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Lecture 12

Moment of Inertia: Theorems

There are two key theorems that you will frequently need to use in order to calculate momentsof inertia: The Parallel and Perpendicular Axis Theorems:

12.1 Perpendicular Axis Theorem (For a Plane Lamina)

We consider our plane lamina focusing on the mass element mi

O

v

m

r

x

y

z

i

Recall that the moment of inertia about the z axis was calcuated as,

Iz =i

mi(x2

i + y2

i ) =i

mix2

i +i

miy2

i (12.1)

but

imix2

i is just the moment of inertia, Iy about the y axis (since the lamina lies in thexy plane and, thus, zi = 0 for all of the mass elements). Similarly,

imiy

2

i is the momentof inertia Ix about the x axis.Therefore, we arrive at the Perpendicular Axis Theorem,

Iz = Ix + Iy (12.2)

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LECTURE 12. MOMENT OF INERTIA: THEOREMS 58

Example 12.1 Perpendicular Axis Theorem Applied to a Circular Disc Consider a thin cir-

cular disc lying in the xy plane. We showed last lecture that,

Iz =1

2ma2 (12.3)

Now, from symmetry it is clear that Ix = Iy. Thus, from Equation 12.2, we conclude that:

Ix = Iy =1

4ma2 (12.4)

12.2 Parallel-Axis Theorem (for Rigid Bodies)

Consider the expression for the moment of inertia about the z axis:

Iz =i

mi(x2

i + y2

i )

Let us transform to center-of-mass coordinates:

xi = xcm + x

i

yi = ycm + y

i

Such that Iz becomes,

Iz =imi

(x

i)2

+ (y

i)2

+imix2

cm + y2

cm

+ 2xcmimix

i + 2ycmimiy

i

From the definition of the center-of-mass, the last two terms must vanish. Therefore,we are left with,

Iz =i

mi

(xi)

2+ (yi)

2

+

i

mi

x2cm

+ y2cm

The first term in the above equation is the moment of inertia about an axis parallel tothe z axis and passing through the center-of-mass, call it Icm.

The second term is just the total mass multiplied by the square of the distance betweenthe center of mass and the z axis, which we will denote by l,

l2 = x2cm

+ y2cm

We these identifications, we arrive at the Parallel Axis Theorem:

I = Icm + Ml2 (12.5)

which is valide for all rigidbodies (including 3D solid objects and 2D lamina).

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LECTURE 12. MOMENT OF INERTIA: THEOREMS 59

ir

cmr

ir

x

y

z

O

CM

mi

x

y

z

z

aa

O

y

Example 12.2 Parallel Axis Theorem Applied to a Circular Disc

We first consider the moment of inertia about the z axis in the above figure (that is,an axis which is parallel to the z axis and perpendicular to the plane of the disc.

Since the z axis passes through the center of mass, we have Icm = 12ma2.

The distance from any axis which is perpendicular to the plane of the disc and adjacentto the edge of the disc is simply the radius, here given as a. Thus, l = a.

We conclude that,

I=1

2ma2 + ma2 =

3

2ma2 (12.6)

Referring back to the figure above, we may also consider the moment of inertia aboutan axis which is in the plane of the disc and tangent to the edge of the disc (labelled asthey axis).

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LECTURE 12. MOMENT OF INERTIA: THEOREMS 60

The moment of inertia about the y axis was calculated earlier to be Iy =1

4ma2. Thus,

since the distance from the edge to they

axis is alsoa

,l

=a

and we find that,

I=1

4ma2 + ma2 =

5

4ma2 (12.7)