1. 10.9 Mass Moment of Inertia
Mass moment of inertia of a body is the property
that measures the resistance of the body to
angular acceleration
Mass moment of inertia is defined as
the integral of the second moment
about an axis of all the elements of
mass dm which compose the body
Example
Consider rigid body

2. 10.9 Mass Moment of Inertia
For body’s moment of inertia about the z axis,
I = ∫ r 2 dm
m
Here, the moment arm r is the perpendicular
distance from the axis to the arbitrary element
dm
Since the formulation involves r, the value of I is
unique for each axis z about which it is
computed
The axis that is generally chosen for analysis,
passes through the body’s mass center G

3. 10.9 Mass Moment of Inertia
Moment of inertia computed about this axis
will be defined as IG
Mass moment of inertia is always positive
If the body consists of material having a
variable density ρ = ρ(x, y, z), the element
mass dm of the body may be expressed as
dm = ρ dV
Using volume element for integration,
I = ∫ r 2 ρdV
V

4. 10.9 Mass Moment of Inertia
In the special case of ρ being a constant,
I = ρ ∫ r 2 dV
V
When element volume chosen for integration has
differential sizes in all 3 directions, dV = dx dy dz
Moment of inertia of the body determined by
triple integration
Simplify the process to single integration
by choosing an element volume with
a differential size or thickness in 1
direction such as shell or disk elements

5. 10.9 Mass Moment of Inertia
Procedure for Analysis
Consider only symmetric bodies having surfaces
which are generated by revolving a curve about
an axis
Shell Element
For a shell element having height z, radius y and
thickness dy, volume dV = (2πy)(z)dy

6. 10.9 Mass Moment of Inertia
Procedure for Analysis
Shell Element
Use this element to determine the moment of
inertia Iz of the body about the z axis since the
entire element, due to its thinness, lies at the
same perpendicular distance r = y from the z
axis
Disk Element
For disk element having radius y, thickness dz,
volume dV = (πy2) dz

7. 10.9 Mass Moment of Inertia
Procedure for Analysis
Disk Element
Element is finite in the radial direction and
consequently, its parts do not lie at the same
radial distance r from the z axis
To perform integration using this element,
determine the moment of inertia of the
element about the z axis and then integrate
this result

8. 10.9 Mass Moment of Inertia
Example 10.11
Determine the mass moment of inertia of the
cylinder about the z axis. The density of the
material is constant.

9. 10.9 Mass Moment of Inertia
Solution
Shell Element
For volume of the element,
dV = (2πr )(h )dr
For mass,
dm = ρdV = ρ (2πrh dr )
Since the entire element lies at the
same distance r from the z axis, for
the moment of inertia of the element,
dI z = r 2 dm = ρ 2πhr 3

10. 10.9 Mass Moment of Inertia
Solution
Integrating over entire region of the cylinder,
R ρπ 4
I z = ∫ r dm = ρ 2πh ∫ r dr =
2 3
R h
m 0 2
For the mass of the cylinder
R
m = ∫ dm = ρ 2πh ∫ rdr = ρπhR 2
m 0
So that
1
I z = mR 2
2

11. 10.9 Mass Moment of Inertia
Example 10.12
A solid is formed by revolving the shaded area
about the y axis. If the density of the material is
5 Mg/m3, determine the mass moment of inertia
about the y axis.

12. 10.9 Mass Moment of Inertia
Solution
Disk Element
Element intersects the curve at the arbitrary point (x,
y) and has a mass
dm = ρ dV = ρ (πx2)dy
Although all portions of the
element are not located at the
same distance from the y axis,
it is still possible to determine
the moment of inertia dIy about
the y axis

13. 10.9 Mass Moment of Inertia
Solution
In the previous example, it is shown that the
moment of inertia for a cylinder is
I = ½ mR2
Since the height of the cylinder is not
involved, apply the about equation for a disk
dI y =
1
2
1
[( ) ]
(dm) x 2 = ρ πx 2 dy x 2
2
For moment of inertia for the entire solid,
5π 1 5π 1
Iy =
2 ∫
0
x 4 dy =
2 ∫
0
y 8 dy = 0.873Mg.m 2 = 873kg.m 2

14. 10.9 Mass Moment of Inertia
Parallel Axis Theorem
If the moment of inertia of the body about
an axis passing through the body’s mass
center is known, the moment of inertia about
any other parallel axis may be determined by
using parallel axis theorem
Considering the body where the z’ axis
passes through the mass center G, whereas
the corresponding parallel z axis lie at a
constant distance d away

15. 10.9 Mass Moment of Inertia
Parallel Axis Theorem
Selecting the differential mass element dm,
which is located at point (x’, y’) and using
Pythagorean theorem,
r 2 = (d + x’)2 + y’2
For moment of inertia of body about the z axis,
m m
[ 2
]
I = ∫ r 2 dm = ∫ (d + x') + y '2 dm
m
( 2 2
)
= ∫ x' + y ' dm + 2d ∫ x' dm + d
m
2
∫ dm
m
First integral represent IG

16. 10.9 Mass Moment of Inertia
Parallel Axis Theorem
Second integral = 0 since the z’ axis passes
through the body’s center of mass
Third integral represents the total mass m of
the body
For moment of inertia about the z axis,
I = IG + md2

17. 10.9 Mass Moment of Inertia
Radius of Gyration
For moment of inertia expressed using k,
radius of gyration,
I
I = mk 2
or k=
m
Note the similarity between the definition of
k in this formulae and r in the equation dI =
r2 dm which defines the moment of inertia of
an elemental mass dm of the body about an
axis

18. 10.9 Mass Moment of Inertia
Composite Bodies
If a body is constructed from a number of
simple shapes such as disks, spheres, and
rods, the moment of inertia of the body
about any axis z can be determined by
adding algebraically the moments of inertia
of all the composite shapes computed about
the z axis
Parallel axis theorem is needed if the center
of mass of each composite part does not lie
on the z axis

19. 10.9 Mass Moment of Inertia
Example 10.13
If the plate has a density of 8000kg/m3 and a
thickness of 10mm, determine its mass moment of
inertia about an axis perpendicular to the page and
passing through point O.

20. 10.9 Mass Moment of Inertia
Solution
The plate consists of 2 composite parts, the
250mm radius disk minus the 125mm
radius disk
Moment of inertia about O is determined by
computing the moment of inertia of each of
these parts about O and then algebraically
adding the results

21. 10.9 Mass Moment of Inertia
Solution
Disk
For moment of inertia of a disk about an axis
perpendicular to the plane of the disk,
1 2
I G = mr
2
Mass center of the disk is located 0.25m from
point O md = ρ dVd = 8000[π (0.25)2 (0.01)] = 15.71kg
(I O )d = 1 md rd2 + md d 2
2
1
= (15.71)(0.25)2 + (15.71)(0.25)2 = 1.473kg.m 2
2

22. 10.9 Mass Moment of Inertia
Solution
Hole
[ ]
mh = ρ hVh = 8000 π (0.125) (0.01) = 3.93kg
2
1
(I O )h = mh rh2 + mh d 2
2
1
= (3.93)(0.125)2 + (3.93)(0.25)2 = 0.276kg.m 2
2
For moment of inertia of plate about point O,
I O = (I O )d − (I O )h
= 1.473 − 0.276 = 1.20kg.m 2

23. 10.9 Mass Moment of Inertia
Example 10.14
The pendulum consists of two
thin robs each having a mass of
100kg. Determine the
pendulum’s mass moment of
inertia about an axis passing
through (a) the pin at point O,
and (b) the mass center G of the
pendulum.

24. 10.9 Mass Moment of Inertia
Solution
Part (a)
For moment of inertia of rod OA about an axis
perpendicular to the page and passing through the end
point O of the rob,
1
IO = ml 2
3
Hence,
1 1
(IOA )O = ml 2 = (100)(3)2 = 300kg.m2
3 3
Using parallel axis theorem,
1 2
IG = ml
12
(IOA )O = 1 ml 2 + md 2 = 1 (100)(3)2 + (100)(1.5)2 = 300kg.m2
12 12

25. 10.9 Mass Moment of Inertia
Solution
For rod BC,
1 1
(I BC )O = ml 2 + md 2 = (100)(3) + (100)(3)
2 2
12 12
= 975kg.m 2
For moment of inertia of pendulum about O,
I O = 300 + 975 = 1275kg.m 2

26. 10.9 Mass Moment of Inertia
Solution
Part (b)
Mass center G will be located relative to pin at O
For mass center,
∑ ~m 1.5m(100kg ) + 3m(100kg )
y
y= = = 2.25m
∑m 100kg + 100kg
Mass of inertia IG may be computed in the same
manner as IO, which requires successive applications
of the parallel axis theorem in order to transfer the
moments of inertias of rod OA and BC to G

27. 10.9 Mass Moment of Inertia
Solution
Apply the parallel axis theorem for IO,
I O = I G + md 2 ;
125kg.m 2 = I G + (200)(2.25)
2
I G = 262.5kg.m 2