9781118230725.pdf

274 CHAPTER 10 ROTATION

Table 10-2 Some Rotational Inertias

Axis

Hoop about

central axis

Axis

Annular cylinder

(or ring) about

central axis

R

I=MR^2 (a) I= M(R 12 +R^22 ) (b)

R 2

R 1

Thin rod about

axis through center

perpendicular to

length

I= ML^2 (e)

L

Axis

Axis Axis

Hoop about any

diameter

Slab about

perpendicular

axis through

center

I= MR^2 (h) I= M(a^2 +b^2 ) (i)

R

b

a

Axis

Solid cylinder

(or disk) about

central axis

I= MR^2 (c)

R

L

Axis

Solid cylinder

(or disk) about

central diameter

I= MR^2 + ML^2 (d)

R

L

Axis

Thin

spherical shell

about any

diameter

I= MR^2 (g)

2 R

Solid sphere

about any

diameter

I= MR^2 (f)

2 R

Axis

1 __ 2 __ 1

2

14 (^25)
23 (^12)
121
121
__ 121
Figure 10-12A rigid body in cross section,
with its center of mass at O. The parallel-
axis theorem (Eq. 10-36) relates the
rotational inertia of the body about an axis
throughOto that about a parallel axis
through a point such as P, a distance h
from the body’s center of mass.
dm
r
P
h
a
b
x– a
y– b
com
O
Rotation axis
through
center of mass
Rotation axis
throughP
y
x
We need to relate the rotational inertia
around the axis at P to that around the
axis at the com.
through the center of mass (remember these two axes must be parallel). Then the
rotational inertia Iabout the given axis is
IIcomMh^2 (parallel-axis theorem). (10-36)
Think of the distance has being the distance we have shifted the rotation axis
from being through the com. This equation is known as the parallel-axis theorem.
We shall now prove it.
Proof of the Parallel-Axis Theorem
LetObe the center of mass of the arbitrarily shaped body shown in cross section
in Fig. 10-12. Place the origin of the coordinates at O.Consider an axis through O
perpendicular to the plane of the figure, and another axis through point Pparal-
lel to the first axis. Let the xandycoordinates of Pbeaandb.
Letdmbe a mass element with the general coordinates xandy.The rota-
tional inertia of the body about the axis through Pis then, from Eq. 10-35,
which we can rearrange as
(10-37)
From the definition of the center of mass (Eq. 9-9), the middle two integrals of
Eq. 10-37 give the coordinates of the center of mass (multiplied by a constant)
I (x^2 y^2 )dm 2 axdm 2 bydm(a^2 b^2 )dm.
Ir^2 dm[(xa)^2 (yb)^2 ]dm,