The Parallel-Axis Theorem

Learning Goal:

To apply the parallel-axis theorem to calculate moment of inertia.

When solving problems that involve rotational motion, the moment of inertia of each object is important. Calculating the moments of inertia of various objects, even symmetrical ones, can be a lengthy and tedious process. Although it is important to be able to calculate moments of inertia from the definition, I=∫mr2dm, in most cases it is useful simply to recall the moment of inertia of a particular type of object. The moments of inertia of common shapes (such as a uniform rod, a uniform or a hollow cylinder, a uniform or a hollow sphere) are well known and readily accessible in any mechanics textbook. However, in reality, an object has not one but an infinite number of moments of inertia. One of the distinctions between the moment of inertia and mass (the latter being the measure of translational inertia) is that a body’s moment of inertia depends on the axis of rotation. The moments of inertia found in textbooks are usually calculated with respect to an axis passing through the object’s center of mass. However, in many problems the axis of rotation does not pass through the center of mass. Does that mean that a lengthy process of finding the moment of inertia is necessary?

In many cases, the moment of inertia can be calculated rather easily using the parallel-axis theorem. Mathematically, it is expressed as I=IG+md2, where IG is the moment of inertia about an axis passing through the center of mass, m is the object’s total mass, and I is the moment of inertia about another axis that is parallel to the IG's axis and that is located a distance d from the center of mass.

As shown, a dumbbell of length 2L consists of two small spheres, each of mass m; the spheres are connected by a light rod. Note that, unless otherwise stated, all axes are perpendicular to the page’s plane.

 

 

Part A

Using the definition of the moment of inertia, calculate IG, the moment of inertia about the dumbbell’s center of mass.

Express your answer in terms of m and L.

IG =

2mL2

 

Part B

Using the definition of the moment of inertia, calculate IB, the dumbbell’s moment of inertia about an axis through point B, which coincides with the center of one of the spheres.

Express your answer in terms of m and L.

IB =

4mL2

 

Part C

Calculate IB for the dumbbell using the parallel-axis theorem.

Express your answer in terms of IG, m, and L.

IB =

IG+2mL2

 

Part D

Using the definition of the moment of inertia, calculate IC, the moment of inertia about an axis through point C, which is located a distance L from the center of mass.

Express your answer in terms of m and r.

IC =

4mL2

 

Part E

Calculate IC for the dumbbell using the parallel-axis theorem.

Express your answer in terms of IG, m, and L.

IC =

IG+2mL2

As shown, an irregularly shaped object has a mass m. Its moment of inertia measured with respect to axis A (parallel to the page's plane), which passes through the center of mass, is given by IA=0.64md2. Axes B, C, D, and E are parallel to axis A; the separation between adjacent axes is d, as shown in the diagram.

In subsequent questions, the subscript indicates the axis with respect to which the moment of inertia is measured; for example, IC is the moment of inertia about axis C .

Part F

Which moment of inertia is the smallest?

IA is correct

 

Part G

Which moment of inertia is the largest?

 

IE

Part H

Which moments of inertia if any are equal?

 

IB and IC

Part I

Which moment of inertia equals 4.64md2?

 

ID

Part J

Although not shown in the figure, axis X is parallel to the axes shown. It is known that IX=6md2. Which of the following is a possible location for axis X?

 

between axes D and E