Lecture 12
Moment of Inertia: Theorems
There are two key theorems that you will frequently need to use in order to calculate moments
of 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
y
v
mi
r
φ
ω
x
O
z
Recall that the moment of inertia about the z axis was calcuated as,
X
X
X
mi yi2
mi x2i +
Iz =
mi (x2i + yi2 ) =
i
i
(12.1)
i
P
but i mi x2i is just the moment of inertia, Iy about the y axis (since
Pthe lamina lies in the
xy plane and, thus, zi = 0 for all of the mass elements). Similarly, i mi yi2 is the moment
of inertia Ix about the x axis.
Therefore, we arrive at the Perpendicular Axis Theorem,
Iz = Ix + Iy
(12.2)
LECTURE 12. MOMENT OF INERTIA: THEOREMS
58
Example 12.1 Perpendicular Axis Theorem Applied to a Circular Disc Consider a thin circular disc lying in the xy plane. We showed last lecture that,
1
Iz = ma2
2
(12.3)
Now, from symmetry it is clear that Ix = Iy . Thus, from Equation 12.2, we conclude that:
1
Ix = Iy = ma2
4
12.2
(12.4)
Parallel-Axis Theorem (for Rigid Bodies)
• Consider the expression for the moment of inertia about the z axis:
X
Iz =
mi (x2i + yi2 )
i
• Let us transform to center-of-mass coordinates:
xi = xcm + x0i
yi = ycm + yi0
Such that Iz becomes,
´ X ¡
X ³
X
X
¢
0 2
0 2
2
Iz =
mi (xi ) + (yi ) +
mi x2cm + ycm
+ 2xcm
mi x0i + 2ycm
mi yi0
i
i
i
i
• From the definition of the center-of-mass, the last two terms must vanish. Therefore,
we are left with,
Ã
!
i
X h
X
£
¤
2
2
2
Iz =
mi (x0i ) + (yi0 ) +
mi x2cm + ycm
i
i
• The first term in the above equation is the moment of inertia about an axis parallel to
the 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 between
the center of mass and the z axis, which we will denote by l,
2
l2 = x2cm + ycm
• We these identifications, we arrive at the Parallel Axis Theorem:
I = Icm + M l2
which is valide for all rigid bodies (including 3D solid objects and 2D lamina).
(12.5)
LECTURE 12. MOMENT OF INERTIA: THEOREMS
59
y
m
i
r’i
ri
CM
rcm
O
x
z
y
y’
a
a
z’
O
x
z
Example 12.2 Parallel Axis Theorem Applied to a Circular Disc
• We first consider the moment of inertia about the z 0 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 = 12 ma2 .
• The distance from any axis which is perpendicular to the plane of the disc and adjacent
to the edge of the disc is simply the radius, here given as a. Thus, l = a.
• We conclude that,
1
3
I = ma2 + ma2 = ma2
2
2
(12.6)
• Referring back to the figure above, we may also consider the moment of inertia about
an axis which is in the plane of the disc and tangent to the edge of the disc (labelled as
the y 0 axis).
LECTURE 12. MOMENT OF INERTIA: THEOREMS
60
• The moment of inertia about the y axis was calculated earlier to be Iy = 14 ma2 . Thus,
since the distance from the edge to the y axis is also a, l = a and we find that,
5
1
I = ma2 + ma2 = ma2
4
4
¨
(12.7)