8/10/2019 Exemplo 9.5, Tipler e Mosca

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Try It Yourself

Calculating the Moment of Inertia S E C T I O N 9 - 3 | 297

THE PARALLEL-AXIS THEOREM

We can often simplify the calculation of moments of inertia for various objects byusing the parallel-axis theorem, which relates the moment of inertia about an axisthrough the center of mass to the moment of inertia about a second, parallel axis(Figure 9-10). Let I be the moment of inertia, and let be the moment of inertiaabout a parallel axis through the center of mass. In addition, let M be the total massof the object and let h be the distance between the two axes. The parallel-axistheorem states that

9-14PA R A L L E L- A X I S T H E O R E M

Example 9-2 and the Practice Problem following it illustrate a special case of thistheorem with and

Example 9-5Applying the Parallel-Axis Theorem

A thin uniform rod of mass M and length L on the x axis (Figure 9-11) has one end at the ori-gin. Using the parallel-axis theorem, find the moment of inertia about the axis, which isparallel to the y axis, and through the center of the rod.

PICTURE Here you know that about one end (see Example 9-4) and want to findUse the parallel-axis theorem with

SOLVE

Cover the column to the right and try these on your own before looking at the answers.

h 12 L.Icm .I 13 ML2

y

Icm 4ma2.h a,M 4m,

I Icm Mh2

Icm

h

cmcm

F I G U R E 9 - 1 1

y

x

L2

cm

y

Steps Answers

1. Apply the parallel-axis theorem to write I about the end interms of Icm . I y I y M(

12 L)2

I Icm

Mh 2

2. Substitute, using for for and solve for Icm .I y,I y, Icm13 ML2

112 ML2Icm I y Mh

2 13 ML2 M(

12 L)2

CHECK Calculate the moment of inertia by direct integration. This calculation is the sameas the calculation in Example 9-4 except that the integration limits are from to Theresult is

which is the same as the step-2 result.

TAKING IT FURTHER The step-2 result is only 25% of the result gotten in Example 9-4,where the uniform rod is rotated about an axis through one end.

PROOF OF THE PARALLEL-AXIS THEOREM

To prove the parallel-axis theorem, we start with an object (Figure 9-12) that is ro-tating about a fixed axis, one that does not pass through the center of mass. The ki-netic energy K of such an object is given by (Equation 9-12), where I is the mo-ment of inertia about the fixed axis. We saw in Chapter 8 (Equation 8-7) that the ki-netic energy of a system can be written as the sum of its translational kinetic en-ergy ( ) and the kinetic energy relative to the center of mass. For an objectthat is rotating, the kinetic energy relative to its center of mass is where I

cm

1

2I

cmv 2,

12 Mv 2cm

12 Iv 2

I x2 dm ML

L>2L>2

x2 dx ML

13

x3 ` L>2

L>2 M3L aL

3

8

L3

8 b 112 ML2

12 L.

12 L

*

F I G U R E 9 - 1 2

h

cmv cm

v cm = h

F I G U R E 9 - 1 0 An object rotating aboutan axis parallel to an axis through the centerof mass and a distance h from it.