 2000, W. E. Haisler
Mass Moment of Inertia
Mass Moment of Inertia Examples
The moment of inertia about the a-axis is given by
I
  r 2 dm   r 2  dV
a
n
n
where rn is the direction normal to the a-axis.
For the circular disk, we have
I
z
  r 2  rdrd   ( x 2  y 2 )  dxdy
In this case, the integral is much
easier to evaluate in polar
coordinates.
1
 2000, W. E. Haisler
2
Mass Moment of Inertia
Example: Consider a circular cylinder of diameter D, length L,
and mass m.
z
y
r
G
y L
z
dA=2rdr
dV=dAdz
x
x
D
Show that Iz (about an axis parallel to the bar's length through G)
is given by: Iz = mD2/8.
2
I

r
G z  dm
We can write dm   dV . In cylindrical coordinates, a volume
element is given by dV  dAdz  rdrd dz  rdr (2 ) L  2 Lrdr .
 2000, W. E. Haisler
3
Mass Moment of Inertia
The density is given by   m / V . The volume of the rod is
V  AL   ( D / 2)2 L . Thus the density is   m / V  4m .
 D2L
The perpendicular distance from the z-axis to any point of the
volume element is r. With this, the mass moment of inertia about
the z-axis becomes
G Iz

rn2
dm  
D/2 2
0

4
m
r 
2 Lrdr

  D2L 
D/2
4
D/2 3
8
m
8m r

r
dr


2
D 0
D2 4
0
 mD 2 / 8
(cylinder )
We can also show that G I y  m(3D 2 / 4  L2 ) /12
(cylinder ) .
 2000, W. E. Haisler
4
Mass Moment of Inertia
Long, slender rod of length L
Suppose the diameter D of the cylinder is very
small compared to its length L ( D L ), i.e., a
very slender rod of length L.
z
y L
G
Using the results for a cylinder and letting
D  0, we obtain:
x
B
G Iz
0
and G I y  mL /12 ( slender rod )
2
We could also derive G I y in the following way.
If the rod has a mass of m and length of L, then
the differential mass is given by dm=(m/L)dz.
Then
G Iy  
L/2
L / 2
z dm  
2
L/2
L / 2
z 2 (m / L)dz  mL2 /12
Long Slender Rod
dm
G
x
B
dz
z
y L
 2000, W. E. Haisler
5
Mass Moment of Inertia
Moment of inertia about one end of slender rod:
Suppose we wanted the mass moment of
inertia about a y-axis located at point B (the
end of the rod). The last integral becomes:
z
dm
L 2
L 2
z dm  z (m / L)dz
0
0
BIy  

 mL2 / 3
dz
G
L
z
y
Notice: The moment of inertia about some
x
B
point other then the center of mass will have a
large value. In this case, the moment of inertia about B ( B I y ) is
4 times larger then the value about the center of mass G ( G I y ).
An easier way to obtain this is by using the parallel axis theorem
discussed below.
 2000, W. E. Haisler
Mass Moment of Inertia
Suppose a mass of M is added to each end of the
slender rod.
For G I z , we assume that M is a point mass so
that the result is the same as for a long slender
rod, i.e., G I z  0 .
For G I y , we note that each lumped mass is
located a distance d=L/2 from G and has a
moment of inertia about G equal to
2
2
2
I

r
dm

d
M

(
L
/
2)
M . Since, moment of inertia is
G y  n
additive, we add the additional moment of inertia of the two
lumped masses to the previous result to obtain:
G
I
y
 mL2 /12  ( L / 2)2 M  ( L / 2) 2 M  mL2 /12  ML2 / 2
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 2000, W. E. Haisler
7
Mass Moment of Inertia
Parallel Axis Theorem
Consider a plane body with the x-y Cartesian coordinate system
located at the center of mass G as shown below. Consider
another Cartesian coordinate system (x', y') located at B such that
the x'-axis is parallel to the x-axis. The perpendicular distance
between the two parallel axes is "d." Assume we know the mass
moment of inertia about the x-axis ( G I x ) and wish to determine
the moment of inertia about the parallel x'-axis ( I x ' ).
y’
y
y’ = y - d
B
d
G
x’
x
 2000, W. E. Haisler
Mass Moment of Inertia
8
We first note that the moment of intertia about the x-axis through
the center of gravity is given by G I x   G rn2 dm   y 2dm .
Recall that rn is normal to the x-axis and is thus equal to y. We
wish to obtain the moment of inertia about the parallel x'-axis
through point B (i.e., about a parallel axis that is a distance d
from the center of gravity G). This is given by I   ( y ')2 dm .
x'
We note that y' = y - d so that I becomes
x'
I x '   ( y ')2dm   ( y  d ) 2dm
  y 2dm   2 y (d )dm   d 2dm
The first integral in the above equation is the moment of inertia
about the x-axis (through G): G I x   y 2dm . For the middle
integral, the constant 2d may be factored outside the integral so
 2000, W. E. Haisler
Mass Moment of Inertia
9
that it becomes 2d  ( y )dm  0 . This integral is zero since y is
measured from the center of mass. For the third integral, d is a
constant so the integral becomes md 2 . Thus we obtain
I x '  G I x  md 2
This is called the parallel axis theorem or the transfer theorem.
It allows one to transfer the moment of inertia about an axis
through the center of gravity (the x-axis) to a parallel axis (the xaxis) that is a distance "d" from the center of gravity.
The parallel axis theorem applies to any two parallel axes so long
as one of them passes through the center of mass.
 2000, W. E. Haisler
Mass Moment of Inertia
y
10
x
Two rigid slender bars and a
2 kg
disk
O
disk are used to make the rigid
1 kg
3 kg
structure pinned at O. The
2m
horizontal bar has a length of 3
2m
1m
m and a mass of 1 kg; the
0.5 m
vertical bar has a length of 2 m
and a mass of 2 kg. The disk has a diameter of 0.5 m and a mass of 3 kg.
Determine the mass moment of inertia about a z-axis through point O.
2
2
2
I

mR
/
2

3
kg
(0.5
m
)
/
2

0.375
kgm
G z
2
2
I

mR
/
2

md
thus, O z
 3kg (0.5m) 2 / 2  3kg (2.25m) 2  15.5625 kgm 2
Horizontal Rod:
2
2
2
2
2
I

mL
/12

md

1
kg
(3
m
)
/12

1
kg
(0.5
m
)

1
kgm
O z
Vertical Rod:
2
2
2
2
2
I

mL
/12

md

2
kg
(2
m
)
/12

2
kg
(1
m
)

2.667
kgm
O z
Total: Summing the three components gives: O I z  19.229 kgm2
Disk: