8.2 Relate moment to the rate of angular momentum | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Angular Momentum

EN PH 131 study topic 8.2

A net force changes linear momentum; a net moment changes angular momentum. The central relationship is that the net external moment about a point equals the rate of change of angular momentum about that same point. The point and observer matter, so state them before calculating. This lesson uses an inertial frame and, for planar problems, takes counterclockwise as positive. Angular momentum and moment then point perpendicular to the plane, and their signed scalar components track direction.

What you will learn

Define angular momentum and moment about a specified point using a consistent coordinate system.
Relate the net external moment to the rate of change of angular momentum.
Apply the relationship to a particle and to a rigid body rotating about a fixed axis.
Check signs, directions, units, and assumptions in angular-momentum calculations.

1. Define angular momentum and moment

Choose an observer fixed in an inertial reference frame and an origin

O

. For a particle of mass

m

, position vector

\mathbf{r}

measured from

O

, and velocity

\mathbf{v}

measured in the chosen frame, angular momentum about

O

is

\mathbf{H}_O=\mathbf{r}\times m\mathbf{v}

. Position is measured from the origin; velocity is measured relative to the frame. Angular momentum depends on both.

The moment of a force about

O

is

\mathbf{M}_O=\mathbf{r}\times\mathbf{F}

, where

\mathbf{r}

extends from

O

to the force’s point of application. In a planar problem, choose counterclockwise as positive. A force that tends to turn the system counterclockwise has positive moment; one that tends to turn it clockwise has negative moment.

For a system of particles, add the angular momenta of its particles. Apply the moment–rate relationship to the net external moment on the system. Internal forces between particles cancel in the total moment when they occur as equal and opposite forces along the same line.

\mathbf{H}_O=\sum_i \mathbf{r}_i\times m_i\mathbf{v}_i

State the origin and inertial frame before calculating angular momentum.
Use the same point $O$ for angular momentum and moment.
For planar motion, counterclockwise angular quantities are positive and clockwise quantities are negative.

2. The moment–rate relationship

For a system observed in an inertial frame, the net external moment about a fixed origin equals the time derivative of the system’s angular momentum about that origin. It is the rotational counterpart of the relationship between net force and the rate of change of linear momentum. An external moment changes angular momentum; it does not have to point in the direction of the angular momentum itself.

In a planar problem, angular momentum and moment are perpendicular to the plane. Their signed components can be used directly. If the net moment is zero, angular momentum about the chosen origin stays constant. If the moment varies with time, integrate it over the time interval to find the change in angular momentum.

For a rigid body rotating about a fixed axis through

O

, the angular momentum component along that axis is

H_O=I_O\omega

, where

I_O

is the body’s mass moment of inertia about the axis and

\omega

is signed angular velocity. If

I_O

is constant, differentiation gives

M_O=I_O\alpha

. This simplified form is for fixed-axis rotation; the moment–rate relationship is the more general starting point.

\sum \mathbf{M}_O^{\mathrm{ext}}=\frac{d\mathbf{H}_O}{dt}

Net external moment changes angular momentum about the same origin.
Zero net moment means constant angular momentum about that origin.
For fixed-axis rotation with constant moment of inertia, use $H_O=I_O\omega$ and $M_O=I_O\alpha$ .

3. A reliable solution process

First define the system and observer. State whether the system is one particle or a whole body, identify the inertial frame, and specify the origin about which moments and angular momentum are calculated. If the problem describes a time interval, identify its initial and final states.

Sketch the geometry when useful. Mark the origin, position vector, and force or motion directions. Choose a positive direction; counterclockwise is a convenient choice for planar rotation. List the knowns and unknowns, then select the moment–rate equation or its fixed-axis form.

Calculate each moment with its sign. In a planar problem, moment magnitude is the perpendicular distance from the origin to the force’s line of action multiplied by force magnitude. A force whose line of action passes through the origin has zero moment about that origin. Use SI units: angular momentum has units of

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

and moment has units of

\mathrm{N\,m}

, equivalent to

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

.

Finally, check whether the sign predicts the stated direction of change, whether dimensions match, and whether zero moment or zero elapsed time gives a sensible result. Do not assume angular momentum is conserved unless the net external moment about the chosen origin is zero.

\Delta H_O=\int_{t_1}^{t_2}\sum M_O^{\mathrm{ext}}\,dt

A moment’s sign follows its turning tendency about the chosen origin.
A force line through the origin has zero moment about that origin.
Check units and the zero-moment limit before accepting a result.

4. Interpreting the result

The equation relates a moment to a rate of change, not directly to angular momentum itself. A positive moment means signed angular momentum is increasing in the counterclockwise-positive sense. It does not necessarily mean the body is already rotating counterclockwise: a clockwise rotation can be slowing while a counterclockwise moment acts.

For a particle, use the cross product or the perpendicular lever arm and its sign. For a rigid body rotating about a fixed axis, use the moment of inertia and angular acceleration when the fixed-axis conditions are met. These are course-level applications of the same angular-momentum relationship.

Moment determines the change in angular momentum, not its existing value.
Angular velocity and angular acceleration can have opposite signs.

Worked example

A force changes a particle’s angular momentum

A

2.0\,\mathrm{kg}

particle is at

(0.30,0.40)\,\mathrm{m}

relative to fixed origin

O

in an inertial frame. Its initial velocity is

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

. The only external force is a constant

(0,5.0)\,\mathrm{N}

force, acting for

0.20\,\mathrm{s}

. Find its angular momentum about

O

initially and after the force has acted.

Particle and applied force

Position relative to O is (0.30, 0.40) m; arrows show the force and initial velocity directions.

Set the frame and signs
The system is the particle, and the observer is fixed in the stated inertial frame. Take $O$ as the origin, $+x$ right, $+y$ up, and counterclockwise angular momentum as positive.
Find the initial angular momentum
Use position measured from $O$ and the initial momentum. The planar component is $xmv_y-ymv_x$ , so the initial value is negative, meaning clockwise.
$H_{O,i}=-(0.40)(2.0)(3.0)=-2.4\,\mathrm{kg\,m^2/s}$
Find the moment and its effect
The upward force at positive $x$ produces a counterclockwise moment. Because it is the only external force, this is the net external moment. It is constant, so angular momentum changes by moment multiplied by elapsed time.
$M_O=(0.30)(5.0)=1.5\,\mathrm{N\,m},\quad H_{O,f}=H_{O,i}+M_O\Delta t=-2.4+1.5(0.20)=-2.1\,\mathrm{kg\,m^2/s}$

Answer: Initially,

H_O=-2.4\,\mathrm{kg\,m^2/s}

; after

0.20\,\mathrm{s}

,

H_O=-2.1\,\mathrm{kg\,m^2/s}

. It remains clockwise, but its magnitude decreases.

Check: The positive moment makes angular momentum increase algebraically from

-2.4

toward zero, as found. Also,

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

.

Worked example

A fixed-axis body under constant moment

A planar rotor turns about a fixed axle through

O

. Its moment of inertia is

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

, and its initial angular velocity is

-4.0\,\mathrm{rad/s}

. A constant counterclockwise net external moment of

2.4\,\mathrm{N\,m}

acts for

1.5\,\mathrm{s}

. Find its angular acceleration and final angular velocity.

Rotor about a fixed axle

The rotor turns about a fixed axis. It initially rotates clockwise, slows, stops, and then rotates counterclockwise.

Define the system
The system is the rotor, viewed from an inertial frame. The axis through $O$ is fixed. Take counterclockwise as positive; the initial clockwise angular velocity is negative.
Use the fixed-axis relationship
The moment of inertia is constant, so the net external moment equals $I_O$ times angular acceleration. The moment is positive, giving positive angular acceleration.
$\alpha=\frac{M_O}{I_O}=\frac{2.4}{0.80}=3.0\,\mathrm{rad/s^2}$
Advance the angular velocity
For constant angular acceleration, angular velocity changes by $\alpha$ multiplied by elapsed time. Keep the initial negative sign when adding this change.
$\omega_f=\omega_i+\alpha\Delta t=-4.0+(3.0)(1.5)=0.50\,\mathrm{rad/s}$

Answer: The angular acceleration is

3.0\,\mathrm{rad/s^2}

counterclockwise, and the final angular velocity is

0.50\,\mathrm{rad/s}

counterclockwise. The rotor reverses direction during the interval.

Check: It reaches zero angular velocity at

t=4.0/3.0\,\mathrm{s}

, which is before

1.5\,\mathrm{s}

, so reversal is consistent. The units

\mathrm{N\,m}/(\mathrm{kg\,m^2})

reduce to

\mathrm{s^{-2}}

, equivalent to

\mathrm{rad/s^2}

.

Worked example

Testing a proposed force from angular-momentum change

A

1.5\,\mathrm{kg}

particle is at position

(0.20,0)\,\mathrm{m}

from fixed origin

O

in an inertial frame. Its initial velocity is

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

. A constant horizontal force acts for

0.50\,\mathrm{s}

. It is explicitly the only external force on the particle during this interval. At the end, the stated angular momentum about

O

is

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

, counterclockwise. Can a force magnitude and direction satisfy these conditions?

Particle with horizontal force

The particle is at (0.20, 0) m. Its only external force is horizontal; the force direction is unknown.

State the initial and final quantities
The system is the particle; the observer and origin are fixed in an inertial frame. Take $+x$ right, $+y$ up, and counterclockwise positive. Use the given position and velocity to find initial angular momentum.
$H_{O,i}=(0.20)(1.5)(2.0)=0.60\,\mathrm{kg\,m^2/s}$
Find the net external moment
The only external force is horizontal and acts at $y=0$ . Its line of action passes through $O$ , so its moment about $O$ is zero, whatever its horizontal magnitude or direction.
$M_O=yF_x=0$
Test the final state
With zero net external moment about $O$ , angular momentum about $O$ must remain constant. Compare the required change with that consequence; the supplied final value cannot be reached under the stated conditions.
$H_{O,f}-H_{O,i}=0.90-0.60=0.30\,\mathrm{kg\,m^2/s}\ne 0$

Answer: No. The conditions are inconsistent: the only external force has zero moment about

O

, so angular momentum must remain

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

, not

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

. No horizontal force magnitude or direction can satisfy the stated final condition.

Check: This follows directly from the moment–rate relationship: zero net external moment requires constant angular momentum about the chosen origin.

Common mistakes and how to avoid them

Using a moment about one point with angular momentum about another.

Correction: Choose one origin and use it consistently in both quantities.

Treating a positive moment as proof that the body is already rotating counterclockwise.

Correction: A positive moment means angular momentum is increasing in the counterclockwise-positive sense; the current angular momentum may still be negative.

Assuming angular momentum is conserved just because the system is a particle or rigid body.

Correction: Conservation about the selected origin requires zero net external moment about that origin.

Using

M_O=I_O\alpha

without checking the fixed-axis condition and constant moment of inertia.

Correction: Use the general moment–rate relationship unless the conditions for the simplified fixed-axis form are met.

Lesson summary

Angular momentum about an origin is formed from position relative to that origin and momentum.
Net external moment about a fixed origin equals the rate of change of angular momentum about that same origin.
If net moment is zero, angular momentum about that origin is constant.
For fixed-axis rotation with constant moment of inertia, the relationship becomes $M_O=I_O\alpha$ .
Check the reference frame, origin, sign, units, and conservation conditions.

Check your understanding

Question 1

A system has zero net external moment about fixed point

O

over a time interval. What follows?

Its angular momentum about $O$ is constant.
Its angular velocity must be zero.
Its linear momentum must be zero.
Its angular momentum about every possible point is zero.

Show answer and explanation

Its angular momentum about

O

is constant.

The moment–rate relationship gives zero time rate of change of angular momentum about

O

, so that angular momentum remains constant. It need not be zero.

Question 2

A particle is at

(0,0.25)\,\mathrm{m}

relative to

O

. A

4.0\,\mathrm{N}

force acts to the right. With counterclockwise positive, what is its moment about

O

?

$+1.0\,\mathrm{N\,m}$
$-1.0\,\mathrm{N\,m}$
$0\,\mathrm{N\,m}$
$+4.25\,\mathrm{N\,m}$

Show answer and explanation

-1.0\,\mathrm{N\,m}

The force at positive

y

tends to turn the particle clockwise. The signed moment is

M_O=-yF_x=-(0.25)(4.0)=-1.0\,\mathrm{N\,m}

.

Question 3

A fixed-axis rotor has constant

I_O=0.50\,\mathrm{kg\,m^2}

and net moment

-1.0\,\mathrm{N\,m}

. What is its angular acceleration?

$-2.0\,\mathrm{rad/s^2}$
+2.0 $\mathrm{rad/s^2}$
$-0.50\,\mathrm{rad/s^2}$
+0.50 $\mathrm{rad/s^2}$

Show answer and explanation

-2.0\,\mathrm{rad/s^2}

For fixed-axis rotation,

\alpha=M_O/I_O=(-1.0)/(0.50)=-2.0\,\mathrm{rad/s^2}

. The negative sign means clockwise angular acceleration.

Key terms

Angular momentum: A measure of rotational motion about a specified origin; for a particle it is position from the origin crossed with linear momentum.
Moment: The turning effect of a force about a specified point, with sign set by the chosen rotational direction.
Inertial frame: A reference frame in which Newton’s laws apply without adding effects from an accelerating observer.
Moment of inertia: For a rigid body about a specified axis, a measure of how its mass is distributed relative to that axis.

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8.2 · Relate moment to the rate of angular momentum

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 study topic 8.2

What you will learn

1. Define angular momentum and moment

2. The moment–rate relationship

3. A reliable solution process

4. Interpreting the result

A force changes a particle’s angular momentum

A fixed-axis body under constant moment

Testing a proposed force from angular-momentum change

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