11 Angular Momentum
11.2 Angular Momentum
Learning Objectives
By the end of this section, you will be able to:
- Describe the vector nature of angular momentum
- Find the total angular momentum and torque about a designated origin of a system of particles
- Calculate the angular momentum of a rigid body rotating about a fixed axis
- Calculate the torque on a rigid body rotating about a fixed axis
- Use conservation of angular momentum in the analysis of objects that change their rotation rate
Why does Earth keep on spinning? What started it spinning to begin with? Why doesn’t Earth’s gravitational attraction not bring the Moon crashing in toward Earth? And how does an ice skater manage to spin faster and faster simply by pulling her arms in? Why does she not have to exert a torque to spin faster?
The answer is in a new conserved quantity, since all of these scenarios are in closed systems. This new quantity, angular momentum, is analogous to linear momentum. In this chapter, we first define and then explore angular momentum from a variety of viewpoints. First, however, we investigate the angular momentum of a single particle. This allows us to develop angular momentum for a system of particles and for a rigid body that is cylindrically symmetric.
Angular Momentum of a Single Particle
Figure shows a particle at a position
Angular Momentum of a Particle
The angular momentum

The intent of choosing the direction of the angular momentum to be perpendicular to the plane containing
where
As with the definition of torque, we can define a lever arm
We see that if the direction of
If we take the time derivative of the angular momentum, we arrive at an expression for the torque on the particle:
Here we have used the definition of
Note the similarity with the linear result of Newton’s second law,
Problem-Solving Strategy: Angular Momentum of a Particle
- Choose a coordinate system about which the angular momentum is to be calculated.
- Write down the radius vector to the point particle in unit vector notation.
- Write the linear momentum vector of the particle in unit vector notation.
- Take the cross product
and use the right-hand rule to establish the direction of the angular momentum vector. - See if there is a time dependence in the expression of the angular momentum vector. If there is, then a torque exists about the origin, and use
to calculate the torque. If there is no time dependence in the expression for the angular momentum, then the net torque is zero.
Example
Angular Momentum and Torque on a Meteor
A meteor enters Earth’s atmosphere (Figure) and is observed by someone on the ground before it burns up in the atmosphere. The vector

Strategy
We resolve the acceleration into x– and y-components and use the kinematic equations to express the velocity as a function of acceleration and time. We insert these expressions into the linear momentum and then calculate the angular momentum using the cross-product. Since the position and momentum vectors are in the xy-plane, we expect the angular momentum vector to be along the z-axis. To find the torque, we take the time derivative of the angular momentum.
Solution
The meteor is entering Earth’s atmosphere at an angle of
We write the velocities using the kinematic equations.
- The angular momentum is
At
, the angular momentum of the meteor about the origin isThis is the instant that the observer sees the meteor.
- To find the torque, we take the time derivative of the angular momentum. Taking the time derivative of
as a function of time, which is the second equation immediately above, we haveThen, since
, we haveThe units of torque are given as newton-meters, not to be confused with joules. As a check, we note that the lever arm is the x-component of the vector
in Figure since it is perpendicular to the force acting on the meteor, which is along its path. By Newton’s second law, this force isThe lever arm is
Thus, the torque is
Significance
Since the meteor is accelerating downward toward Earth, its radius and velocity vector are changing. Therefore, since
Check Your Understanding
A proton spiraling around a magnetic field executes circular motion in the plane of the paper, as shown below. The circular path has a radius of 0.4 m and the proton has velocity
Show Answer
From the figure, we see that the cross product of the radius vector with the momentum vector gives a vector directed out of the page. Inserting the radius and momentum into the expression for the angular momentum, we have
Angular Momentum of a System of Particles
The angular momentum of a system of particles is important in many scientific disciplines, one being astronomy. Consider a spiral galaxy, a rotating island of stars like our own Milky Way. The individual stars can be treated as point particles, each of which has its own angular momentum. The vector sum of the individual angular momenta give the total angular momentum of the galaxy. In this section, we develop the tools with which we can calculate the total angular momentum of a system of particles.
In the preceding section, we introduced the angular momentum of a single particle about a designated origin. The expression for this angular momentum is
Similarly, if particle i is subject to a net torque
The sum of the individual torques produces a net external torque on the system, which we designate
Figure states that the rate of change of the total angular momentum of a system is equal to the net external torque acting on the system when both quantities are measured with respect to a given origin. Figure can be applied to any system that has net angular momentum, including rigid bodies, as discussed in the next section.
Example
Angular Momentum of Three Particles
Referring to Figure(a), determine the total angular momentum due to the three particles about the origin. (b) What is the rate of change of the angular momentum?

Strategy
Write down the position and momentum vectors for the three particles. Calculate the individual angular momenta and add them as vectors to find the total angular momentum. Then do the same for the torques.
Solution
- Particle 1:
Particle 2:
,Particle 3:
,We add the individual angular momenta to find the total about the origin:
- The individual forces and lever arms are
Therefore:
Significance
This example illustrates the superposition principle for angular momentum and torque of a system of particles. Care must be taken when evaluating the radius vectors
Angular Momentum of a Rigid Body
We have investigated the angular momentum of a single particle, which we generalized to a system of particles. Now we can use the principles discussed in the previous section to develop the concept of the angular momentum of a rigid body. Celestial objects such as planets have angular momentum due to their spin and orbits around stars. In engineering, anything that rotates about an axis carries angular momentum, such as flywheels, propellers, and rotating parts in engines. Knowledge of the angular momenta of these objects is crucial to the design of the system in which they are a part.
To develop the angular momentum of a rigid body, we model a rigid body as being made up of small mass segments,

Using the right-hand rule, the angular momentum vector points in the direction shown in part (b). The sum of the angular momenta of all the mass segments contains components both along and perpendicular to the axis of rotation. Every mass segment has a perpendicular component of the angular momentum that will be cancelled by the perpendicular component of an identical mass segment on the opposite side of the rigid body. Thus, the component along the axis of rotation is the only component that gives a nonzero value when summed over all the mass segments. From part (b), the component of
The net angular momentum of the rigid body along the axis of rotation is
The summation
This equation is analogous to the magnitude of the linear momentum
Example
Angular Momentum of a Robot Arm
A robot arm on a Mars rover like Curiosity shown in Figure is 1.0 m long and has forceps at the free end to pick up rocks. The mass of the arm is 2.0 kg and the mass of the forceps is 1.0 kg. See Figure. The robot arm and forceps move from rest to

Strategy
We use Figure to find angular momentum in the various configurations. When the arm is rotating downward, the right-hand rule gives the angular momentum vector directed out of the page, which we will call the positive z-direction. When the arm is rotating upward, the right-hand rule gives the direction of the angular momentum vector into the page or in the negative z-direction. The moment of inertia is the sum of the individual moments of inertia. The arm can be approximated with a solid rod, and the forceps and Mars rock can be approximated as point masses located at a distance of 1 m from the origin. For part (c), we use Newton’s second law of motion for rotation to find the torque on the robot arm.
Solution
- Writing down the individual moments of inertia, we haveRobot arm:
Forceps: Mars rock: Therefore, without the Mars rock, the total moment of inertia isand the magnitude of the angular momentum is
The angular momentum vector is directed out of the page in the
direction since the robot arm is rotating counterclockwise. - We must include the Mars rock in the calculation of the moment of inertia, so we have
and
Now the angular momentum vector is directed into the page in the
direction, by the right-hand rule, since the robot arm is now rotating clockwise. - We find the torque when the arm does not have the rock by taking the derivative of the angular momentum using Figure
But since , and understanding that the direction of the angular momentum and torque vectors are along the axis of rotation, we can suppress the vector notation and findwhich is Newton’s second law for rotation. Since
, we can calculate the net torque:
Significance
The angular momentum in (a) is less than that of (b) due to the fact that the moment of inertia in (b) is greater than (a), while the angular velocity is the same.
Check Your Understanding
Which has greater angular momentum: a solid sphere of mass m rotating at a constant angular frequency
Show Solution
Visit the University of Colorado’s Interactive Simulation of Angular Momentum to learn more about angular momentum.
Summary
- The angular momentum
of a single particle about a designated origin is the vector product of the position vector in the given coordinate system and the particle’s linear momentum. - The angular momentum
of a system of particles about a designated origin is the vector sum of the individual momenta of the particles that make up the system. - The net torque on a system about a given origin is the time derivative of the angular momentum about that origin:
. - A rigid rotating body has angular momentum
directed along the axis of rotation. The time derivative of the angular momentum gives the net torque on a rigid body and is directed along the axis of rotation.
Conceptual Questions
Can you assign an angular momentum to a particle without first defining a reference point?
For a particle traveling in a straight line, are there any points about which the angular momentum is zero? Assume the line intersects the origin.
Show Solution
All points on the straight line will give zero angular momentum, because a vector crossed into a parallel vector is zero.
Under what conditions does a rigid body have angular momentum but not linear momentum?
If a particle is moving with respect to a chosen origin it has linear momentum. What conditions must exist for this particle’s angular momentum to be zero about the chosen origin?
Show Solution
The particle must be moving on a straight line that passes through the chosen origin.
If you know the velocity of a particle, can you say anything about the particle’s angular momentum?
Problems
A 0.2-kg particle is travelling along the line
A bird flies overhead from where you stand at an altitude of 300.0 m and at a speed horizontal to the ground of 20.0 m/s. The bird has a mass of 2.0 kg. The radius vector to the bird makes an angle
Show Solution
The magnitude of the cross product of the radius to the bird and its momentum vector yields
A Formula One race car with mass 750.0 kg is speeding through a course in Monaco and enters a circular turn at 220.0 km/h in the counterclockwise direction about the origin of the circle. At another part of the course, the car enters a second circular turn at 180 km/h also in the counterclockwise direction. If the radius of curvature of the first turn is 130.0 m and that of the second is 100.0 m, compare the angular momenta of the race car in each turn taken about the origin of the circular turn.
A particle of mass 5.0 kg has position vector
Show Solution
a.
b.
Use the right-hand rule to determine the directions of the angular momenta about the origin of the particles as shown below. The z-axis is out of the page.
Suppose the particles in the preceding problem have masses
Show Solution
a.
Two particles of equal mass travel with the same speed in opposite directions along parallel lines separated by a distance d. Show that the angular momentum of this two-particle system is the same no matter what point is used as the reference for calculating the angular momentum.
An airplane of mass
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a.
At a particular instant, a 1.0-kg particle’s position is
A particle of mass m is dropped at the point
a.
b.
(a) Calculate the angular momentum of Earth in its orbit around the Sun. (b) Compare this angular momentum with the angular momentum of Earth about its axis.
A boulder of mass 20 kg and radius 20 cm rolls down a hill 15 m high from rest. What is its angular momentum when it is half way down the hill? (b) At the bottom?
Show Solution
a.
b.
A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius 2.0 m and mass 10,000 kg, and two antennas projecting out from the center of mass of the main body that can be approximated with rods of length 3.0 m each and mass 10 kg. The antenna’s lie in the plane of rotation. What is the angular momentum of the satellite?
A propeller consists of two blades each 3.0 m in length and mass 120 kg each. The propeller can be approximated by a single rod rotating about its center of mass. The propeller starts from rest and rotates up to 1200 rpm in 30 seconds at a constant rate. (a) What is the angular momentum of the propeller at
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a.
b.
A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of
The blades of a wind turbine are 30 m in length and rotate at a maximum rotation rate of 20 rev/min. (a) If the blades are 6000 kg each and the rotor assembly has three blades, calculate the angular momentum of the turbine at this rotation rate. (b) What is the torque require to rotate the blades up to the maximum rotation rate in 5 minutes?
Show Solution
a.
b.
A roller coaster has mass 3000.0 kg and needs to make it safely through a vertical circular loop of radius 50.0 m. What is the minimum angular momentum of the coaster at the bottom of the loop to make it safely through? Neglect friction on the track. Take the coaster to be a point particle.
A mountain biker takes a jump in a race and goes airborne. The mountain bike is travelling at 10.0 m/s before it goes airborne. If the mass of the front wheel on the bike is 750 g and has radius 35 cm, what is the angular momentum of the spinning wheel in the air the moment the bike leaves the ground?
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Glossary
- angular momentum
- rotational analog of linear momentum, found by taking the product of moment of inertia and angular velocity