8.1 Calculate particle angular momentum about a point | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Angular Momentum

EN PH 131 Engineering Mechanics: Dynamics — Study topic 8.1

Angular momentum describes how a particle’s motion is arranged relative to a chosen point. The same particle can have different angular momentum about different points because its position relative to the point changes. In this lesson, an observer measures motion in an inertial reference frame, where Newton’s laws apply in their usual form. We use fixed horizontal and vertical axes, with positive

x

to the right and positive

y

upward. For planar motion, positive angular momentum points out of the page and corresponds to counterclockwise rotation about the reference point; negative angular momentum points into the page and corresponds to clockwise rotation. The task is to calculate angular momentum from the particle’s state at an instant, not to infer it from a force or assume it is conserved.

What you will learn

Define a particle’s angular momentum about a stated point using its position and linear momentum.
Calculate the magnitude and signed direction of angular momentum in planar motion.
Use coordinates or perpendicular distance to calculate and check the result.

1. Define the particle, observer, and reference point

Treat the object as a particle: its mass is concentrated at a point, and its size and rotation are not part of this calculation. The system is that particle. An observer describes its position and velocity in a chosen inertial frame. State the point

O

about which angular momentum is required; it may be the origin of the coordinates, but the definition works for any specified point.

The position vector from

O

to the particle is

\mathbf r

. Its velocity in the chosen frame is

\mathbf v

, and its linear momentum is

\mathbf p=m\mathbf v

, where

m

is the particle’s mass. Angular momentum about

O

is the cross product of position relative to

O

and linear momentum. The position vector matters: changing the reference point changes

\mathbf r

, and can change the answer.

In a plane, the result has only a component perpendicular to that plane. With

x

right and

y

up, report its signed component along the positive

z

direction. A positive value is counterclockwise by the right-hand rule; a negative value is clockwise. The sign describes direction, while the absolute value gives magnitude.

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

Specify the reference point and the frame in which velocity is measured.
Use position relative to that point, not the particle’s distance travelled.
For planar motion, the signed $z$ component gives the direction.

2. Calculate the planar component

A cross product depends on the angle between the position vector and velocity. Its magnitude is the product of their magnitudes and the absolute value of the sine of their included angle. Thus, only the component of velocity perpendicular to the position vector contributes. A velocity directed straight toward or away from

O

produces zero angular momentum about

O

.

For coordinates

\mathbf r=x\mathbf i+y\mathbf j

and

\mathbf v=v_x\mathbf i+v_y\mathbf j

, the signed planar component is

m(xv_y-yv_x)

. This coordinate form is often the safest method: retain the signs of the coordinate and velocity components, then evaluate the expression. Do not take absolute values until after finding the signed component. If using the angle formula, the angle is the usual included angle between the vectors, so its sine gives a magnitude; assign clockwise or counterclockwise direction separately.

The dimensions are mass times length times velocity, or

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

. This is angular momentum, not torque: torque involves a force and has units

\mathrm{N\,m}

. The two quantities are related in broader dynamics, but calculating angular momentum here requires only the particle’s position, mass, and velocity.

|H_{O,z}|=mrv|\sin\theta|

Use the perpendicular part of velocity when reasoning from magnitudes.
Use signed components to determine clockwise or counterclockwise direction.
Check that the units reduce to $\mathrm{kg\,m^2/s}$ .

3. A reliable calculation and interpretation

Begin by identifying the particle and the reference point, then record the particle’s mass, position relative to that point, and velocity in the stated frame. Draw the geometry when it helps establish the position or velocity direction. Choose axes and the positive rotation direction before substituting values.

Next, calculate the signed component with coordinates, or calculate the magnitude using the included angle and then assign a direction. These approaches must agree. If the velocity is parallel or antiparallel to the position vector, the result is zero. If the particle is farther from the point while its mass and perpendicular speed stay fixed, the magnitude increases in direct proportion to that distance.

A particle’s angular momentum is not automatically constant. This lesson’s calculation is instantaneous: it uses the particle’s state at the instant described. Do not assume conservation unless a separate problem provides the conditions and asks for a change over time. As a final check, confirm that the reference point, units, sign, and stated direction all match the calculation.

[H_{O,z}]=\mathrm{kg\,m^2/s}

The state is instantaneous: use position and velocity at the same instant.
A zero result is expected for purely radial motion relative to the point.
Do not infer conservation from the definition alone.

Worked example

Coordinate calculation with a negative result

A particle of mass

2.5\,\mathrm{kg}

is at position

(3,1)\,\mathrm{m}

relative to fixed point

O

. Its velocity is

(4,-2)\,\mathrm{m/s}

. Calculate its angular momentum about

O

.

Particle state in the plane

The particle is located relative to O at (3, 1) m; its velocity points right and down.

Set the frame and signs
Use fixed point $O$ as the coordinate origin, with $x$ right and $y$ up. The observer measures the particle’s given position and velocity in this frame. Positive angular momentum is out of the page, or counterclockwise.
Substitute the signed components
The known components are $x=3\,\mathrm{m}$ , $y=1\,\mathrm{m}$ , $v_x=4\,\mathrm{m/s}$ , and $v_y=-2\,\mathrm{m/s}$ . The negative vertical velocity must remain negative in the coordinate formula.
$H_{O,z}=2.5[3(-2)-1(4)]$
Interpret the result
The component is negative, so the angular momentum points into the page and is clockwise about $O$ . Its units are mass times length times velocity.
$H_{O,z}=-25\,\mathrm{kg\,m^2/s}$

Answer:

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

clockwise about

O

.

Check: The position and velocity components give

xv_y-yv_x=-6-4<0

, confirming clockwise direction. The units are

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

.

Worked example

Use a perpendicular distance for straight-line motion

A

1.2\,\mathrm{kg}

particle moves horizontally to the left at

3.0\,\mathrm{m/s}

. At the instant of interest, it is at

(0.60,0.80)\,\mathrm{m}

relative to fixed point

O

. Calculate its angular momentum about

O

.

Straight path past O

The particle moves left on the stated horizontal path; the velocity arrow indicates its direction at the stated position.

Identify the instantaneous state
The particle is the system; the observer uses fixed point $O$ and axes with positive $x$ right and positive $y$ up. The initial and final states are not needed because the question asks for angular momentum at this instant. The velocity components are $v_x=-3.0\,\mathrm{m/s}$ and $v_y=0$ .
Calculate the signed component
Use the particle’s position relative to $O$ and retain the negative sign of its leftward velocity. The positive result means counterclockwise angular momentum.
$H_{O,z}=1.2[0.60(0)-0.80(-3.0)]=2.88\,\mathrm{kg\,m^2/s}$
Check using the perpendicular offset
The perpendicular distance from $O$ to this horizontal path is $0.80\,\mathrm{m}$ . Multiplying mass, speed, and that distance gives the same magnitude; leftward motion above $O$ produces the positive, counterclockwise sign.
$mvd_\perp=1.2(3.0)(0.80)=2.88\,\mathrm{kg\,m^2/s}$

Answer:

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

counterclockwise about

O

.

Check: The coordinate and perpendicular-distance calculations agree. If the path were moved to pass through

O

, its perpendicular offset would be zero and so would the angular momentum.

Worked example

Circular motion about the centre

A particle of mass

0.75\,\mathrm{kg}

moves counterclockwise on a circle of radius

1.5\,\mathrm{m}

centred at fixed point

O

. Its speed is

4.0\,\mathrm{m/s}

. Calculate its angular momentum about

O

.

Circular path about O

At the top of the stated circular path, the tangential velocity points left.

Choose the reference point and direction
The particle is the system and the observer uses an inertial frame with $O$ at the centre of the stated circular path. Positive angular momentum is counterclockwise. At the top of the circle, the radius from $O$ points up and the tangential velocity points left, so they are perpendicular.
Calculate the magnitude
Because the position and velocity are perpendicular, the sine of their included angle has magnitude one. Multiply the mass, radius, and speed.
$|H_{O,z}|=mrv=0.75(1.5)(4.0)=4.5\,\mathrm{kg\,m^2/s}$
Assign the sign
The stated motion is counterclockwise as viewed in the plane, so the component is positive under the selected convention.
$H_{O,z}=+4.5\,\mathrm{kg\,m^2/s}$

Answer:

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

counterclockwise about

O

.

Check: The result has the required units. If the speed or radius tends to zero, the angular momentum tends to zero; reversing the direction of travel reverses its sign.

Common mistakes and how to avoid them

Using the distance travelled instead of the position relative to the reference point.

Correction: Use the vector from the stated point to the particle at the instant of interest.

Reporting only a positive magnitude and omitting the rotation direction.

Correction: Retain the sign of

m(xv_y-yv_x)

, then describe positive as counterclockwise and negative as clockwise for the stated axes.

Using the full speed as if all of it contributed, even when the velocity is not perpendicular to the position vector.

Correction: Use the perpendicular velocity component, or include the absolute sine of the angle between position and velocity to calculate magnitude.

Assuming angular momentum is always conserved.

Correction: The definition calculates angular momentum at an instant; conservation requires additional conditions not assumed here.

Lesson summary

Angular momentum about a point uses the particle’s position relative to that point and its linear momentum in the chosen frame.
For planar motion, calculate $H_{O,z}=m(xv_y-yv_x)$ ; positive is counterclockwise and negative is clockwise when $x$ is right and $y$ is up.
The units are $\mathrm{kg\,m^2/s}$ . Radial motion relative to the point gives zero angular momentum.

Check your understanding

Question 1

A particle moves directly away from point

O

along a straight line through

O

. What is its angular momentum about

O

at that instant?

Zero
Positive, because the particle moves away from O
Negative, because the particle moves away from O
It cannot be calculated without the particle’s acceleration

Show answer and explanation

Zero

The position and velocity are parallel, so the sine of their included angle is zero. The velocity has no component perpendicular to the position vector.

Question 2

With

x

right and

y

up, a particle has positive

x

, zero

y

, and positive

v_y

. What is the sign of its angular momentum about the origin?

Positive, corresponding to counterclockwise
Negative, corresponding to clockwise
Zero, because $y=0$
The sign cannot be determined without the mass

Show answer and explanation

Positive, corresponding to counterclockwise

The signed component is

m(xv_y-yv_x)=mxv_y

, which is positive for positive mass,

x

, and

v_y

.

Key terms

Angular momentum about a point: The cross product of the particle’s position relative to the point and its linear momentum.
Reference frame: The coordinate system and observer used to describe position and motion.
Perpendicular distance: The shortest distance from the reference point to the particle’s straight line of motion.
Signed component: A value whose positive or negative sign identifies direction relative to the chosen positive axis.

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8.1 · Calculate particle angular momentum about a point

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 Engineering Mechanics: Dynamics — Study topic 8.1

What you will learn

1. Define the particle, observer, and reference point

2. Calculate the planar component

3. A reliable calculation and interpretation

Coordinate calculation with a negative result

Use a perpendicular distance for straight-line motion

Circular motion about the centre

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through EN PH 131

About this lesson and its review