Lecture on rigid dynamics Kinetics of a Rigid Body: Force and Acceleration Objective • Moment of Inertia of a body • Parallel Axis Theorem • Radius of Gyration • Moment of

Lecture on

rigid dynamics

Sri.S. N. Mishra

RIGID DYNAMICS

Planar Kinetics of a Rigid Body:

Force and Acceleration

Objective

• Moment of Inertia of a body

• Parallel Axis Theorem

• Radius of Gyration

• Moment of Inertia of Composite Bodies

Moment and Angular Acceleration

• When M 0, rigid body experiences

angular acceleration

• Relation between M and a is analogous to

relation between F and a

aIma M,F

Moment of Inertia

Mass = Resistance

Moment of Inertia

• This mass analog is called the

moment of inertia, I, of the object

– r = moment arm

– SI units are kg m2

m

dmrI 2

dzdydxrI 2

dVrI

dVdm

2

:density volume theis where, Using

Shell Element

Disk Element

dyzydV )2(

dzydV )( 2

Example 17-1

)2( hdrrdVdm

)(2

1

22 224

0

32 hRRhRdrrhdmrI

R

m

hRm 2

2

2

1RmI z

RodThin

12

1 2MLI

L

end)at (axis RodThin

3

1 2MLI

L

Disk Solid

2

1 2MRI

R

Cylinder Hollow

)(2

1 2

2

2

1 RRMI

R2

R2

Cylinder Hollow dThin Walle

2MRI

R

a

b

center)(through Plater Rectangula

)(12

1 22 baMI

a

b

edge)(about Plater RectangulaThin

3

1 2MaI

Sphere Solid

5

2 2MRI

R

Sphere Hollow dThin Walle

3

2 2MRI

R

Moments of inertia for some common geometric solids

Parallel Axis Theorem

• The moment of inertia about any axis parallel to

and at distance d away from the axis that

passes through the centre of mass is:

• Where

– IG= moment of inertia for mass centre G

– m = mass of the body

– d = perpendicular distance between the parallel axes.

2mdII GO

Radius of Gyration

Frequently tabulated data related to moments of inertia will be

presented in terms of radius of gyration.

m

IkormkI 2

Mass Center

m

myy

~

Example

ft5.1)2.32/10()2.32/10(

)2.32/10(2)2.32/10(1~

m

myy

10 Ib

Moment of Inertia of Composite bodies

1. Divide the composite area into simple body.

2. Compute the moment of inertia of each simple body about its

centroidal axis from table.

3. Transfer each centroidal moment of inertia to a parallel reference axis

4. The sum of the moments of inertia for each simple body about the

parallel reference axis is the moment of inertia of the composite

body.

5. Any cutout area has must be assigned a negative moment; all others

are considered positive.

Moment of inertia of a hollow cylinder

• Moment of Inertia of a

solid cylinder

• A hollow cylinder

I = 1/2 mR2

= -m1

R1 R2

m2

I = 1/2 m1R12 - 1/2 m2R2

2 = 1/2 M (R12 - R2

2 )

M

Example 17-3

2

2

1rmIG kgmm

m

kgVm ddd 71.15)]01.0()25.0([8000 2

3

2

22

22

kg.m473.1

)25.0)(71.15()25.0)(71.15(2

1

2

1)(

mkgmkg

dmrmI dddOd

kgmmm

kgVm hhh 93.3)]01.0()125.0([8000 2

3

2

22

22

kg.m276.0

)25.0)(93.3()125.0)(93.3(2

1

2

1)(

mkgmkg

dmrmI hhhOh

2.2.1276.0473.1

)()(

mkg

III OhOdO

2

22

22

kg.m276.0

)25.0)(93.3()125.0)(93.3(2

1

2

1)(

mkgmkg

dmrmI hhhOh

kgmmm

kgVm ddd 71.15)]01.0()25.0([8000 2

3

kgmmm

kgVm hhh 93.3)]01.0()125.0([8000 2

3

2

2

3ddzz rmI )

2

1( 22

dmrm hhh

2)25.0)(71.15(

2

3zzI ))25.0)(93.3()125.0)(93.3(

2

1( 22 mkgmkg

Example 17-4

222 slug.ft414.0)ft2)(ft/s2.32

Ib10(

3

1

3

1)( mlI OOA

2

2222

slug.ft346.1

)2)(2.32

10()2)(

2.32

10(

12

1

12

1)(

mdmlI OBC

2slug.ft76.1346.1414.0 OI

m

myy

~

ft5.1)2.32/10()2.32/10(

)2.32/10(2)2.32/10(1~

m

myy 2

2

2

slug.ft362.0

)5.1)(2.32

20(76.1

G

G

GO

I

I

mdII