8.5 Analyze angular momentum for a system of particles | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Angular Momentum

Summing particle contributions and tracking the effect of external moments

Angular momentum describes how the motion of a particle or system is distributed around a chosen point. For a system of particles, the total angular momentum is the vector sum of the individual particles’ angular momenta. This lesson focuses on planar motion, where the calculation often reduces to one signed component perpendicular to the plane. The choice of reference point matters: state it before calculating, and use it consistently when evaluating positions and moments.

What you will learn

Calculate a particle’s angular momentum about a specified point using its position and linear momentum.
Find a system’s angular momentum by adding the angular momenta of its particles.
Relate the change in system angular momentum to the moment of external forces.
Use conservation of angular momentum when the net external moment about the chosen point is zero.

1. Define the system, observer, and reference point

A system of particles is the group of particles you choose to study. The system may be treated as closed over a time interval when no particles enter or leave it. Choose an observer at rest in an inertial reference frame, such as a fixed laboratory frame, and specify a point

O

about which angular momentum is to be measured.

For planar motion, set the

x

and

y

axes in the plane and take positive angular momentum to point in the positive

z

direction, out of the page. By the right-hand rule, positive

z

corresponds to counterclockwise rotation in the plane. A clockwise contribution is negative.

For particle

i

, let

\mathbf r_i

be its position measured from

O

, and let

\mathbf v_i

be its velocity in the chosen frame. Its linear momentum is

m_i\mathbf v_i

. Angular momentum depends on both this momentum and its position relative to

O

.

\mathbf H_O=\sum_i \mathbf r_i\times m_i\mathbf v_i

State the system, observer, frame, reference point, and positive direction before calculating.
Angular momentum is measured about a point; changing that point can change the result.

2. Calculate and add particle contributions

The cross product in the defining equation gives a vector perpendicular to the plane of

\mathbf r_i

and

\mathbf v_i

. For planar motion, only its

z

component is needed. If

\mathbf r_i=(x_i,y_i,0)

and

\mathbf v_i=(v_{xi},v_{yi},0)

, the signed contribution is

x_i m_i v_{yi}-y_i m_i v_{xi}

.

This expression shows why direction matters. A particle moving parallel to its position vector from

O

has zero angular momentum about

O

. A particle moving across that position vector generally contributes a nonzero amount. The system total is the algebraic sum of all contributions, so clockwise and counterclockwise contributions can partly or fully cancel.

The units are mass times distance times speed, or

\mathrm{kg\,m^2/s}

. This is distinct from linear momentum, whose units are

\mathrm{kg\,m/s}

. In planar problems, a positive scalar result means the angular-momentum vector points out of the page; a negative result means it points into the page.

H_{O,z}=\sum_i m_i(x_i v_{yi}-y_i v_{xi})

Use each particle’s position from the selected point, not its distance from another particle.
Keep the sign of each cross product; do not add magnitudes without direction.

3. Connect angular momentum to external moments

For a system of particles, internal forces occur in action–reaction pairs. When the system and the forces are modeled so that internal force moments cancel, the rate of change of total angular momentum about a fixed point equals the net external moment about that point. This is the system form of the angular-momentum principle.

Over a time interval, the angular impulse of the external forces equals the change in angular momentum. For a constant net external moment, the change is the moment multiplied by the elapsed time. If the net external moment about the chosen point is zero throughout the interval, angular momentum about that point is conserved.

Conservation is not a claim that each particle’s angular momentum stays fixed. Particles can exchange angular momentum through internal interactions while the system total remains constant. Also, zero external force alone does not guarantee zero external moment about every point; check the moment about the point used in the calculation.

\mathbf H_{O,2}-\mathbf H_{O,1}=\int_{t_1}^{t_2}\sum\mathbf M_O^{\mathrm{ext}}\,dt

Use external moments about the same point used to calculate angular momentum.
Conserve system angular momentum only when the net external moment about that point is zero over the interval.

4. A reliable solution plan

First identify the system and the initial and final states. Record the observer’s frame, the point about which angular momentum is measured, the coordinate axes, and the positive rotational direction. A small particle diagram is useful when it makes positions and velocity directions clear.

Next list known masses, positions, velocities, external moments, and the requested quantity. Select the angular-momentum principle that matches the information: add particle contributions for a state, use angular impulse for a change over time, or use conservation when the external moment is zero. Keep position and velocity components in the same coordinate system.

Finally, solve with signed components, substitute SI units, and check the result. Confirm the sign using the right-hand rule, check dimensions, and test whether the answer makes sense if a particle’s velocity or perpendicular distance from the point is zero.

\sum\mathbf M_O^{\mathrm{ext}}=\mathbf 0\;\Rightarrow\;\mathbf H_{O,1}=\mathbf H_{O,2}

A clear sign convention makes both the calculation and the direction check easier.
Check dimensions and limiting cases, not just arithmetic.

Worked example

1. Sum two particle contributions

Two particles move in a fixed laboratory frame. Relative to point

O

, particle 1 has mass

2.0\,\mathrm{kg}

, position

(1.0,0)\,\mathrm{m}

, and velocity

(0,3.0)\,\mathrm{m/s}

. Particle 2 has mass

1.0\,\mathrm{kg}

, position

(0,2.0)\,\mathrm{m}

, and velocity

(-4.0,0)\,\mathrm{m/s}

. Find the system angular momentum about

O

.

Particle positions and momenta

Positions are measured from O; positive angular momentum is out of the page.

Set the system and signs
The system is the two particles, observed in the fixed laboratory frame. Measure both positions from $O$ , use the stated $x,y$ coordinates, and take counterclockwise angular momentum as positive.
Evaluate each signed contribution
Particle 1 moves upward at a positive $x$ position, so its contribution is counterclockwise. Particle 2 moves left at a positive $y$ position, which also gives a counterclockwise contribution about $O$ .
$H_{1,z}=2(1)(3)=6,\qquad H_{2,z}=1[0-2(-4)]=8$
Add the contributions
The system angular momentum is the signed sum. Both contributions point in the positive $z$ direction, so they add.
$H_{O,z}=6+8=14\,\mathrm{kg\,m^2/s}$

Answer:

14\,\mathrm{kg\,m^2/s}

in the positive

z

direction, out of the page.

Check: Each term has units

\mathrm{kg\,m^2/s}

. The upward velocity at positive

x

and the leftward velocity at positive

y

both give positive cross products, matching the reported direction.

Worked example

2. Apply angular impulse

A system has angular momentum

-2.0\,\mathrm{kg\,m^2/s}

about fixed point

O

at

t=0

. A constant net external moment of

+6.0\,\mathrm{N\,m}

acts about

O

for

0.50\,\mathrm{s}

. Find its angular momentum at the end of the interval. Positive is counterclockwise.

External moment on system

The moment is about fixed point O; counterclockwise is positive.

Identify the interval and signs
The system is observed in a fixed inertial frame from $t=0$ to $t=0.50\,\mathrm{s}$ . The initial angular momentum is clockwise, hence negative; the constant external moment is counterclockwise, hence positive.
Use angular impulse
The external angular impulse equals the final angular momentum minus the initial angular momentum. Because the moment is constant, its integral over time is its value multiplied by the interval.
$H_{O,2}=H_{O,1}+M_O^{\mathrm{ext}}\Delta t$
Substitute and interpret
The angular impulse is positive and larger in magnitude than the initial negative angular momentum, so the final result is positive.
$H_{O,2}=-2.0+(6.0)(0.50)=1.0\,\mathrm{kg\,m^2/s}$

Answer:

1.0\,\mathrm{kg\,m^2/s}

counterclockwise.

Check: The moment-time product has units

\mathrm{N\,m\,s} = \mathrm{kg\,m^2/s}

. If the interval were zero, the result would equal the stated initial angular momentum.

Worked example

3. Use conservation to find a velocity

A two-particle system is observed in a fixed frame at two instants. About fixed point

O

, the initial positions are

(1.0,0)\,\mathrm{m}

for particle 1 and

(0,1.0)\,\mathrm{m}

for particle 2. Their masses are

2.0\,\mathrm{kg}

and

1.0\,\mathrm{kg}

, and their initial velocities are

(0,2.0)\,\mathrm{m/s}

and

(-3.0,0)\,\mathrm{m/s}

. At the final instant, the positions are unchanged, particle 1 has velocity

(0,1.0)\,\mathrm{m/s}

, and particle 2 has velocity

(u,0)

. The net external moment about

O

is zero during the interval. Find

u

.

Final-state particle data

The stated positions and velocities are final-state data; no path between states is implied.

State the conservation condition
The system is the two particles, the observer uses the fixed frame, and angular momentum is measured about fixed point $O$ . Since the net external moment about $O$ is zero, the total angular momentum about $O$ is the same at the initial and final instants.
$H_{O,1}=H_{O,2}$
Calculate the initial total
Particle 1 contributes $2(1)(2)=4\,\mathrm{kg\,m^2/s}$ . Particle 2 contributes $1[0-1(-3)]=3\,\mathrm{kg\,m^2/s}$ . Both are positive.
$H_{O,1}=4+3=7\,\mathrm{kg\,m^2/s}$
Write the final total and solve
At the final instant, particle 1 contributes $2(1)(1)=2\,\mathrm{kg\,m^2/s}$ . Particle 2 contributes $1[0-1(u)]=-u$ . Equating the total to the initial value determines the signed velocity.
$7=2-u\;\Rightarrow\;u=-5.0\,\mathrm{m/s}$

Answer: Particle 2 has velocity

5.0\,\mathrm{m/s}

to the left, so

u=-5.0\,\mathrm{m/s}

.

Check: The final contributions are

2

and

-(-5)=5\,\mathrm{kg\,m^2/s}

, totaling the conserved

7\,\mathrm{kg\,m^2/s}

. The negative velocity is consistent with the required clockwise contribution from particle 2.

Common mistakes and how to avoid them

Using the speed alone and reporting every particle’s angular momentum as positive.

Correction: Use the signed cross product of position from the chosen point and linear momentum. The direction may be clockwise or counterclockwise.

Measuring a particle’s position from another particle instead of from the stated reference point.

Correction: Use position vectors based on the same point about which angular momentum is being calculated.

Assuming angular momentum is conserved because the net external force is zero.

Correction: Check the net external moment about the chosen point. Conservation requires that moment to be zero over the interval.

Applying conservation to each particle separately.

Correction: The conservation statement applies to the total angular momentum of the system; individual particle contributions may change.

Lesson summary

A particle’s angular momentum about a point is its position from that point crossed with its linear momentum.
For a system, add all particle contributions with their signs and directions.
The net external moment about a fixed point governs the change in system angular momentum about that point.
If the net external moment is zero, total angular momentum about that point is conserved.

Check your understanding

Question 1

A particle is at

(2,0)\,\mathrm{m}

relative to

O

and moves at

(0,-3)\,\mathrm{m/s}

. Its mass is

1\,\mathrm{kg}

. What is its angular momentum about

O

?

$-6\,\mathrm{kg\,m^2/s}$
$+6\,\mathrm{kg\,m^2/s}$
$0\,\mathrm{kg\,m^2/s}$
correctIndex ө

Show answer and explanation

-6\,\mathrm{kg\,m^2/s}

The signed component is

m(xv_y-yv_x)=1[2(-3)-0]=-6

. The negative sign indicates clockwise angular momentum.

Question 2

Which condition allows total angular momentum about fixed point

O

to be conserved over an interval?

The net external moment about $O$ is zero throughout the interval.
The net external force is zero, regardless of its moment about $O$ .
Every particle has constant velocity.
correctIndex 0

Show answer and explanation

The net external moment about

O

is zero throughout the interval.

The angular-momentum principle connects the change in total angular momentum about

O

to the net external moment about

O

.

Question 3

A system has initial angular momentum

+5\,\mathrm{kg\,m^2/s}

and experiences a constant external moment of

-2\,\mathrm{N\,m}

for

1.0\,\mathrm{s}

. What is its final angular momentum?

$+7\,\mathrm{kg\,m^2/s}$
$+3\,\mathrm{kg\,m^2/s}$
$-3\,\mathrm{kg\,m^2/s}$
correctIndex 1

Show answer and explanation

+3\,\mathrm{kg\,m^2/s}

The external angular impulse is

(-2)(1.0)=-2\,\mathrm{kg\,m^2/s}

. Adding it to the initial value gives

+3\,\mathrm{kg\,m^2/s}

.

Key terms

Angular momentum: A vector measure of a particle’s or system’s motion about a chosen point.
Angular impulse: The time integral of the net external moment about a selected point.
External moment: The turning effect of forces exerted on the chosen system by objects outside it, measured about a specified point.
System of particles: The identified group of particles whose combined motion is being analyzed.

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8.5 · Analyze angular momentum for a system of particles

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Summing particle contributions and tracking the effect of external moments

What you will learn

1. Define the system, observer, and reference point

2. Calculate and add particle contributions

3. Connect angular momentum to external moments

4. A reliable solution plan

1. Sum two particle contributions

2. Apply angular impulse

3. Use conservation to find a velocity

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through EN PH 131

About this lesson and its review