8.3 Apply angular impulse–momentum | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Angular Momentum

Relating a system’s angular momentum change to the moment of external forces over time

Angular impulse–momentum is useful when forces act over a time interval and the resulting change in rotation is more direct to find than the motion at every instant. The central idea is that the angular momentum at the end of an interval equals the angular momentum at its start plus the accumulated moment of external forces. First specify the system and the fixed point about which moments and angular momentum are measured. Then choose a positive rotation direction and apply the equation with consistent signs. The method applies to particles and, in introductory planar dynamics, to rigid bodies rotating about a fixed axis.

What you will learn

Define a system, inertial observer, reference point, axes, and positive rotation direction for an angular impulse–momentum problem.
Calculate angular momentum about a fixed point for a particle or a rigid body rotating about a fixed axis.
Apply the angular impulse–momentum equation using a constant or time-varying external moment.
Check signs, units, assumptions, and limiting cases in a solution.

1. Set up the system and reference point

A system is the particle or body whose motion you are studying. An observer measures its motion from a reference frame; in these examples, take that frame to be fixed to the ground and treat it as inertial. Choose a fixed point

O

in that frame as the reference point for angular momentum and moments. In a planar problem, use

x

to the right,

y

upward, and positive angular direction counterclockwise, unless the problem defines another choice.

For a particle, angular momentum about

O

depends on its position relative to

O

and its linear momentum. In vector form it is the cross product of position and momentum. In a planar calculation, its signed value points along the axis perpendicular to the plane: counterclockwise is positive and clockwise is negative. For a rigid body rotating about a fixed axis through

O

, angular momentum along that axis is its mass moment of inertia about the axis multiplied by its angular velocity.

Angular impulse is the time accumulation of the net external moment about the chosen point. Moments from external forces and couples contribute; internal forces within the selected system do not appear in the net external moment. The point

O

must remain fixed in the inertial frame for the forms used in this lesson. If the net moment is zero throughout the interval, angular momentum about

O

is unchanged.

H_{O2}=H_{O1}+\int_{t_1}^{t_2}\sum M_O\,dt

State the system, fixed reference point, inertial frame, axes, and positive rotation direction before calculating.
Use signed moments: counterclockwise positive under the convention used here.
Include the moments of all external forces and applied couples about the chosen point.

2. Choose the matching angular-momentum model

For one particle of mass

m

, position vector

\mathbf r

measured from

O

, and velocity

\mathbf v

, use

\mathbf H_O=\mathbf r\times m\mathbf v

. In planar motion, its signed component is

m(xv_y-yv_x)

. This expression uses the particle’s position from

O

, not the distance it has travelled along its path.

For a rigid body rotating in a plane about a fixed axis through

O

, use

H_O=I_O\omega

, where

I_O

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

\omega

is its signed angular velocity. The planar rigid-body model is appropriate when the axis is fixed and the rotation is about that axis. Do not replace

I_O

with the body’s mass: the units and physical meaning are different.

When the net moment is constant over the interval, its angular impulse is the moment multiplied by the elapsed time. When it varies with time, integrate its signed value over the interval. A moment–time graph can also be used: the signed area under the graph is the angular impulse. With a changing moment, the average moment multiplied by the interval gives the same impulse.

\mathbf H_O=\mathbf r\times m\mathbf v,\qquad H_O=I_O\omega

Particle model: angular momentum comes from position crossed with linear momentum.
Fixed-axis rigid-body model: angular momentum about the axis is $I_O\omega$ .
Angular impulse has units of newton-metres times seconds.

3. Apply the equation and check the result

Write the initial and final states at the ends of the stated time interval. Add the signed angular impulse from every external moment to the initial angular momentum to obtain the final angular momentum. For a fixed-axis rigid body, substitute

I_O\omega

for angular momentum; if the body and axis do not change, the same

I_O

applies at both ends.

If a force is given rather than a moment, calculate its moment about

O

using its perpendicular lever arm, or the planar cross-product rule. A force whose line of action passes through

O

has zero moment about

O

, even if its magnitude is large. An applied couple contributes its signed couple moment directly.

Before accepting an answer, check units and direction. Angular momentum has units

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

, and angular impulse has the same units because

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

. A positive impulse should increase signed angular momentum; a negative one should decrease it. If the net moment is zero, the final angular momentum must equal the initial value. These checks often reveal a sign or lever-arm error.

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

Use the same point and sign convention for initial momentum, moments, and final momentum.
A force through the reference point has no moment about that point.
Check units, direction, zero-moment behaviour, and whether the stated initial condition is recovered when the time interval is zero.

Worked example

Particle acted on by a constant moment

A particle of mass

2.0\,\mathrm{kg}

moves in the plane. Relative to fixed point

O

, its initial position is

(0.30,0.20)\,\mathrm{m}

and its initial velocity is

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

. During the next

0.40\,\mathrm{s}

, the net external moment about

O

is constant at

+1.5\,\mathrm{N\,m}

. Find its final angular momentum about

O

.

Particle and positive moment

The particle’s position is measured from fixed point O; the shown positive moment is counterclockwise.

Define the states and sign
The system is the particle, and the observer uses a ground-fixed inertial frame. Take $O$ as the fixed reference point, with $x$ right, $y$ up, and counterclockwise positive. The initial state is given; the final state is after $0.40\,\mathrm{s}$ . The unknown is final angular momentum about $O$ .
Calculate initial angular momentum
For planar motion, use the signed component $m(xv_y-yv_x)$ . Here the initial velocity has no $y$ component, so the initial value is clockwise and therefore negative.
$H_{O1}=2.0[(0.30)(0)-(0.20)(1.0)]=-0.40\,\mathrm{kg\,m^2/s}$
Add angular impulse
The constant positive moment acts for $0.40\,\mathrm{s}$ , so its angular impulse is positive. Add it to the initial signed angular momentum.
$H_{O2}=H_{O1}+M_O\Delta t=-0.40+(1.5)(0.40)=0.20\,\mathrm{kg\,m^2/s}$

Answer: The final angular momentum is

+0.20\,\mathrm{kg\,m^2/s}

about

O

, in the counterclockwise sense.

Check: The angular impulse is

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

. It changes the initial negative value to a positive final value, consistent with a counterclockwise net moment.

Worked example

Fixed-axis rotor under a constant couple

A rotor with mass moment of inertia

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

about its fixed axis initially rotates clockwise at

3.0\,\mathrm{rad/s}

. A constant counterclockwise couple of

2.0\,\mathrm{N\,m}

acts for

1.5\,\mathrm{s}

. Find its final angular velocity.

Rotor about a fixed axis

The rotor turns about a fixed axis; counterclockwise angular quantities are positive.

Set the signed initial state
The system is the rotor, observed from a ground-fixed inertial frame. The reference is its fixed axis. Set counterclockwise positive, so the stated clockwise initial angular velocity is negative. The rotor’s inertia about the axis remains constant.
$\omega_1=-3.0\,\mathrm{rad/s}$
Apply angular impulse–momentum
For rotation about a fixed axis, angular momentum is $I_O\omega$ . The couple is positive, so its impulse increases signed angular momentum.
$I_O\omega_2=I_O\omega_1+M_O\Delta t$
Solve for final angular velocity
Substitute the given inertia, initial angular velocity, couple, and duration. The result is positive, so the rotor ends turning counterclockwise.
$\omega_2=-3.0+\frac{(2.0)(1.5)}{0.80}=0.75\,\mathrm{rad/s}$

Answer: The final angular velocity is

0.75\,\mathrm{rad/s}

counterclockwise.

Check: The angular impulse is

3.0\,\mathrm{N\,m\,s}

. The initial angular momentum is

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

, so the final value is

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

, equal to

I_O\omega_2

.

Worked example

A time-varying moment on a fixed-axis body

A body with fixed-axis mass moment of inertia

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

starts from rest. Its net external moment is

M_O(t)=4t\,\mathrm{N\,m}

for

0\le t\le2.0\,\mathrm{s}

, where

t

is in seconds. Find its angular velocity at

2.0\,\mathrm{s}

.

Body with increasing moment

The net moment grows with time and remains counterclockwise over the stated interval.

Define the interval and initial condition
The system is the rigid body and the observer uses a ground-fixed inertial frame. Measure angular momentum about the fixed axis, with counterclockwise positive. The initial angular velocity at $t=0$ is zero; the desired final state is at $t=2.0\,\mathrm{s}$ .
$H_{O1}=I_O\omega_1=0$
Integrate the moment
Because the moment varies with time, its angular impulse is its signed time integral. It stays positive, so the integral adds positive angular momentum.
$\int_0^{2.0}4t\,dt=\left[2t^2\right]_0^{2.0}=8.0\,\mathrm{N\,m\,s}$
Find the final angular velocity
Set the final angular momentum equal to the initial value plus the calculated angular impulse, then divide by the fixed-axis inertia.
$\omega_2=\frac{8.0}{1.2}=6.67\,\mathrm{rad/s}$

Answer: At

2.0\,\mathrm{s}

, the body’s angular velocity is approximately

6.67\,\mathrm{rad/s}

counterclockwise.

Check: The integral has units

\mathrm{N\,m\,s}

, matching angular momentum. The positive moment produces a positive final angular velocity. If the interval were zero, the impulse and the change in angular velocity would both be zero.

Common mistakes and how to avoid them

Using the magnitude of angular momentum while ignoring whether it is clockwise or counterclockwise.

Correction: Choose a positive rotation direction and use signed angular momentum and moments throughout the equation.

Taking a force’s moment using its full distance from the reference point even when the force is not perpendicular to that distance.

Correction: Use the perpendicular lever arm or the cross product; a force through the reference point has zero moment about it.

Using a changing reference point between the initial and final states.

Correction: Measure both angular momenta and all moments about the same fixed point in the inertial frame.

Treating angular impulse as force times time, without accounting for the lever arm.

Correction: Angular impulse is the time integral of moment. If starting from a force, first find its moment about the chosen point.

Lesson summary

Angular impulse–momentum states that final angular momentum equals initial angular momentum plus the net external angular impulse.
For a particle, use position crossed with linear momentum; for a rigid body rotating about a fixed axis, use $H_O=I_O\omega$ .
Keep the reference point fixed and the sign convention consistent, integrate variable moments, and verify units and direction.

Check your understanding

Question 1

A body rotating about a fixed axis has zero net external moment during an interval. What follows about its angular momentum about that axis?

It remains constant over the interval.
It must become zero.
It increases in the counterclockwise direction.
It changes by the body’s mass multiplied by the interval.

Show answer and explanation

It remains constant over the interval.

With zero net moment, angular impulse is zero, so final angular momentum equals initial angular momentum.

Question 2

A constant clockwise moment acts for a positive time interval. With counterclockwise defined as positive, what is the sign of its angular impulse?

Positive
Negative
Zero
It depends only on the body’s mass

Show answer and explanation

Negative

A clockwise moment is negative under this convention, and multiplying by a positive duration gives a negative angular impulse.

Question 3

Which units are equivalent to angular momentum?

$\mathrm{N\,m\,s}$
$\mathrm{N\,s}$
$\mathrm{kg\,m/s}$
$\mathrm{N\,m}$

Show answer and explanation

\mathrm{N\,m\,s}

\mathrm{N\,m\,s}

reduces to

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

, the units of angular momentum.

Key terms

Angular momentum: A measure of rotational motion about a specified point or axis; for a particle it is position crossed with linear momentum.
Angular impulse: The time integral of the net external moment about a specified point.
Moment: The rotational effect of a force about a point, determined by the force and its perpendicular lever arm.
Mass moment of inertia: A measure of how a body’s mass is distributed relative to a specified rotation axis; in fixed-axis planar rotation it relates angular velocity to angular momentum.

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8.3 · Apply angular impulse–momentum

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Relating a system’s angular momentum change to the moment of external forces over time

What you will learn

1. Set up the system and reference point

2. Choose the matching angular-momentum model

3. Apply the equation and check the result

Particle acted on by a constant moment

Fixed-axis rotor under a constant couple

A time-varying moment on a fixed-axis body

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