6161103 10.9 mass moment of inertia
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10.9 Mass Moment of Inertia10.9 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 � 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
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
� For body’s moment of inertia about the z axis,
� Here, the moment arm r is the perpendicular distance from the axis to the arbitrary element
∫= mdmrI 2
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
10.9 Mass Moment of Inertia10.9 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 � 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,
∫= VdVrI ρ2
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
� In the special case of ρ being a constant,
� When element volume chosen for integration has differential sizes in all 3 directions, dV = dx dy dz
∫=V
dVrI 2ρ
� 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
Procedure for Analysis� Consider only symmetric bodies having surfaces
which are generated by revolving a curve about an axis
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Shell Element� For a shell element having height z, radius y and
thickness dy, volume dV = (2πy)(z)dy
10.9 Mass Moment of Inertia10.9 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 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
10.9 Mass Moment of Inertia10.9 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 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
10.9 Mass Moment of Inertia10.9 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.material is constant.
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Solution
Shell Element
� For volume of the element,
( )( )2 drhrdV π=
� For mass,
� Since the entire element lies at the
same distance r from the z axis, for
the moment of inertia of the element,
( )( )
( )
32 2
2
hrdmrdI
drrhdVdm
z πρ
πρρ
==
==
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Solution
� Integrating over entire region of the cylinder,
4
0
32
22 hRdrrhdmrI
R
mz === ∫∫ρπ
πρ
� For the mass of the cylinder
� So that
2
2
0
0
2
1
2
22
mRI
hRrdrhdmm
hRdrrhdmrI
z
R
m
mz
=
===
===
∫∫
∫∫
ρππρ
πρ
10.9 Mass Moment of Inertia10.9 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 5 Mg/m3, determine the mass moment of inertia
about the y axis.
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
SolutionDisk Element� Element intersects the curve at the arbitrary point (x,
y) and has a mass dm = ρ dV = ρ (πx2)dydm = 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
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Solution
� In the previous example, it is shown that the moment of inertia for a cylinder is
I = ½ mR2I = ½ mR
� Since the height of the cylinder is not involved, apply the about equation for a disk
� For moment of inertia for the entire solid,
( )[ ]
∫∫ ====
==
1
0
2281
0
4
222
.873.873.02
5
2
5
2
1)(
2
1
mkgmMgdyydyxI
xdyxxdmdI
y
y
ππ
πρ
10.9 Mass Moment of Inertia10.9 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 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
10.9 Mass Moment of Inertia10.9 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,
� First integral represent IG
( )[ ]( ) ∫∫∫
∫∫+++=
++==
mmm
mm
dmddmxddmyx
dmyxddmrI
222
222
'2''
''
10.9 Mass Moment of Inertia10.9 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 � Third integral represents the total mass m of the body
� For moment of inertia about the z axis,
I = IG + md2
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Radius of Gyration
� For moment of inertia expressed using k, radius of gyration,
IkormkI == 2
� 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
mkormkI ==
10.9 Mass Moment of Inertia10.9 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 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
10.9 Mass Moment of Inertia10.9 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 inertia about an axis perpendicular to the page and
passing through point O.
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Solution
� The plate consists of 2 composite parts, the 250mm radius disk minus the 125mm radius diskradius 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
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Solution
Disk
� For moment of inertia of a disk about an axis perpendicular to the plane of the disk, perpendicular to the plane of the disk,
� Mass center of the disk is located 0.25m from point O ( ) ( )[ ]
( )
( )( ) ( )( ) 222
22
2
2
.473.125.071.1525.071.152
12
1
71.1501.025.08000
2
1
mkg
dmrmI
kgVm
mrI
ddddO
ddd
G
=+=
+=
===
=
πρ
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Solution
Hole
( ) ( )[ ]( )
2
1
93.301.0125.08000 kgVm hhh === πρ
� For moment of inertia of plate about point O,
( )
( )( ) ( )( )
( ) ( )2
222
22
.20.1276.0473.1
.276.025.093.3125.093.32
12
1
mkg
III
mkg
dmrmI
hOdOO
hhhhO
=−=
−=
=+=
+=
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Example 10.14
The pendulum consists of two
thin robs each having a mass of
100kg. Determine the 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.
10.9 Mass Moment of Inertia10.9 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,point O of the rob,
� Hence,
� Using parallel axis theorem,
( ) ( )( )
( ) ( )( ) ( )( ) 22222
2
222
2
.3005.11003100121
121
121
.300310031
31
31
mkgmdmlI
mlI
mkgmlI
mlI
OOA
G
OOA
O
=+=+=
=
===
=
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Solution
� For rod BC,
( ) ( )( ) ( )( )2222 3100310012
1
12
1mdmlI OBC +=+=
� For moment of inertia of pendulum about O,2
2
.1275975300
.9751212
mkgI
mkg
O =+=
=
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Solution
Part (b)
� Mass center G will be located relative to pin at O
� For mass center, � For mass center,
� 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
mkgkg
kgmkgm
m
myy 25.2
100100
)100(3)100(5.1~=
+
+=
∑
∑=
10.9 Mass Moment of Inertia10.9 Mass Moment of Inertia
Solution
� Apply the parallel axis theorem for IO,
2;mdII GO +=
( )( )2
22
.5.262
25.2200.125
;
mkgI
Imkg
mdII
G
G
GO
=
+=
+=