# Module 2 -- Angular Momentum of a Rigid Body Rotating about a Fixed Axis

#### Learning Objectives for this Module

At the end of this module you should be able to:

• Calculate the angular momentum of a system of particles as the sum of the angular momentum of the individual particles.
• Calculate the angular momentum of a planar rigid object rotating about a fixed axis.
• Calculate the angular momentum of a rigid object rotating about an axis of symmetry.

In this module we will consider objects that rotate about a fixed axis. We will restrict the study to two types of objects:

1. Objects that are symmetric and rotate about an axis of symmetry as those in figure 1.

2. Objects that have planar symmetry and are rotating about an axis perpendicular to the plane where they are contained as shown in figure 2.

We will start with objects composed by point masses and then we will generalized our results to rigid objects which mass is distributed continuously within the body.

### Angular momentum of a system of point particles

 The angular momentum of a system of particles is obtained by adding the angular momentum of each of the constituents.

#### Example 1. Two balls attached by a rod

Lets consider a rigid object consisting of two small balls of masses m1 and m2 connected by a rod of length 2R and negligible mass. The object is placed on a horizontal and frictionless table contained in the xy - plane. The object is set to rotate counterclockwise as viewed from above the xy - plane with an angular speed ω. The axis of rotation passes through the center of the rod, point Q in the figure.

Question: What is the direction and the magnitude of the angular momentum vector about point Q?

#### Example 2. Two balls attached by a rod - another point of view.

Consider the same rigid object of the previous example. We will now calculate the angular momentum of the object about a point S located in the axis of rotation at a distance d below the center of the circle as shown in the figure below.

Question 1: At the instant shown in the figure above, which of the arrows below best represents the direction of the angular momentum of each of the particles about point S?
Question 2: What is the direction and the magnitude of the total angular momentum about point S if both particles have the same mass, m1 = m2 = m?

#### Symmetric Objects

Implications of the results obtained in Examples 1 and 2:

After answering the questions asked in the two examples above we conclude:

If the masses are equal, then the angular momentum about points Q and S are equal.

$\vec{L}_Q = \vec{L}_S$

This result implies that the angular momentum is independent of the location of the point as long as the point is in the axis of rotation. In this situation, we can refer to the angular momentum to be calculated about the axis instead of being calculated about a point.

$\vec{L}_{axis} = I_{axis}\vec\omega$

where Iaxis is the moment of inertia of the object calculated about the axis. (Here we are using the label axis because the z-axis used in the example is arbitrary. )

Symmetric Objects Rotating about an Axis of Symmetry

The result above is a direct consequence of a symmetrical distribution of the masses with respect to the axis of rotation. In this situation the axis of rotation is an axis of symmetry of the object and the angular momentum about any point in the axis is parallel to ω.

 For a symmetric object rotating about an axis of symmetry, the angular momentum calculated about any point in the axis is: $\vec{L}_{axis} = I_{axis}\vec\omega$

Symmetric Objects Rotating about an Axis that is Not an Axis of Symmetry

If a symmetric objects rotates about an axis that is not an axis of symmetry, then:

1. the angular momentum vector of the object about a point in the axis is NOT parallel to ω,

$\vec{L}_Q \not\parallel \vec{\omega}$

2. the component of the angular momentum along the axis of rotation is proportional to ω and given by:

LQ,axis = Iaxisω

This result is derived below. We can see from this expression that the component along the axis of rotation is independent of the point. We can then drop the subscript Q to have that:

 For a symmetric body rotating about an axis that is not an axis of symmetry, the component of the angular momentum about the axis of rotation is: Laxis = Iaxisω

### Planar Objects

When a plane object, as those in the figure below, rotates about an axis perpendicular to the plane where it is contained, then the angular momentum about the point of intersection between the axis and the body, point Q in the figure, is parallel toω and is equal to:

$\vec{L}_Q = I_{axis}\vec\omega$

where Iaxis is the moment of inertia measured about the axis of rotation.

In the examples shown in the figure, the object is contained in the xy-plane and the axis of rotation is parallel to ω, perpendicular to the xy-plane. Note that the axis of rotation in the examples below does not necessarily pass through the center of mass of the system, therefore the Iaxis must be obtained using the parallel axis theorem.

### From point particles to rigid objects

The above results obtained for a system of discrete point particles can be generalized to rigid objects by calculating the moment of inertia as an integral. This is illustrated in the following example:

#### Illustrative Examples

A hoop: Consider a hoop of mass M and radius R contained in the xy-plane as shown below. The mass is distributed uniformly throughout the hoop.

If the hoop is rotating with angular velocity ω, calculate the angular momentum of the hoop:

1. about the axis of symmetry (left figure),
2. about an axis perpendicular to the xy-plane, passing through point Q (right figure).

NOTE: Finding the angular momentum about a fixed axis for symmetric or planar objects is reduced to calculate the proper expression of the moment of inertia of the object about the axis of rotation. For that purpose, we either solve the integral or we use a table of moments of inertia like the one presented in the module describing moment of inertia.