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Mass Moment of Inertia

Engineering Mechanics- Distributed Forces

Encounter in the engineering problems involving rotational motion of rigid bodies

in dynamics

Moment of Inertia of a Mass

Angular acceleration about the axisAA of the

small mass Dm due to the application of a

couple is related to r2Dm.

r2Dm = moment of inertiaof the

massDm with respect to the

axisAA

For a body of mass m the resistance to rotation

about the axisAA is

inertiaofmomentmassdmr

mrmrmrI

2

23

22

21

The radius of gyration for a concentrated mass

with equivalent mass moment of inertia is

m

IkmkI 2

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Moment of Inertia of a Mass

Moment of inertia with respect to they

coordinate axis is

dmxzdmrIy 222

Similarly, for the moment of inertia with

respect to thex andz axes,

dmyxI

dmzyI

z

x

22

22

Engineering Mechanics- Distributed Forces

Parallel Axis Theorem For the rectangular axes with origin at O and parallel

centroidal axes,

dmzydmzzdmyydmzy

dmzzyydmzyIx2222

2222

22

22 zymII xx

2222

yxmII

xzmII

zz

yy

Generalizing for any axisAAand a parallel centroidal

axis,

2mdII

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Moments of Inertia of Thin Plates For a thin plate of uniform thickness tand homogeneous

material of density r, the mass moment of inertia with

respect to axisAA contained in the plate is

IA A r2dm t r2 dA

t IA A ,area

Similarly, for perpendicular axisBB which is also

contained in the plate,

IB B t IB B ,area

For the axis CC which is perpendicular to the plate,

IC C t JC,area t IA A ,areaIB B ,area IA A IB B

Engineering Mechanics- Distributed Forces

Moments of Inertia of Thin Plates

For the principal centroidal axes on a rectangular plate,

21213

121 mabatItI area,AAAA

21213

121 mbabtItI area,BBBB

22121 bamIII mass,BBmass,AACC

For centroidal axes on a circular plate,

2414

41 mrrtItII area,AABBAA

2

21 mrIII BBAACC

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Mass Moment of Inertia of a Rectangular Prism

Engineering Mechanics- Distributed Forces

Find mass moment of

inertia about Z axis

Engineering Mechanics- Distributed Forces

Mass Moment of Inertia of a Sphere

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Moments of Inertia of Common Geometric Shapes

Engineering Mechanics- Distributed Forces

Sample Problem: Composite Section

Determine the moments of inertia

of the steel forging with respect to

thexyz coordinate axes, knowing

that the specific weight of steel is

7896 kg/m3.

SOLUTION:

With the forging divided into a prism and

two cylinders, compute the mass and

moments of inertia of each component

with respect to thexyz axes using the

parallel axis theorem.

Add the moments of inertia from the

components to determine the total moments

of inertia for the forging.

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Sample Problem: Composite Section

kg0744.0

1031

kg/m7896

:cylindereach

362

3

m

m

Vm

2

222

121

222

121

kgm5394.0

5.20744.03130744.0

3

xmLamIy

2 2 2 2112

2 2 2 2112

3

0 0744 3 1 3 0 0744 2 5 2

0 837 2kgm

zI m a L m x y

. . .

.

2

22

21

22

21

kgm5394.0

20744.010744.0

ymmaIx

cylinders :cm2cm,5.2,cm3,cm1 yxLaSOLUTION: Compute the moments of inertia

of each component with respect

to thexyz axes.

Engineering Mechanics- Distributed Forces

Sample Problem: Composite Section

kg1895.0

m10622kg/m7896

:prism

363

m

Vm

prism (a = 2 cm, b = 6 cm, c = 2 cm):

2

22

12122

121

gmk6320

2618950

.

.cbmII zx

2

22

12122

121

gmk1260

2218950

.

.acmIy

Add the moments of inertia from the componentsto determine the total moments of inertia.

372.02632.0 xI2

kgm376.1xI

5394.02126.0 yI

2kgm2048.1yI

837.02632.0 zI2

kgm306.2zI

Engineering Mechanics- Distributed Forces