Terminology
Moment of Inertia
ME 202
•
Moment of inertia (MOI) = second
mass moment
•
Instead of multiplying mass by
distance to the first power (which
gives the first mass moment), we
multiply it by distance to the
second power.
2
Definitions
Moment of inertia of a body with mass, m, about
the x axis:
(
)
I x = ∫ y 2 + z 2 dm
Transfer Theorem - 1
•
We can “transfer” the moment of
inertia from one axis to another,
provided that the two axes are
parallel.
•
In other words, if we know the
moment of inertia about one axis,
we can compute it about any
other axis parallel to the first axis.
m
Moment of inertia of a body with mass, m, about
the y axis:
(
)
I y = ∫ x 2 + z 2 dm
m
Moment of inertia of a body with mass, m, about
the z axis:
(
)
I z = ∫ x 2 + y 2 dm
m
3
4
Transfer Theorem - 2
If the moment of inertia of a body with
mass m about an axis x’ through the
mass center is I Gx ′ , and the
perpendicular distance from the x’ axis
to the (parallel) axis x is d, then the
moment of inertia of the body about
the x axis is
Transfer Theorem - 3
•
The moment of inertia to which the
transfer term is added is always the
one for an axis through the mass
center.
•
The moment of inertia about an
axis through the mass center is
smaller than the moment of inertia
about any other parallel axis.
2
I x = I Gx′ + md
!
transfer term
5
6
Transfer Theorem - 4
Radius of Gyration
We can transfer from any axis to a
parallel axis through the mass
center by subtracting the transfer
term.
I Gx′ = I x − md 2
7
•
The radius of gyration about the x axis is
denoted by k x .
(
)
I x = ∫ y 2 + z 2 dm = kx2 m
m
•
By definition, the radius of gyration of a
mass m about the x axis is
kx =
8
Ix
m
Find IA (about the axis through
A that is normal to the plane of
the figure).
Into how many parts should we
divide the structure?
Composite Masses
Since the moment of inertia is an
integral, and since the integral over a
sum of several masses equals the sum of
the integrals over the individual masses,
we can find the moment inertia of a
composite mass by adding the
moments of inertia of its parts. (Regions
with no mass can be subtracted.)
Use two bars, each with length
γ = 3 lb / ft
For a straight,
slender, uniform
bar:
L = 3 ft
2
IA =
1
1
⎛ L⎞
⎛ 2L ⎞
mL2 + m ⎜ ⎟ + mL2 + m ⎜
⎝
⎠
⎝ 3 ⎟⎠
12
2
12
2
⎛ 1 1 1 4 ⎞ γ L 2 3 + 9 + 3 + 16
= mL2 ⎜ + + + ⎟ =
L
⎝ 12 4 12 9 ⎠
g
36
1
I G = mL2
12
31 γ L3 31 ( 3 lb / ft ) ( 3 ft )
=
= 2.17 slug ⋅ ft 2
36 g
36 32.2 ft / sec 2
3
IA =
I A = I G + md 2
9
Find Iz.
10
ρ = 200 kg / m 3
Into how many parts should we
divide the structure?
For a right, uniform,
circular cone:
ρ = 200 kg / m 3
Is the missing peak the same size
as the hole?
For a right, uniform,
circular cone:
IG =
m = ρV
3
mr 2
10
1
V = π r 2h
3
hpeak
1000
⎛ 1⎞
=
⇒ hpeak = ⎜ ⎟ 1000 mm
⎝ 3⎠
200 800 − 200
Use three parts: large, full cone;
small missing peak of large cone;
hole.
I z = I large − I peak − I hole
3
3
3
I z = mlarge R 2 − mpeak r 2 − mhole r 2
10
10
10
11
3
mr 2
10
1
V = π r 2h
3
IG =
m = ρV
3
3
3
mlarge R 2 − mpeak r 2 − mhole r 2
10
10
10
3
I z = ρ Vlarge R 2 − Vpeak r 2 − Vhole r 2
10
π
= ρ R 4 hlarge − r 4 hpeak − r 4 hhole
10
π
= ρ ⎡⎣ R 4 hlarge − r 4 hpeak + hhole ⎤⎦
10
Iz =
(
)
(
)
(
12
)
ρ = 200 kg / m 3
π ⎡ 4
ρ ⎣ R hlarge − r 4 hpeak + hhole ⎤⎦
10
⎧
π
⎤⎫
4⎛ 4
4 ⎡⎛ 1
⎞
⎞
= ( 200 kg / m 3 ) ⎨( 0.8 m ) ⎜ m ⎟ − ( 0.2 m ) ⎢⎜ m ⎟ + 0.2 m ⎥ ⎬
⎝
⎠
⎝
⎠
10
3
⎣ 3
⎦⎭
⎩
Iz =
(
)
I z = 34.3 kg ⋅ m 2
13
Rod: 3 kg/m
Plate with hole: 12 kg/m2
Find IG (about an axis normal
to the plane of the figure).
First, find G.
Second, compute the MOI.
Into how many parts should we divide the
structure?
Three pieces: two rods and one plate.
14
Moment of Inertia
2
Although the FMM is a vector, the MOI is not. Mass and the square of a distance are both scalars. The dimensions of the MOI are mass times the square of length. Typical units are slug times foot squared or kilogram times meter squared.
3
In these integrals, each element of mass, dm, is multiplied by the square of its distance from the axis about
which the MOI is computed. Results of these integrals depend on the shape of the body and the distribution
of the mass within it. For a body with what is called “spherical symmetry,” all three of the MOI are equal. A
uniform sphere and a uniform cube are both among the bodies that have spherical symmetry.
4
Computing the MOI using the integrals on the previous slide can be tedious. But many books contain tables
showing the MOI for various three-dimensional bodies. The MOI are typically given about three perpendicular axes that pass through the bodies’ mass centers. The transfer theorem (aka the “parallel axis” theorem)
makes it easy to use the result from a table to compute the MOI about an axis that does not pass through a
body’s mass center but is parallel to an axis for which the MOI is given in the table.
5
You should memorize this equation. And you should understand that it can be used between any two parallel
axes, provided that one of them passes through the mass center of the body in question.
Page 1 of 4
6
You should memorize the fact that the MOI about a given axis through the mass center is the smallest MOI
that can be computed for any axis parallel to the given axis.
7
This is simply a rearrangement of the equation on slide 5, showing how to use the transfer theorem to compute the MOI about an axis through the mass center when we already know the MOI about a parallel axis
that does not pass through the mass center.
8
The position of a body’s mass center is where we would find all of the body’s mass if we considered the body
to be a particle. In a similar way, the radius of gyration is where we would find all of the body’s mass if we
considered the body to be an infinitely thin cylinder with radius k. For the MOI about the x axis, the cylinder
would be parallel to and centered on the x axis. Since all of the cylinder’s mass would be at the same distance from the axis, the squared distance in the first integral on slide 3 is the radius of gyration and can be
taken outside the integral. That leads to the final equation on this slide. If we have any two of the quantities
mass, radius of gyration and MOI, we can compute the third using this definition.
9
Because a MOI is an integral, the techniques used to locate a body’s mass center or to compute the FMM for
a body made up of several simple shapes can also be applied to compute the body’s MOI. The MOI of a hole
can be subtracted. But adding and subtracting are correct only when all of the MOI of the body’s pieces are
computed about the same axis.
Page 2 of 4
10 We cannot account for the exact shape of the region where the rods intersect, because we have no information about that.
Two equations are needed for this calculation. One is the MOI for a thin, uniform rod about an axis perpendicular to the rod and passing through the rod’s mass center. The other equation is the parallel axis theorem.
Each rod’s MOI is computed using the two equations, and the two MOI are added. Note that the mass of
each rod is its weight per unit length divided by g and multiplied by L. In the final equation, we must expand
the lb unit to slug-ft/sec2 before simplifying.
11 A simple calculation shows that the conical hole is not the same size as the missing peak of the larger cone.
Formulas for a right, circular cone can be found in books or at web sites. In the final equation, the radius of
the base of the large cone is denoted by R, and the radius of the bases of the missing peak and hole is denoted by r. It remains to substitute for the masses and volumes, simplify and compute. These operations are
shown on the next two slides.
12 The final equation shows that each term is proportional to the fourth power of a radius.
13 This is the numerical result for the final equation from the previous slide.
Page 3 of 4
14 For many composite shapes, it is necessary to find the mass center before computing the MOI. If the mass
center’s location were known, it would be a simple matter to transfer the MOI for each piece of the body to
the mass center. But in transferring the MOI for each piece to the body’s overall mass center, the transfer distance depends on the location of the mass center. Because the location is not given, we must first find it.
That requires that we compute the FMM of each piece relative to a common point, add the FMM and divide
by the total mass.
Page 4 of 4