6161103 10.9 mass moment of inertia

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


� 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


� 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


� 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


Shell Element� For a shell element having height z, radius y and

thickness dy, volume dV = (2πy)(z)dy


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


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


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.


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

==

==


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


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.


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


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


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



� 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''

''



� 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


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 ==


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


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.


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


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


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

=−=

−=

=+=

+=


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.


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

=+=+=

=

===

=


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 =+=

=


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~=

+

+=

∑

∑=


Solution

� Apply the parallel axis theorem for IO,

2;mdII GO +=

( )( )2

22

.5.262

25.2200.125

;

mkgI

Imkg

mdII

G

G

GO

=

+=

+=

6161103 10.9 mass moment of inertia

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