
Which quantity remains constant, and why?

Solution:

As $r$ decreases, $ω$ increases so that $L$ remains constant.
The angular momentum $L = m r^{2} ω$ is conserved because no external torque acts (tension in the string is radial, so $τ = 0$ ).

A car’s wheels lock and skid (no rotation) on an icy patch, then regain traction and spin again at the same forward speed.

Questions:

Compare the angular momentum of a wheel about its axle before lock, during skid, and after regaining traction.
What external torques act during each phase?

Solution:

Before lock: $L = I ω_{rolling}$ . During skid: $ω = 0 ⟹ L = 0$ . After traction: wheels spin so $L = I ω_{new}$ (same as before if no slip).
Before lock: static friction provides torque to spin wheel but net torque about axle is zero for steady rolling. During skid: kinetic friction exerts a torque opposing wheel spin (bringing $L \to 0$ ). After traction: static friction again enforces rolling without slipping (no net torque if speed constant).

Motion of the wheel before lock, during skid and after traction

A satellite moves in an elliptical orbit around Earth. As it swings inward toward perigee, its speed increases; as it moves outward to apogee, its speed decreases.

Questions:

Why does the satellite’s angular momentum about Earth remain constant?
Which component of its velocity contributes to $L$ ?

Solution:

The only force is gravity, which acts along the line from Earth to satellite (radial). A radial force produces zero torque about Earth, so $L$ is conserved.
Only the transverse (perpendicular to the radius vector) component $v_{⊥}$ contributes, since $L = m r v_{⊥}$ .

A heavy flywheel spins freely. A brake pad is pressed suddenly against its rim, applying a large torque for a very short time, then released.

Questions:

How does the flywheel’s angular momentum change during the brake pulse?
Why does a large torque for a short time produce the same $Δ L$ as a smaller torque for a longer time?

Solution:

The brake’s frictional torque (opposite rotation) reduces $L$ by $Δ L = τ Δ t$ until it stops or slows by the impulse delivered.
Angular impulse depends on the product $τ Δ t$ (the area under the torque-time curve), so a higher torque for shorter duration can equal the impulse of a lower torque for longer duration.

Example problem 1

A uniform rod of length $L = 1.2$ m and mass $M = 2.0$ kg is free to rotate about its center. A mass $m = 0.5$ kg moving at $v = 4.0$ m/s strikes the rod perpendicularly at $r = 0.30$ m and sticks. Find immediately after collision:

a. Angular momentum $L$ .

b. Angular speed $ω_{f}$ .

Solution:

The rod is initially at rest, so the system’s initial angular momentum comes from the incoming mass:

$L = m v r = 0.5 \times 4.0 \times 0.30 = 0.60 kg \cdot m^{2} /s .$

Moment of inertia:

$I_{rod} = \frac{1}{12} M L^{2} = \frac{1}{12} \times 2.0 \times (1.2)^{2} = 0.24, I_{m} = m r^{2} = 0.5 \times 0.3 0^{2} = 0.045,$

$I_{total} = 0.24 + 0.045 = 0.285, ω_{f} = \frac{L}{I _{total}} = \frac{0.60}{0.285} \approx 2.11 rad/s .$

Example problem 2

A turntable ( $I = 0.50$ kg·m $^{2}$ ) spins at $ω_{i} = 3.0$ rad/s. A $0.20$ kg lump of clay drops onto the rim at $r = 0.25$ m. Find final $ω_{f}$ .

Solution:

Compute the initial angular momentum and the added moment of inertia from the clay:

$L_{i} = I ω_{i} = 0.50 \times 3.0 = 1.50, I_{c} = m r^{2} = 0.20 \times 0.2 5^{2} = 0.0125,$

Then use conservation of angular momentum, $L_{i} = I_{f} ω_{f}$ :

$I_{f} = 0.50 + 0.0125 = 0.5125, ω_{f} = \frac{1.50}{0.5125} \approx 2.93 rad/s .$

Example problem 3

Compare angular impulse $Δ L$ when

$τ_{0} acts for T versus 2 τ_{0} acts for T /2.$

First case:

$J_{1} = τ_{0} \times T .$
Second case:

$J_{2} = (2 τ_{0}) \times \frac{T}{2} = τ_{0} T .$
Conclusion: $J_{1} = J_{2}$ , so both deliver the same $Δ L$ - impulse depends on the area under the $τ$ - $t$ graph.

Comparing impulse generated for the two scenarios

Example problem 4

A uniform solid disk of mass $M$ and radius $R$ is mounted on a frictionless axle through its center so it can rotate freely in a horizontal plane. Initially the disk is at rest. A small clay ball of mass $m$ is projected horizontally, strikes and sticks to the rim of the disk at point $P$ , and comes instantaneously to rest relative to the disk (see figure). After the collision, an external constant torque $τ_{ext}$ is applied to the disk-ball system for a time interval $t_{1}$ . Finally, a small frictional torque $τ_{f}$ (constant in magnitude, opposite to the direction of rotation) acts until the system comes to rest again.

Disk-ball system as stated in the question

(a) Angular speed $ω_{1}$ after collision

Initial angular momentum (disk at rest, ball of momentum $m v$ at radius $R$ ):

$L_{i} = m v R .$
Final angular momentum of disk+ball:

$L_{f} = I_{tot} ω_{1} .$
Conservation $L_{i} = L_{f}$ $⟹$

$ω_{1} = \frac{m v R}{I _{tot}} = \frac{m v R}{\frac{1}{2} M R ^{2} + m R ^{2}} = \frac{m v}{\frac{1}{2} MR + m R} .$

(b) Speed $ω_{2}$ after external torque $τ_{ext}$ applied for time $t_{1}$

Angular impulse from constant torque:

$J_{ext} = τ_{ext} t_{1} .$
Change in angular momentum $⟹$

$I_{tot} ω_{2} - I_{tot} ω_{1} = J_{ext} .$
Solve for $ω_{2}$ :

$ω_{2} = ω_{1} + \frac{τ _{ext} t _{1}}{I _{tot}} .$

(c) Work done by $τ_{ext}$

Since $τ_{ext}$ is constant and the angular displacement from $ω_{1}$ to $ω_{2}$ is $Δ θ$ , the work can be found via the change in rotational kinetic energy:

$W_{ext} = Δ K_{rot} = \frac{1}{2} I_{tot} ω_{2}^{2} - \frac{1}{2} I_{tot} ω_{1}^{2} .$

(d) Angular impulse from friction $τ_{f}$ until rest

Friction acts with constant magnitude $τ_{f}$ opposite rotation.
To bring $ω_{2}$ to zero, it must deliver an impulse $J_{f}$ satisfying

$I_{tot} 0 - I_{tot} ω_{2} = J_{f} ⟹ J_{f} = - I_{tot} ω_{2} .$
Its magnitude is $∣ J_{f} ∣ = I_{tot} ω_{2}$ .

(e) Energy dissipated by friction

All the rotational kinetic energy at $ω_{2}$ is lost:

$E_{lost} = \frac{1}{2} I_{tot} ω_{2}^{2} .$

Angular momentum of a particle\

Defined as $L = r \times (m v)$
Magnitude: $L = m r v sin ϕ$
- $L = 0$ if motion is purely radial ( $ϕ = 0$ )

Angular momentum of a rigid body\

For rotation about fixed axis: $L = I ω$
Moment of inertia: $I = \sum_{i} m_{i} r_{i}^{2}$
- Common forms:
  - Thin hoop: $I = M R^{2}$
  - Solid disk: $I = \frac{1}{2} M R^{2}$
  - Solid sphere: $I = \frac{2}{5} M R^{2}$
  - Rod (center): $I = \frac{1}{12} M L^{2}$
  - Rod (end): $I = \frac{1}{3} M L^{2}$

Conservation of angular momentum\

No net external torque $⟹$ total $L$ is constant
Use $L = I ω$ (rigid body) or $L = r \times p$ (point mass)
Internal torques cancel; only external torques change $L$

Angular impulse-momentum theorem\

Angular impulse: $J_{θ} = τ Δ t$
Change in angular momentum: $Δ L = J_{θ}$
Applies for constant or variable torque (sum/integrate over time)

Reasoning practice\

Decreasing radius ( $r$ ) with no external torque $⟹$ angular speed ( $ω$ ) increases so $L$ stays constant
Locked, skidding, and rolling wheels: $L$ changes only when external torque (friction) acts
Satellite orbit: $L$ conserved since gravity is radial (no torque); only transverse velocity contributes to $L$
Brake on flywheel: $Δ L = τ Δ t$ ; impulse depends on torque-time product, not just torque or time alone

Example problems\

Conservation of $L$ to find post-collision angular speed: $L = m v r$ , $I_{total}$ includes all masses
Adding mass to a rotating object: $L_{initial} = I_{final} ω_{final}$
Equal angular impulses: $τ_{0} T$ and $2 τ_{0} (T /2)$ give same $Δ L$
Disk-ball system:
- (a) $ω_{1} = \frac{m v R}{I _{tot}}$
- (b) $ω_{2} = ω_{1} + \frac{τ _{ext} t _{1}}{I _{tot}}$
- © Work: $W_{ext} = \frac{1}{2} I_{tot} (ω_{2}^{2} - ω_{1}^{2})$
- (d) Friction impulse: $∣ J_{f} ∣ = I_{tot} ω_{2}$
- (e) Energy lost to friction: $E_{lost} = \frac{1}{2} I_{tot} ω_{2}^{2}$

Key reminders\

Always specify rotation axis and sign convention
State explicitly when using conservation of angular momentum
$L$ is rotational analogue of linear momentum; changes only with net external torque

Next | 8.1 Simple harmonic motion

Angular momentum and angular impulse

In this sub-chapter we cover:

The concept of angular momentum
The algebraic definition of angular momentum
Concept and application of the angular impulse-momentum theorem.

Angular momentum of a particle

Consider a particle of mass $m$ at position $r$ (measured from an origin $O$ ) moving with velocity $v$ . Its angular momentum about $O$ is defined as

$L = r \times (m v),$

and its magnitude is

$L = m r v sin ϕ,$

where $ϕ$ is the angle between $r$ and $v$ .

A useful special case is when the particle moves directly toward or away from the origin. Then the velocity is purely radial ( $ϕ = 0$ ), so $sin ϕ = 0$ and therefore $L = 0$ .

Angular momentum of a rigid body

$L_{i} = m_{i} (r_{i} ω) r_{i} = m_{i} r_{i}^{2} ω .$

Adding the contributions from all mass elements gives

$L = i \sum L_{i} = ω i \sum m_{i} r_{i}^{2} = I ω,$

where

$I = i \sum m_{i} r_{i}^{2}$

is the moment of inertia about that axis.

Common forms of $I$ :

$Thin hoop: Solid disk: Solid sphere: Rod about center: Rod about end: I I I I I = M R^{2}, = \frac{1}{2} M R^{2}, = \frac{2}{5} M R^{2}, = \frac{1}{12} M L^{2}, = \frac{1}{3} M L^{2} .$

Conservation of angular momentum

If a system experiences no net external torque about a chosen axis, then the system’s total angular momentum about that axis stays constant.

Equivalently,

$i \sum τ_{ext, i} = 0 ⟹ L_{initial} = L_{final} .$

Here, depending on the situation, you’ll typically use

$L = I ω (for a rigid body about a fixed axis), or L = r \times p (for a point mass).$

“No net external torque” means all torques due to forces from outside the system sum to zero.
Internally generated torques (action-reaction pairs within the system) can’t change the system’s total $L$ .
This is the rotational analogue of Newton’s first law for linear momentum. Illustration 1:

Angular impulse-momentum theorem

Torque plays the same role in rotation that force plays in translation.

A constant force $F$ acting for time $Δ t$ gives a linear impulse $J = F Δ t$ and changes momentum by $Δ p = J$ .
A constant torque $τ$ acting for time $Δ t$ gives an angular impulse

$J_{θ} = τ Δ t,$

and changes angular momentum by

$Δ L = L_{f} - L_{i} = J_{θ} .$

Reasoning practice

You whirl a small stone on a string in a horizontal circle. Without external torque, you pull the string in so that the stone’s radius decreases.

Questions:

What happens to the stone’s angular speed $ω$ ?
Which quantity remains constant, and why?

Solution:

As $r$ decreases, $ω$ increases so that $L$ remains constant.
The angular momentum $L = m r^{2} ω$ is conserved because no external torque acts (tension in the string is radial, so $τ = 0$ ).

A car’s wheels lock and skid (no rotation) on an icy patch, then regain traction and spin again at the same forward speed.

Questions:

Compare the angular momentum of a wheel about its axle before lock, during skid, and after regaining traction.
What external torques act during each phase?

Solution:

Before lock: $L = I ω_{rolling}$ . During skid: $ω = 0 ⟹ L = 0$ . After traction: wheels spin so $L = I ω_{new}$ (same as before if no slip).
Before lock: static friction provides torque to spin wheel but net torque about axle is zero for steady rolling. During skid: kinetic friction exerts a torque opposing wheel spin (bringing $L \to 0$ ). After traction: static friction again enforces rolling without slipping (no net torque if speed constant).

A satellite moves in an elliptical orbit around Earth. As it swings inward toward perigee, its speed increases; as it moves outward to apogee, its speed decreases.

Questions:

Why does the satellite’s angular momentum about Earth remain constant?
Which component of its velocity contributes to $L$ ?

Solution:

The only force is gravity, which acts along the line from Earth to satellite (radial). A radial force produces zero torque about Earth, so $L$ is conserved.
Only the transverse (perpendicular to the radius vector) component $v_{⊥}$ contributes, since $L = m r v_{⊥}$ .

A heavy flywheel spins freely. A brake pad is pressed suddenly against its rim, applying a large torque for a very short time, then released.

Questions:

How does the flywheel’s angular momentum change during the brake pulse?
Why does a large torque for a short time produce the same $Δ L$ as a smaller torque for a longer time?

Solution:

The brake’s frictional torque (opposite rotation) reduces $L$ by $Δ L = τ Δ t$ until it stops or slows by the impulse delivered.
Angular impulse depends on the product $τ Δ t$ (the area under the torque-time curve), so a higher torque for shorter duration can equal the impulse of a lower torque for longer duration.

Example problem 1

a. Angular momentum $L$ .

b. Angular speed $ω_{f}$ .

Solution:

The rod is initially at rest, so the system’s initial angular momentum comes from the incoming mass:

$L = m v r = 0.5 \times 4.0 \times 0.30 = 0.60 kg \cdot m^{2} /s .$

Moment of inertia:

$I_{rod} = \frac{1}{12} M L^{2} = \frac{1}{12} \times 2.0 \times (1.2)^{2} = 0.24, I_{m} = m r^{2} = 0.5 \times 0.3 0^{2} = 0.045,$

$I_{total} = 0.24 + 0.045 = 0.285, ω_{f} = \frac{L}{I _{total}} = \frac{0.60}{0.285} \approx 2.11 rad/s .$

Example problem 2

A turntable ( $I = 0.50$ kg·m $^{2}$ ) spins at $ω_{i} = 3.0$ rad/s. A $0.20$ kg lump of clay drops onto the rim at $r = 0.25$ m. Find final $ω_{f}$ .

Solution:

Compute the initial angular momentum and the added moment of inertia from the clay:

$L_{i} = I ω_{i} = 0.50 \times 3.0 = 1.50, I_{c} = m r^{2} = 0.20 \times 0.2 5^{2} = 0.0125,$

Then use conservation of angular momentum, $L_{i} = I_{f} ω_{f}$ :

$I_{f} = 0.50 + 0.0125 = 0.5125, ω_{f} = \frac{1.50}{0.5125} \approx 2.93 rad/s .$

Example problem 3

Compare angular impulse $Δ L$ when

$τ_{0} acts for T versus 2 τ_{0} acts for T /2.$

First case:

$J_{1} = τ_{0} \times T .$
Second case:

$J_{2} = (2 τ_{0}) \times \frac{T}{2} = τ_{0} T .$
Conclusion: $J_{1} = J_{2}$ , so both deliver the same $Δ L$ - impulse depends on the area under the $τ$ - $t$ graph.

Example problem 4

(a) Angular speed $ω_{1}$ after collision

Initial angular momentum (disk at rest, ball of momentum $m v$ at radius $R$ ):

$L_{i} = m v R .$
Final angular momentum of disk+ball:

$L_{f} = I_{tot} ω_{1} .$
Conservation $L_{i} = L_{f}$ $⟹$

$ω_{1} = \frac{m v R}{I _{tot}} = \frac{m v R}{\frac{1}{2} M R ^{2} + m R ^{2}} = \frac{m v}{\frac{1}{2} MR + m R} .$

(b) Speed $ω_{2}$ after external torque $τ_{ext}$ applied for time $t_{1}$

Angular impulse from constant torque:

$J_{ext} = τ_{ext} t_{1} .$
Change in angular momentum $⟹$

$I_{tot} ω_{2} - I_{tot} ω_{1} = J_{ext} .$
Solve for $ω_{2}$ :

$ω_{2} = ω_{1} + \frac{τ _{ext} t _{1}}{I _{tot}} .$

(c) Work done by $τ_{ext}$

Since $τ_{ext}$ is constant and the angular displacement from $ω_{1}$ to $ω_{2}$ is $Δ θ$ , the work can be found via the change in rotational kinetic energy:

$W_{ext} = Δ K_{rot} = \frac{1}{2} I_{tot} ω_{2}^{2} - \frac{1}{2} I_{tot} ω_{1}^{2} .$

(d) Angular impulse from friction $τ_{f}$ until rest

Friction acts with constant magnitude $τ_{f}$ opposite rotation.
To bring $ω_{2}$ to zero, it must deliver an impulse $J_{f}$ satisfying

$I_{tot} 0 - I_{tot} ω_{2} = J_{f} ⟹ J_{f} = - I_{tot} ω_{2} .$
Its magnitude is $∣ J_{f} ∣ = I_{tot} ω_{2}$ .

(e) Energy dissipated by friction

All the rotational kinetic energy at $ω_{2}$ is lost:

$E_{lost} = \frac{1}{2} I_{tot} ω_{2}^{2} .$

Achievable AP Physics 1

7. Torque and rotational mechanics

Our AP Physics 1 course is currently in development and is a work-in-progress.