8.4 Apply conservation of angular momentum | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Angular Momentum

Choose a reference point, check external moment, and compare initial and final states

Angular momentum conservation connects the motion of a system at two different times. It is useful when forces during the change are complicated or unknown, but it applies only when the system’s net external angular impulse about the chosen reference point is zero. The reference point matters: use the same point for the initial and final angular momenta, and check the external moment about that point. This lesson uses planar motion, where angular momentum is represented by a signed quantity perpendicular to the plane.

What you will learn

Define angular momentum about a stated point for a particle or planar rigid body.
Decide whether angular momentum is conserved by checking external angular impulse about that point.
Apply conservation to particles and introductory planar rigid bodies using signed quantities and SI units.
Check results using dimensions, directions, initial and final states, and limiting cases.

Define the system, frame, and angular momentum

Begin by naming the system: it might be one particle, several particles, or a rigid body. Name the observer and reference frame too. In the examples here, the observer is fixed to the ground and uses stationary Cartesian axes. Select a reference point

O

, and use it for both the initial and final states. Take counterclockwise as positive and clockwise as negative.

For a particle, angular momentum about

O

is the moment of its linear momentum. Its value depends on the position measured from

O

and the particle’s velocity. In planar motion, the vector points perpendicular to the plane, so a signed scalar is enough. For several particles, add their signed angular momenta about the same point.

For a rigid body rotating about a fixed axis through

O

, the planar relation uses the body’s mass moment of inertia about that axis. The moment of inertia depends on how the body’s mass is distributed and on the selected axis. Do not change the reference point or axis partway through a calculation.

\mathbf{H}_O=\mathbf{r}\times m\mathbf{v},\quad H_O=I_O\omega

State the system, ground-fixed observer, reference point, coordinate axes, and positive rotation direction.
Particle angular momentum depends on position and velocity relative to the reference point.
Use signed values in planar problems: counterclockwise is positive under the convention used here.

The conservation rule and its conditions

The angular impulse–momentum relationship says that the change in a system’s angular momentum about

O

equals the angular impulse of the external moments about

O

. Internal forces between parts of the system are not external. External forces may still act; what matters is the total effect of their moments about the chosen point.

If the net external moment about

O

is zero throughout the interval, angular momentum about

O

is constant. More generally, the total external angular impulse can be zero even if moments act during the interval. In either case, explain why the condition is reasonable before using conservation. A force whose line of action passes through

O

has zero moment about

O

.

For a particle moving tangentially around

O

, the angular momentum magnitude is the mass times radius times tangential speed. For a rigid body rotating about a fixed axis, angular momentum is its moment of inertia times angular velocity. When the system or its mass distribution changes, include every part of the chosen system in the appropriate state.

H_{O,1}+\int_{t_1}^{t_2}\sum M_{O,\mathrm{ext}}\,dt=H_{O,2}

Conservation is stated about a reference point and requires zero net external angular impulse about that point.
A force can be present without changing angular momentum about $O$ if its moment about $O$ is zero.
Use the same system boundary and reference point in the initial and final states.

A dependable solution method

Write down the initial and final states and the unknown. Draw a diagram when it helps show the reference point, position, force direction, or rotation direction. Then choose axes and a sign convention. For planar problems, positive angular momentum points out of the page when counterclockwise is positive.

Check the moments of external forces about

O

. If a force’s line of action passes through

O

, its moment about

O

is zero. If the total external angular impulse cannot be neglected, do not set the initial and final angular momenta equal; retain the angular impulse term instead.

Write the conservation equation symbolically before substituting numbers. Keep units consistent: angular momentum has units of kilogram metre squared per second, while moment of inertia has units of kilogram metre squared. Check that the answer’s sign gives the stated direction, that all system parts are included, and that the result behaves sensibly in a limiting case such as an unchanged radius.

[H_O]=\mathrm{kg\,m^2/s}

A useful diagram identifies $O$ and the positions, directions, or axis that determine angular momentum.
Conservation connects two states; it does not by itself describe the forces or motion during the change.
Check dimensions, signs, and whether the predicted change fits the physical setup.

Worked example

A particle is pulled closer to an axis

A

0.50\,\mathrm{kg}

particle moves in a horizontal plane around a fixed, smooth point

O

. At the initial state it is

0.80\,\mathrm{m}

from

O

and has tangential speed

2.4\,\mathrm{m/s}

counterclockwise. A radial mechanism pulls it inward to

0.40\,\mathrm{m}

. Assume the mechanism exerts no moment about

O

. Find the final tangential speed and direction.

Radial pull toward O

The radial pull acts along the line to O, so its moment about O is zero.

Define the states and signs
Take the particle as the system, use a ground-fixed observer, and measure angular momentum about $O$ . Counterclockwise is positive. The initial and final radii are known; the final tangential speed is unknown.
Check conservation
The pulling force is radial, so its line of action passes through $O$ and its moment about $O$ is zero. With the stated assumption that there is no other moment about $O$ , angular momentum is conserved. The tangential motion is counterclockwise at the initial state, so the signed angular momentum is positive.
$m r_1 v_1=m r_2 v_2$
Solve and check
Solve for the final speed. The mass cancels, and the positive sign means the direction remains counterclockwise.
$v_2=\frac{r_1v_1}{r_2}=\frac{(0.80\,\mathrm{m})(2.4\,\mathrm{m/s})}{0.40\,\mathrm{m}}=4.8\,\mathrm{m/s}$

Answer: The final tangential speed is

4.8\,\mathrm{m/s}

, counterclockwise.

Check: Both states have angular momentum magnitude

mrv=0.96\,\mathrm{kg\,m^2/s}

. The radius is halved, so the tangential speed doubles while the signed angular momentum remains positive.

Worked example

A particle sticks to a rotating disk

A disk with mass moment of inertia

0.60\,\mathrm{kg\,m^2}

rotates counterclockwise at

2.0\,\mathrm{rad/s}

about a fixed, frictionless axle through its centre

O

. A

0.20\,\mathrm{kg}

particle approaches tangentially at a distance of

0.30\,\mathrm{m}

from

O

and sticks to the disk. Its incoming motion contributes positive angular momentum about

O

, and its speed just before impact is

5.0\,\mathrm{m/s}

. Find the angular velocity just after impact.

Disk and incoming particle

The particle approaches tangentially at the marked radius; its incoming angular momentum and the disk’s rotation are both positive about O.

Define the system and states
Take the disk and particle together as the system, viewed from the ground, and take moments about the axle at $O$ . Counterclockwise is positive. The initial state includes the rotating disk and incoming particle; the final state is the disk with the particle attached.
Apply angular momentum conservation
During the short impact, the axle force has zero moment about $O$ , and other external angular impulse about $O$ is assumed negligible. The particle’s tangential motion contributes positive angular momentum. After it sticks, its contribution is included in the final moment of inertia.
$I_D\omega_1+mrv=(I_D+mr^2)\omega_2$
Substitute and solve
Both initial contributions are positive. Divide their sum by the combined final moment of inertia to obtain the shared angular velocity.
$\omega_2=\frac{(0.60\,\mathrm{kg\,m^2})(2.0\,\mathrm{rad/s})+(0.20\,\mathrm{kg})(0.30\,\mathrm{m})(5.0\,\mathrm{m/s})}{0.60\,\mathrm{kg\,m^2}+(0.20\,\mathrm{kg})(0.30\,\mathrm{m})^2}=2.43\,\mathrm{rad/s}$

Answer: The disk and particle rotate together at approximately

2.43\,\mathrm{rad/s}

counterclockwise.

Check: The initial angular momentum is

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

and the final moment of inertia is

0.618\,\mathrm{kg\,m^2}

, giving

2.43\,\mathrm{rad/s}

. The final speed exceeds the disk’s initial speed because the particle adds positive angular momentum.

Worked example

A particle acted on by a central force

A

0.40\,\mathrm{kg}

particle is acted on by a force that always points along the line from a fixed point

O

to the particle. At one instant, its distance from

O

is

0.25\,\mathrm{m}

and its velocity is perpendicular to that line, with speed

6.0\,\mathrm{m/s}

. Later, its distance is

0.50\,\mathrm{m}

and its velocity is again perpendicular to the line. Find its later speed.

Central force about O

The force acts along the radius, so it has zero moment about O.

Set the reference and states
Use the particle as the system and a ground-fixed observer. Take angular momentum about $O$ , with counterclockwise positive. At both stated instants the velocity is perpendicular to the radius, so the angular momentum magnitude is $mrv$ . The later speed is unknown.
Check the external moment
The force is central: its line of action passes through $O$ . Its moment about $O$ is therefore zero, so angular momentum about $O$ is conserved between the stated instants.
$m r_1v_1=m r_2v_2$
Calculate the later speed
Cancel the mass and solve for the later speed. The larger radius requires a smaller perpendicular speed to preserve angular momentum.
$v_2=\frac{r_1v_1}{r_2}=\frac{(0.25\,\mathrm{m})(6.0\,\mathrm{m/s})}{0.50\,\mathrm{m}}=3.0\,\mathrm{m/s}$

Answer: The later speed is

3.0\,\mathrm{m/s}

. Its angular-momentum direction is the same as at the first instant.

Check: The angular momentum magnitudes at both instants are

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

. If the radius were unchanged, conservation would require the same perpendicular speed.

Common mistakes and how to avoid them

Assuming angular momentum is conserved whenever an object rotates.

Correction: Name the reference point and check the net external moment or angular impulse about that point.

Using only a force’s size to decide whether it changes angular momentum.

Correction: Check its moment arm: a force whose line of action passes through

O

has zero moment about

O

.

Adding angular momentum magnitudes without considering direction.

Correction: Choose a positive rotational direction and use signed contributions consistently.

Using only the disk’s moment of inertia after a particle sticks to it.

Correction: Include the attached particle’s contribution to the final moment of inertia about the same axis.

Lesson summary

Angular momentum is measured about a specified point and depends on position and momentum relative to that point.
Conservation applies when the net external angular impulse about the chosen point is zero.
In planar problems, use a consistent sign convention for clockwise and counterclockwise contributions.
For tangential particle motion, angular momentum magnitude is $mrv$ ; for a rigid body rotating about a fixed axis, it is $I\omega$ .
Verify the external-moment assumption, units, direction, and inclusion of all parts of the system.

Check your understanding

Question 1

A particle’s angular momentum about

O

is conserved over an interval. Which condition justifies this conclusion?

The particle’s speed is constant.
The net external angular impulse about $O$ is zero.
The particle’s path is circular.
The particle experiences no forces.

Show answer and explanation

The net external angular impulse about

O

is zero.

Conservation follows from zero net external angular impulse about the chosen point. Constant speed or circular motion alone does not establish that condition.

Question 2

A particle moves tangentially at

3.0\,\mathrm{m/s}

at radius

0.20\,\mathrm{m}

, then at radius

0.60\,\mathrm{m}

. If angular momentum about

O

is conserved, what is its later tangential speed?

$1.0\,\mathrm{m/s}$
$3.0\,\mathrm{m/s}$
$6.0\,\mathrm{m/s}$
$9.0\,\mathrm{m/s}$

Show answer and explanation

1.0\,\mathrm{m/s}

Conservation gives

r_1v_1=r_2v_2

, so the later speed is

(0.20\times3.0)/0.60=1.0\,\mathrm{m/s}

.

Question 3

A force acts along the line joining a particle to

O

. What is its moment about

O

?

Zero
Always positive
Always negative
Equal to the particle’s angular momentum

Show answer and explanation

Zero

The force’s line of action passes through

O

, so its moment arm about

O

is zero.

Key terms

Angular momentum: A measure of rotational motion about a chosen point, found for a particle from its position relative to that point and its linear momentum.
Moment: The turning effect of a force about a point, determined by the force and its perpendicular distance from that point.
Moment of inertia: A measure of how a body’s mass is distributed relative to a chosen axis; it appears in the planar relation between angular momentum and angular velocity.
Angular impulse: The accumulated effect of external moments over a time interval; it equals the change in angular momentum about the same point.

Continue through EN PH 131

View the complete EN PH 131 University of Alberta EN PH 131: Engineering Mechanics: Dynamics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows University of Alberta EN PH 131: Engineering Mechanics: Dynamics, study topic 8.4. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question

8.4 · Apply conservation of angular momentum

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Choose a reference point, check external moment, and compare initial and final states

What you will learn

Define the system, frame, and angular momentum

The conservation rule and its conditions

A dependable solution method

A particle is pulled closer to an axis

A particle sticks to a rotating disk

A particle acted on by a central force

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