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10-5 CALCULATING THE ROTATIONAL INERTIA 275

and thus must each be zero. Because x^2 y^2 is equal to R^2 , where Ris the dis-

tance from Otodm, the first integral is simply Icom, the rotational inertia of the

body about an axis through its center of mass. Inspection of Fig. 10-12 shows that

the last term in Eq. 10-37 is Mh^2 , where Mis the body’s total mass. Thus,

Eq. 10-37 reduces to Eq. 10-36, which is the relation that we set out to prove.

Checkpoint 5

The figure shows a book-like object (one side is

longer than the other) and four choices of rotation

axes, all perpendicular to the face of the object.

Rank the choices according to the rotational inertia

of the object about the axis, greatest first.

( 1 )( 2 )( 3 )( 4 )

left and Lfor the particle on the right. Now Eq. 10-33

gives us

Im(0)^2 mL^2 mL^2. (Answer)

Second technique:Because we already know Icomabout an

axis through the center of mass and because the axis here is

parallel to that “com axis,” we can apply the parallel-axis

theorem (Eq. 10-36). We find

mL^2. (Answer)

IIcomMh^2 ^12 mL^2 (2m)(^12 L)^2

Sample Problem 10.06 Rotational inertia of a two-particle system

Figure 10-13ashows a rigid body consisting of two particles of
massmconnected by a rod of length Land negligible mass.

(a) What is the rotational inertia Icomabout an axis through the
center of mass, perpendicular to the rod as shown?

KEY IDEA

Because we have only two particles with mass, we can find
the body’s rotational inertia Icomby using Eq. 10-33 rather
than by integration. That is, we find the rotational inertia of
each particle and then just add the results.

Calculations:For the two particles, each at perpendicular
distance from the rotation axis, we have

(Answer)

(b) What is the rotational inertia Iof the body about an axis
through the left end of the rod and parallel to the first axis
(Fig. 10-13b)?

KEY IDEAS

This situation is simple enough that we can find Iusing
either of two techniques. The first is similar to the one used
in part (a). The other, more powerful one is to apply the
parallel-axis theorem.

First technique:We calculate Ias in part (a), except here
the perpendicular distance riis zero for the particle on the

^12 mL^2.

Imiri^2 (m)(^12 L)^2 (m)(^12 L)^2

1

2 L

Additional examples, video, and practice available at WileyPLUS

m m

(a)

L L

com

Rotation axis

through

center of mass

m m

(b)

L

com

Rotation axis through

end of rod

__ 12 1 __ 2

Here the rotation axis is through the com.

Here it has been shifted from the com

without changing the orientation. We

can use the parallel-axis theorem.

Figure 10-13A rigid body consisting of two particles of mass m

joined by a rod of negligible mass.