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Lecture 12. General theorems of the motion of systems of

material points around their mass center.

1

Theoretical Mechanics

14. General theorems of the motion of systems of material points around their mass center

Let be (S): Mi (mi), ri = OMi, i = 1,..., N a moving system of particles moving in an

inertial (in particular, fixed) Oxyz reference system and consider a Cartesian reference

system of coordinates Cx'y'z' with origin in the center of the masses C of the system (S)

and invariably oriented in space, i.e. Cx'y'z' has only one translation motion with the

velocity of the point C relative to the fixed system of coordinates Oxyz.

Nirrr iCi ,...,1,'

y

x

O

z C

Mi

y‘

rC

ri

z‘

x‘

r‘i

Let be r‘i = CMi, i = 1,..., N. Then we

have

(14.1) Definition:

1. Cx'y'z' is a Koenig frame of reference

2. The motion of the material points system relative

to the mobile system of coordinates Cx'y'z' is called

the motion of the system around the center of gravity

or the motion with respect to a Koenig frame of

reference.



2


iCi vvv '

Deriving (14.1) we have:

(14.2)



3


Indeed, taking into account that and that Poisson’s

formulas are

iCiC

iiC

iiiiiiC

iCi

vvdt

rd

dt

rdr

dt

rd

dt

rd

zdt

kdy

dt

jdx

dt

idk

dt

dzj

dt

dyi

dt

dx

dt

rd

dt

rd

dt

rd

dt

rd

''

''

''

''

''

''

''

''

'

0

''''''' kzjyixr iiii

''

,''

,''

kdt

kdj

dt

jdi

dt

id

Due to the fact that Cx‘y‘z‘ has a motion of translation ( = 0) we have:

(14.3)



4


But the origin of the mobile frame of reference is the centre of mass, C, and thus

we can deduce:

(14.4)

and

N

i

i

N

i

ii mmrmm

CC11

,'1

0

where m is the total mass of the system of particles. Therefore, we obtain:

0'1

N

i

iirm

0'1

N

i

iirm (14.5)



5


The moment of momentum of the system of material points with respect to

the pole O

C

N

j

jj

N

j

jjC

N

j

jjjCC

N

j

jCjjC

N

j

jjjO

vrmvmrvmrvmr

vvmrrvmrK

0

1

0

11

1)2(),1(

1definition

)4.14()5.14(

''''

''

(14.6) Thus we have:

CCCO KrmrK '

where

N

j

jjjC rmrK1

''' (14.7)



6


Equation (14.7) define the moment of momentum of the of the system with respect

to the centre of mass (in the relative motion, with respect to a Koenig frame with

the pole at C.

Equation (14.6) is the first Koenig’s (König) formula.

Theorem (Koenig) The moment of momentum of a mechanical system with respect

to a fixed pole is equal to the sum of the moment of momentum of the same

system with respect to the pole of a Koenig frame of reference and the moment of

momentum of its mass centre at which the whole mass of the mechanical system is

concentrated, with respect to the fixed pole..

Analogously, the kinetic energy is given by:

N

j

jCj

N

j

jj vrmvmT1

2

)2(1

2'

2

1

2

1



7


(14.7)

or

N

j

N

j

Cjjjj

N

j

Cj vvmvmvm1

0

1

2

1

2

)5.14(

''2

1

2

1

CC TmvT '2

1 2

(14.8)

where

N

j

jjC vmT1

2'

2

1'

(14.9)

is the kinetic energy of the system in relative motion (with respect to the mass

centre).



8


Equation (14.8) is the second Koenig‘s formula.

Theorem (Koenig). The kinetic energy of a mechanical system with respect to a fixed

frame of reference is equal to the sum of the kinetic energy of the same system with

respect to a Koenig frame of reference and the kinetic energy of the mass centre at

which the whole mass of the mechanical system is concentrated, with respect to the

fixed frame.

Theorem of the moment of momentum in the motion of a mechanical system

around the centre of mass, C

The theorem of the moment of momentum is: O

O Mdt

Kd

(14.10)

where is the moment of the resultant of the external forces. But,

N

j

jjO FrM1

CCCCCCO KamrKvmr

dt

d

dt

Kd''

)6.14(



9


ROCMM CO

(14.11)

Using the theorem of the centre of mass, maC = R, we have:

CCO KRr

dt

Kd'

Next, we change the pole in the moment formula

(14.12)

And from (14.11) and (14.12) we obtain :

CC

CCC Mdt

KdMROCKRr

'' (14.13)

Theorem of the moment of momentum in the motion of the system around the

center of the mass:

The derivative with respect to time t of the moment of momentum of the system

relative to the center of mass is equal to the resulting moment of external forces

evaluated with respect to the center of the mass.



10


(14.14)

The elementary work of the external forces is:

Theorem of the kinetic energy in the motion of the system around the centre of mass, C

The theorem of the kinetic energy for a system of material points is:

jk

N

j

j

k

jk

N

j

jj rdFrdFLLdT

1

1

11

(int))ext(

Using the second Koenig’s formula we have:

CCCCC rdrmdTTrmddT

''

2

1 2

C

N

j

jjC

N

j

j

N

j

jj

N

j

jCj

N

j

jj rdRrdFrdFrdFrrdFrdF

1111)1(

1

'''

(14.15)

(14.16)



11


Using (14.16) and (14.17) we obtain:

(14.17)

Using the theorem of the centre of mass we get:

CCCCCC

CC rdRrrmdrdRrd

dt

rdmrdR

dt

rdm

CC

N

j

jj

N

j

jj

N

j

jj rrmdrdFrdRrdFrdF

111

'' (14.18)

and using (14.14), (14.15) and (14.18) we have:

N

kj

jjk

N

j

jj rdFrdFLLdT1,1

(int)(ext) '''''

(14.19)



12


where L‘(ext) is the elementary work of the external forces relative to the system of

reference Cx‘y‘z‘, while L‘(int) is the elementary work of the inner forces relative to the

system of reference Cx‘y‘z‘.

Theorem of the kinetic energy in the motion of the system around the centre of mass C:

The differential of the kinetic energy of a system moving about the centre of mass C is

equal with the sum of the elementary work of the external forces and the elementary

work of the inner forces evaluated relative to the system of reference Cx‘y‘z‘.



13


Let be (S) a system of material points (discrete – finite number of points Mi (mi), i =

1,..., N or continuous – rigid body) in an orthogonal frame of referenceOx1x2x3 and

V a simple manifold in R3 (point, line or plane)

Definition: The scalar value I(V) given by:

DV

N

i

ii

dddmd

dmVI

22

1

2

)(

15. Moments of inertia

(15.1)

, (S) discrete system (N points)

, (S) rigid body ( domain D, d is

the element of volume, area or

length)

Is the moment of inertia of the system (S) with respect to the manifold V.

Depending on the manifold the moments can be polar axial or planar.



14


In equation (15.1)

- di is the distance between the point Mi and the manifold V

- d is the distance between the integration point M and the manifold V.

x2

x1

O

x3

Mi(x1i, x2i, x3i)

ri x3i

x1i

x2i

If V = Ox1 or Ox2 or Ox3, we define the moments of inertia about a the axis:



15


V

N

i

iii

dmxx

xxmOxII

2

3

2

2

1

2

3

2

2

111 )( (15.2a)

, discrete system

, rigid body

V

N

i

iii

dmxx

xxmOxII

2

3

2

1

1

2

3

2

1

222 )( (15.2b)

V

N

i

iii

dmxx

xxmOxII

2

2

2

1

1

2

2

2

1

333 )( (15.2c)

, discrete system

, rigid body

, discrete system

, rigid body



16


V

N

i

iii

dmxx

xxmI

21

1

21

12

(15.3a) (15.3b) (15.3c)

In a similar way we can introduce the product of inertia:

V

N

i

iii

dmxx

xxmI

32

1

32

23

V

N

i

iii

dmxx

xxmI

13

1

13

31

One can observe that I12 = I21, I23 = I32, I31 = I13. Using (15.2) and (15.3) it is possible

to build the matrix (tensor) of inertia with respect to the point O

333231

232221

131211

III

III

III

IO(15.4)



17


Steiner‘s theorem:

Let be G the center of massof the material system (S). Consider Δ and ΔG two parallel

axis, such that G∈ ΔG. We note d = dist (Δ, ΔG ). Then for the moment of inertia of the

system with respect to Δ and ΔG we have:

2)()( mdII G

where m is the mass of the system.

(15.5)

We have:

jj rdr "

0' jjrm

Δ ΔG

G

d

Mj

D2 D1

rj r“j

r‘j

zj

0jj zm

Indeed: and

N

i

iirmm

GG1

'1

0

0' jjjj rmprzmg



18


We obtain:

and the axial moment of inertia with respect to Δ is:

2

22

22

)(

"2"

")(

mdI

rmdmdrm

rdmrmI

G

jjjj

jjjj

0'" jjjjj zrmrm

The moment of inertia of a mechanical system with respect to an axis Δ is equal to

the sum of the moment of inertia of the same system with respect to an axis ΔG

parallel to the first one, passing through the centre of mass, and the moment of

inertia of the centre of mass, at which we consider concentrated the mass of the

whole mechanical system, with respect to the axis Δ .



19


Remark: In the case of rotation of rigid system (S) about a fixed axis Δ() with the

angular velocity , the kinetic energy of the system is given by:

2

1

222

22

)(2

1,sin

2

1

2

1

2

1

Iddm

dmvmT

jjj

jjjj

Δ()

Mj dj

2)(2

1 IT



20


Moments of inertia with respect to concurrent axis

Consider Δ(e) an axis passing through the point O, e = (e1, e2, e3) in the system of

reference Ox1x2x3, e12+ e2

2 + e3

2 = 1 (e is the unit vector of the axis Δ).

S

S

dmerr

dmdI

22

2)(

x2

x1

O

x3

M(x1, x2, x3) r

e(e1,e2,e3)

Δ

d

For rigid body (S) we have:



21


dmeexxeexxeexx

exxexxexx

dmexexexeeexxx

S

S

131332322121

2

3

2

2

2

1

2

2

2

3

2

1

2

1

2

3

2

2

2

332211

2

3

2

2

2

1

2

3

2

2

2

1

222

Thus

133132232112

2

333

2

222

2

111 222)( eeIeeIeeIeIeIeII

(15.6)

or in a matrix (tensor) form:

eIeI O

T )( (15.6‘)



22


Matrix IO is symmetric with real element and thus it is diagonalizable (can be reduced

to diagonal form)

Property: In the arbitrary point O exist three directions Δi(e(i)), i = 1,2,3 to which the

moments of inertia have extreme values. Moreover, these directions are orthogonal

(e(i) ┴ e(j), i ≠ j) and considering the matrix in the system of reference defined by these

directions and the origin in O, the matrix of inertia will have:

where A, B, C are the moments of inertia of the system (S) with respect to the three

directions, called principal axis of inertia relative to O.

A, B, C are the eigenvalues of the matrix IO, while e(1), e(2), e(3) are the eigenvectors.

C

B

A

00

00

00



23


The kinetic energy can be ( ) obtain:

where is the versor of

If Ox1x2x3 is determined by the principal axis of inertia relative to O

(I12 = I23 = I31 = 0) we obtain:



24


For the moment of momentum we obtain:

Consider: we have:



25


Therefore:

If Ox1x2x3 is determined by the principal axis of inertia relative to O

(I12 = I23 = I31 = 0) we obtain:



26

Theoretical Mechanics If (S) admits a plane of symmetry then the centre of mass G, belongs to the plane,

two principal axis of inertia belong to the plane and the third is perpendicular on

the plane .

If Δ is an axis of symmetry of the system (S), then the centre of mass belongs to the

axis and Δ is a principal axis of inertia.

Usual moment of inertia:

Calculating the moment of inertia of a thin ring about the central axis



27

Theoretical Mechanics Calculating the moment of inertia of a disk about the central axis



28


Other geometries

-cylinder

-sphere - hollow sphere



29

Theoretical Mechanics If (S) admits a plane of symmetry then the centre of mass G, belongs to the plane,

two principal axis of inertia belong to the plane and the third is perpendicular on

the plane .

If Δ is an axis of symmetry of the system (S), then the centre of mass belongs to the

axis and Δ is a principal axis of inertia.

Example

1.Calculate the mass M and the coordinates of the centre of mass (xG, yG) for a half

disk of radius a and density ρ.

- a x O

y

dr rd

+a

22

1 2

aAM disc



30


rdrddxdydVdVdm ,

0cos1

0 0

2 a

D

G drdrM

xdmM

x

Remark: Due to the symmetry we notice that xG =0.

3

4sin

1

0 0

2 adrdr

Mydm

My

a

D

G



31

Theoretical Mechanics Example

2.Calculate the mass M and the coordinates of the mass centre (xG, yG) for the figure

given bellow (density ρ).

8

3

2

2

2

222

21

aaa

MMM

- a O

y

+a

C1 C2

3

2,

2,

3

4,0

2211

ay

ax

ayx GGGG

Remark: The missing mass can be considered negative.considera masa negativa.

9

14,

6 21

21

21

21 2121a

MM

yMyMy

a

MM

xMxMx

GG

G

GG

G



32


3. Calculate the moment of inertia relative to the centre of inertia C and relative to the

axis Δ for a linear bar of length 2l and density ρ.

dx

y

Δ

C

M(x,0,0)

z

x

l - l

0)(2

0

2

0

D

Cx dmzyI

3)(

2222

0

2 mldxxdmxdmzxI

l

lDD

Cy

dxdldm

We use:

3)(

2222

0

2 mldxxdmxdmyxI

l

lDD

Cz



33


In order to calculate the moment of inertia relative to the axis Δ we use the Steiner‘s

theorem:

0,0,0 D

xy

D

xz

D

zy xydmIxzdmIyzdmI

3

4 222 ml

mlIMdII CyG



34


4.A homogenous bar AB of length 2a and weight P moves under the action of its own

weight, sliding with the ends A and B on the smooth vertical wall Oy and on the

smooth horizontal floor Ox, respectively. Find the angular velocity of the bar and

the pressures NA and NB exerted by the wall and the floor as functions of the angle

between the bar and Ox. At t=0 the angle was = 0. Find the value of for

which the bar breaks off the wall.

The weight P is a potential force

and thus we can use the

theorem of the energy

conservation. The work due to

NA and NB is zero, the reactions

are perpendicular on the

displacements.

We can use the Koenig’s

theorem : x O

y

P

A

C

B

a

NA

NB

vB



35


where T‘C is the kinetic energy of the system in motion around the centre of mass

(inertia).

In the motion of the rigid body about the fixed axis Δ with the angular velocity the

kinetic energy is given by:

CC TmvT '2

1 2

2

2

1 IT

Therefore, we have:

3,

2

1

2

1 222 ml

IImvT CzCzC

Consider: and we get: sin,cos, ayaxmgP CC

22

3

2maT

(i)

(ii)

(iii)

(iv)



36


The elementary work is, L(ext)=-dV:

CBA

ext mgdyOBdNOAdNOCdPL

00

)(

Using the theorem of energy conservation , T+V=h, we obtain:

sinmgamgyVdy

dVmggradVP C

C

The constant h can be obtained using the initial conditions and we have:

hmgamaVT sin3

2 22

0sinsin3

20

22 mgama (v)



37


Thus:

And deriving (vi) we obtain the differential equation for the angular velocity =d/dt.

In order to obtain the reactions we use the equation of the centre of mass:

PNNam BAC

sinsin2

30

2 a

g

0)0(,)0(,cos4

30

a

g

(vi)

(vii)

Eq. (viii) in Cartesian coordinates:

(viii)

mgNym

Nxm

BC

AC

(ix)



38


Using (iii), (vi) and (vii) in (ix) we obtain:

The bar break off the wall when NA = 0 and using (x) we obtain:

0sin2sin3cos4

3 PNA

)sinsin6sin91(4

0

2 mg

NB

(x)

(xi)

0edesprinder sin

3

2arcsin (xii)

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