equation of motion of a variable mass system1

1

SOLO HERMELIN

EQUATIONS OF MOTION OF A VARIABLE MASS SYSTEMSIMPLIFIED PARTICLE

APPROACH

http://www.solohermelin.com

2

SOLO EQUATIONS OF MOTION OF A VARIABLE MASS SYSTEM

• Simplified Particle Approach (this Power Point Presentation)

The equations of motion can be developed using

At a given time t the system has

v (t) – system volume.

m (t) – system mass.

S (t) – system boundary surface.

• Reynolds’ Transport Theorem Approach (see Power Point Presentation)

• Lagrangian Approach (see Power Point Presentation)

3

SOLO EQUATIONS OF MOTION OF A VARIABLE MASS SYSTEMSIMPLIFIED PARTICLE APPROACH

TABLE OF CONTENT

Sir Isaac Newton 1643-1727

• Assumptions

• Inertial Velocity and Acceleration

• Instantaneous Mass Center or Centroid C of the System

• Linear Momentum of the System

• Force Equation

• Moment Relative to a Reference Point O

• Absolute Angular Momentum Relative to a Reference Point O

• External Forces and Moments Applied on the System

• Summary of the Equations of Motion of a Variable Mass System

• References

4

SOLO EQUATIONS OF MOTION OF A VARIABLE MASS SYSTEMSIMPLIFIED PARTICLE APPROACH

Assumptions1. The system at time t contains

N particles.

2. The particle i, of mass dmi, is located at a point (relative to an inertial system – I ).

iR

3. We define a reference point O by the vector (relative to I).OR

4. We obtain the equation of motion for the continuous by taking a very large number N of particles. ∫⇒∑

∞→

=

NN

i 1

We have a system of particles enclosed at the time t by a surface S(t) that bounds the volume v(t). There are no sources or sinks in the volume v(t). The change in the mass of the system is due only to the flow through the surface openings Sopen i (i=1,2,…). In addition the particles are free to move relative to each other.

OiOi RRr

−=:,

The particle relative position to O is given by:

5

SOLO

Assumptions (Continue - 1)

We have

5. The position of the opening ,relative to I, is given by .iopenR

iopenS

( ) ( )( ) ( )ttRttR

tRtR

iflowiopen

iflowiopen

∆+≠∆+

=

&

The position of the mass particle flowing through the opening , relative to I, is given by .

iopenSiflowR

Therefore

( ) ( )I

iflow

I

iopen

td

tRd

td

tRd

≠

and

( ) ( )iopeniflow

I

iopen

I

iflowSi VV

td

tRd

td

tRdV

−=−=:,

is the velocity of flow relative to the opening iopenS

EQUATIONS OF MOTION OF A VARIABLE MASS SYSTEMSIMPLIFIED PARTICLE APPROACH

Table of Content

6

SOLO

The Inertial Velocity and Acceleration of the mass dmi are given by

I

ii td

RdV

=I

i

I

ii td

Rd

td

Vda

2

2

==

Total Mass of the System

( )( )∫∫∑ ==→=

→

∞→

= tvmdmdm

NN

ii dvdmmmdtm

i

ρ1

At a given time t

At the time t + Δ t the mass change is due to the flow through the openings ( ),2,1=iS iopen

( ) ∑∑ ∆+=∆+= openings

iflow

N

ii mmdttm

1

( ) ( ) ( ) ( )∑∑ =∆

∆=

∆−∆+=

→∆→∆openings

iflowopenings

iflow

tttm

t

m

t

tmttmtm

00limlim


Table of Content

The mass rate (flow) , entering / leaving the system, is given by

7

SOLO

Instantaneous Mass Center or Centroid (C) of the System

At the time t + Δ t

By subtracting those two equations, dividing by Δt, and taking the limit, we get

The mass center (Centroid) , of the system, relative to I, at time t, is defined as

( )tRC

( ) ( )( )∫∫∑ ==→=

→

∞→

= tvm

Cdmmd

NN

iiiC dvRdmRtRmmdRtRm

i

ρ

::1

( ) ( )[ ] ( ) ( )∑∑ ∆+∆+∆+=∆+= openings

iflowiflowiflow

N

iiiiCC RRmmdRRtRmtRm

1

( ) ( )[ ] ( )∑+∑=

∆

∑ ∆+∆+∑∆=

∆∆=

=

=

→∆→∆ openingsiflowiflow

N

ii

I

iopeningsiflowiflowiflow

N

iii

t

C

tC Rmmd

td

Rd

t

RRmmdR

t

tRmRm

td

d

1

1

00limlim

Now let add the constraint that at time t the flow at the opening is such that

iopenS

( ) ( )tRtR iflowiopen

=

to obtain( ) ∑∑ −=

= openingsiopeniflow

I

C

N

ii

I

i RmRmtd

dmd

td

Rd

1


Table of Content

8

SOLO

Instantaneous Mass Center or Centroid (C) of the System (continue - 1)

Let develop the right side of this equation

( ) ∑∑ −== openings

iopeniflow

I

C

N

ii

I

i RmRmtd

dmd

td

Rd

1

( )

( )

∑

∑∑∑

∑∑

−=

−−=−+=

=−+=−

openingsCiopeniflow

I

C

openingsCiopenflowi

I

C

openingsiopeniflow

I

C

openingsCiflow

openingsiopeniflow

I

CC

openingsiopeniflow

I

C

rmtd

Rdm

RRmtd

RdmRm

td

RdmRm

Rmtd

RdmRmRmRm

td

d

,

Therefore

( ) ∑∑∑ −=−−== openings

Ciopeniflow

I

C

openingsCiopeniflow

I

CN

ii

I

i rmtd

RdmRRm

td

Rdmmd

td

Rd,

1

The First Moment of Inertia of the System, relative to the point O, is defined as:Oc,

( ) ( ) OCOC

N

iiO

N

iii

N

iiOi

N

iiOiO rmRRmmdRmdRmdRRmdrc ,

1111,, :

=−=∑−∑=∑ −=∑=====


Table of Content

9

SOLO

( ) ∑∑∑ −=−−== openings

Ciopeniflow

I

C

openingsCiopeniflow

I

CN

ii

I

i rmtd

RdmRRm

td

Rdmmd

td

Rd,

1

Linear Momentum of the System

Substitute

At a given time t the Linear Momentum of the system is defined as

( ) ( )( ) ( )

∫∫∑∑ ==→==∞→

→== tmtm I

N

mdmd

N

iii

N

ii

I

i mdVdmtd

RdtPmdVmd

td

RdtP

i

::

11

( ) ( )( ) ( ) ∑∑

∑∑∑

−=−−=

→−=−−==∞→

→=

openingsCiopeniflowC


N

mdmdopenings

Ciopeniflow

I

C

openingsCiopeniflow

I

CN

ii

I

i

rmVmRRmVmtP

rmtd

RdmRRm

td

Rdmmd

td

RdtP

i

,

,1

Differentiate ( )( ) OCOC

N

iiO

N

iii

N

iiOi

N

iiOiO

rmRRm

mdRmdRmdRRmdrc

,

1111,, :

=−=

−=−== ∑∑∑∑====

to obtain ( ) ( ) ( )

+−

+=

−+

−=−+

−=

∑∑

∑

openingsOiflowO

openingsCiflowC

OCopenings

iflow

I

O

I

COC

I

O

I

C

I

O

RmVmRmVm

RRmtd

Rd

td

RdmRRm

td

Rd

td

Rdm

td

tcd

,

to obtain


10

SOLO

Linear Momentum of the System (continue-1)

Substitute

( ) ( ) ∑−=∑ −−=openings

CiopeniflowCopenings

CiopeniflowC rmVmRRmVmtP ,

to obtain

( )

∑+−

∑+=

openingsOiflowO

openingsCiflowC

I

O RmVmRmVmtd

tcd

,

into

( ) ( ) ∑∑ −+=−−+=openings

OiopeniflowO

I

O

openingsOiopeniflowO

I

O rmVmtd

cdRRmVm

td

cdtP ,

,,

At the time t + Δ t the Linear Momentum of the System (including the mass entering/leaving through S) is:

( ) ( ) ( ) ( )∑∑ ∆+∆+∆+=∆+= openings

iflowiflowiflow

N

ii RRmmRRtPtP

1

:

By subtracting those two equations, dividing by Δt, and taking the limit, we get

( )

∑ ∑∑∑

∑∑∑

−−+=+=

∆

−

∆+∆+

∆+

=∆

∆=

=

==

→∆→∆

openingsI

openingsI

Ciopen

iflowopenings

Ciopeniflow

I

C

I

Ciflow

iflow

N

ii

I

i

N

ii

I

i

openingsI

iflow

I

iflow

iflow

N

ii

I

i

I

i

tt

td

rdmrm

td

Rdm

td

Rdm

td

Rdmdm

td

Rd

t

mdtd

Rd

td

Rd

td

Rdmmd

td

Rd

td

Rd

t

tP

td

Pd

,

,2

2

12

2

11

00limlim


11

SOLO

Linear Momentum of the System (continue-2)

We obtain

( ) ∑∑ −−+=i I

Ciflowiflow

openingsCiopeniflow

I

C

I

C

Itd

rdmrm

td

Rdm

td

Rdm

td

tPd ,,2

2

( )

( ) ( ) ( )∑∑

∑∑

∑∑

−−−−+=

→−−+=

++=

→∞

→

=

openingsCiopeniflow


I

C

I

N

mdmdopeningsI

Ciopen

iflowopenings

Ciopeniflow

I

C

I

C

openingsI

Ciflow

iflow

I

CN

ii

I

i

I

VVmRRmVmtd

Vdm

td

tPd

td

rdmrm

td

Rdm

td

Rdm

td

rdm

td

Rdmmd

td

Rd

td

tPd

i

,

,2

2

,

12

2


An equivalent result could be obtained by differentiating

( ) ∑−=openings

Ciopeniflow

I

C rmtd

RdmtP ,

We obtained

Table of Content

12

SOLO

Force Equation

Applying the 2nd Newton’s Law to the particle of mass mi, we obtain:

∑=

+==N

jijiexti

I

ii

I

i fdfdmdtd

Rdmd

td

Vd

1int2

2

where

iextfd

- External forces acting on the mass mi

ijfd int

- Internal forces that particle j exercise on the mass mi

From the 3rd Newton’s Law the internal force that particle j applies on particle i is of equal magnitude but of opposite direction to the force that particle i applies on particle j :

jiij fdfd intint

−=

Therefore

01 1

int

=∑∑

=≠=

N

i

N

ijj

ijfd


13

SOLO

( ) ∞→

→→∑−∑−+=N

mdmdopeningsI

Ciopeniflow

openingsCiopeniflow

I

C

I

C

Iitd

rdmrm

td

Rdm

td

Rdm

td

tPd ,,2

2

Force Equation (continue – 1)

We have ∑∑∑∑∑ =+==

≠===

ext

N

i

N

ijj

ij

N

iiext

N

ii

I

i Ffdfdmdtd

Vd

0

1 1int

11

∑∑=

=N

ii

I

iext md

td

RdF

12

2

Substitute this equation into

to obtain

∑∑∑∑ −−+=++=openings I

Ciopeniflow

openingsCiopeniflow

I

C

I

C

openings I

Ciflowiflow

I

Cext

Itd

rdmrm

td

Rdm

td

Rdm

td

rdm

td

RdmF

td

Pd ,,2

2,

Rearranging we obtain

∑∑∑∑ ++

−+=

openings I

Ciopeniflow

openingsCiopeniflow

openings I

Ciopen

I

Ciflowiflowext

I

C

td

rdmrm

td

rd

td

rdmF

td

Rdm ,

,,,

2

2

2

( ) ∑∑∑∑

−+−+

−+=

openings I

C

I

iopeniflow

openingsCiopeniflow

openingsI

iopen

I

iflowiflowext td

Rd

td

RdmRRm

td

Rd

td

RdmF

2or

( ) ( ) ( )∑∑∑∑ −+−+−+=openings

Ciopeniflowopenings

Ciopeniflowopenings

iopeniflowiflowext

I

C RRmVVmVVmFtd

Vdm

2


Table of Content

14

SOLO

Absolute Angular Momentum Relative to a Reference Point O

The Absolute Momentum Relative to a Reference Point O, of the particle of mass dmi at time t is defined as:

( ) ( ) iiOiiiOiiOiO dmVrdmVRRPdRRHd

×=×−=×−= ,, :

The Absolute Momentum Relative to a Reference Point O, of the mass m (t) is defined as:

( ) ( ) ∑∑∑===

×=×−=×−=N

ii

I

iOi

N

iiiOi

N

iiOiO dm

td

RdrdmVRRPdRRH

1,

11, :

By taking a very large number N of particles, we go from discrete to continuous

∫⇒∑∞→

=

NN

i 1

( )( )

( )( ) ( )

∫ ×=∫ ×−=∫ ×−=tm

Otm

Otv

OO dmVrdmVRRdvVRRH

,, ρ

The Absolute Momentum Relative to a Reference Point O, of the system (including the mass entering (+)/leaving (-) through surface S), at time t + Δt is given by:

( ) ( )∑∑ ∆

∆+×∆++

∆+×∆+=∆+

= openingsiflow

I

iflow

I

iflowOiflowOiflow

N

ii

I

i

I

iOiOiOO m

td

Rd

td

Rdrrdm

td

Rd

td

RdrrHH

,,1

,,,,


15

SOLO

Absolute Angular Momentum Relative to a Reference Point O (continue – 1)

By subtracting

I

O

tI

O

t

H

td

Hd

∆∆

=→∆

,

0

, lim

( ) ( )

t

dmtd

Rdrm

td

Rd

td

Rdrrdm

td

Rd

td

Rdrr

openings

N

ii

I

iiOiflow

I

iflow

I

iflowOiflowOiflow

N

ii

I

i

I

iOiOi

t ∆

×−∆

∆+×∆++

∆+×∆+

=

∑ ∑∑==

→∆

1,,

1,,

0lim

∑∑∑ ×+×+×=== openings

iflow

I

iflowOiflow

N

ii

I

iOiN

ii

I

iOi m

td

Rdrdm

td

Rd

td

rddm

td

Rdr

,1

,

12

2

,

Now let add the constraint that at time t the flow at the opening is such that

iopenS

( ) ( ) ( ) ( )trtrtRtR OiflowOiopeniflowiopen ,,

=→=

to obtain (next page)


( )( )

( )( ) ( )

∫ ×=∫ ×−=∫ ×−=tm

Otm

Otv

OO dmVrdmVRRdvVRRH

,, ρ

dividing by Δt, and taking the limit, we get

from ( ) ( )∑∑ ∆

∆+×∆++

∆+×∆+=∆+

= openingsiflow

I

iflow

I

iflowOiflowOiflow

N

ii

I

i

I

iOiOiOO m

td

Rd

td

Rdrrdm

td

Rd

td

RdrrHH

,,1

,,,,

16

SOLO


∑∑∑ ×+×+×=== openings

iflow

I

iflowOiopen

N

ii

I

i

I

OiN

ii

I

iOi

I

O mtd

Rdrdm

td

Rd

td

rddm

td

Rdr

td

Hd

,1

,

12

2

,,

( )∑ ×−+∑ ×

−+∑ ×=

== openingsiflow

I

iflowOiopen

N

ii

I

i

I

O

I

iN

ii

I

iOi m

td

RdRRdm

td

Rd

td

Rd

td

Rddm

td

Rdr

112

2

,

( )∑ ×−+∑×−∑ ×=== openings

iflow

I

iflowOiopen

N

ii

I

i

I

ON

ii

I

iOi m

td

RdRRdm

td

Rd

td

Rddm

td

Rdr

112

2

,

By taking a very large number N of particles, we go from discrete to continuous

∫⇒∑∞→

=

NN

i 1

( )( )∑ ×−+×−∫ ×=

openingsiflowiflowOiopenO

tmI

O

I

O mVRRPVdmtd

Rdr

td

Hd

2

2

,,

( ) ( ) ∑∑ −+=−−+=openings

OiopeniflowO

I

O


I

O rmVmtd

cdRRmVm

td

cdtP ,

,,

Substitute to obtain

( )( ) ( )∑ ×−+×

∑ −−++∫ ×=

openingsiflowiflowOiopenO

openingsiflowOiopenO

I

O

tmI

O

I

O mVRRVmRRVmtd

cddm

td

Rdr

td

Hd

,

2

2

,,

or

( )( )∑ −×+×+∫ ×=

openingsiflowOiflowOiopenO

I

O

tmI

O

I

O mVVrVtd

cddm

td

Rdr

td

Hd

,,

2

2

,,


17

SOLO


We obtained

( )( )

( )( ) ( )

∫ ×=∫ ×−=∫ ×−=tm

Otm

Otv

OO dmVrdmVRRdvVRRH

,, ρ

Substitute in the previous equation

OIO

O

OO

I

O

I

O

I

OO rtd

rdV

td

rd

td

Rd

td

RdVrRR ,

,,, :&

×++=+==+= ←ω

( )( ) ( )

∫

×++×=∫ ×−= ←

tmOIO

O

OOO

tmOO dmr

td

rdVrdmVRRH ,

,,,

ω

( )( )

( ) ( )∫

×+∫ ××+×

∫= ←tm O

OO

tmOIOOO

tmO dm

td

rdrdmrrVdmr ,,,,,

ω

We obtain

(a) (b) (c)

Let develop those three expressions (a), (b) and (c).

where is the angular velocity vector from I to O.IO←ω


18

SOLO


(a)( ) ( ) ( ) ( ) ( )

( ) OOC

tm

OC

tm

OC

tm

OC

tm

C

tm

O cmRRdmrdmrdmrdmrdmr ,,,,,,

=−===+= ∫∫∫∫∫

Where we used because C is the Center of Mass (Centroid) of the system.( )

0, =∫tm

C dmr

( )OOOOCO

tm

O VcVrmVdmr ×=×=×

∫ ,,,

( )( )

( )[ ]( )

IOOIOtm

OOOOtm

OIBO Idmrrrrdmrr ←←← ⋅=⋅∫ −⋅=∫ ×× ωωω ,,,,,,, 1(b)

where ( )[ ]( )∫ −⋅=tm

OOOOO dmrrrrI ,,,,, 1:

2nd Moment of Inertia Dyadic of all the mass m(t) relative to O

We obtain (a) + (b) + (c)

( )( ) ( )

( )( ) ( )

∫

×+∫ ××+×

∫=∫ ×−= ←tm O

OO

tmOIOOO

tmO

tvOO dm

td

rdrdmrrVdmrvdVRRH ,,,,,, :

ωρ

( )∫

×+⋅+×= ←

tm O

OOIOOOO dm

td

rdrIVc ,,,,

ω


19

SOLO


(c)( )

( ) ( )( )

=∫

+×+=∫

×

tm O

OCCOCC

tm O

OO dm

td

rrdrrdm

td

rdr ,,

,,,

,

( ) ( ) ( ) O

OC

tmC

tm O

CC

tm O

COC

O

OCOC td

rddmrdm

td

rdrdm

td

rdrm

td

rdr ,

0

,,

,,

,,

,

×

∫+∫

×+∫

×+×=

(c1) (c2) (c3)

mtd

rdr

O

OCOC

,,

×(c1) - Change in the relative position of C (varies with time) and O.

(c2)( )

∑×−=∫

×

openingsiflowCiopenOC

tm O

COC mrrdm

td

rdr

,,

,,

(c3)( )∫

×

tm O

CC dm

td

rdr ,,

- Change due to Elasticity, Sloshing, Moving Parts (Rotors, Pistons,..)

If we choose O=C the first two terms (c1), (c2) will be zero, and the third (c3) describes the non-rigidity of the system.


20

SOLO


(c3)

( ) ( )∑+∫

×=∫

× ←

jOjrotorCjrotor

tmFrozenRotors

O

CC

tm O

CC Rj

Idmtd

rdrdm

td

rdr ω

,,

,,

,

where

Consider a system with a number of rigid rotorsI

R

CR

C

( )tS

OR

OOCr

Bx

Bz

shaftr

rotorr

By

Ix

Iy

Iz

Cshaftr

Crotorr

OyOx

Oz

System with Rotors

RjCjrotorI ,

Ojrotor ←ω- Second Moment of Inertia Dyadic of the Rotor j, relative to it’s Centroid

- Angular Velocity Vector of the Rotor j, relative to O


21

SOLO


We obtained

Let differentiate this equation, relative to the inertial system

I

R

CR

C

( )tS

OR

OOCr

Bx

Bz

shaftr

rotorr

By

Ix

Iy

Iz

Cshaftr

Crotorr

OyOx

Oz

∫

×+∑ ⋅+⋅+×= ←←

mFrozemRotor

O

OO

jORjCjrotorIOOOOO md

td

rdrIIVcH

Rj

,,,,,

ωω

OIO

O

O

I

O Htd

Hd

td

Hd,

,,

×+= ←ω

( )

O

mFrozemRotor

O

OO

jORjCjrotor

jORjCjrotorIOOIOO

I

OO

mdtd

rdr

td

d

IIIIVctd

dRjRj

∫

×+

∑ ⋅+∑ ⋅+⋅+⋅+×= ←←←←

,,

0

,,,,,

ωωωω

∫

××+

∑ ⋅×+⋅×+ ←←←←←

mFrozemRotor

O

OOIO

jORjCjrotorIOIOOIO md

td

rdrII

Rj

,,,,

ωωωωω


Table of Content

22

SOLO

Moment Relative to a Reference Point O

Multiplying (vector product) the 2nd Newton’s Law on the particle of mass dmi, by we obtain:OiOi RRr

−=:,

( ) ( ) i

I

iOi

N

jijiextOi dm

td

VdRRfdfdRR

×−=

∑+×−

=1int

from which

( ) ( ) ( )∑ ×−=∑ ∑ ×−+∑ ×−==

≠==

N

ii

I

iOi

N

i

N

ijj

ijtOi

N

iiextOi dm

td

VdRRfdRRfdRR

11 1int

1

We define the moment of external forces, relative to O, on the system, as:

( )∑∑=

×−=N

iiextOiOext fdRRM

1, :


23

SOLO

Moment Relative to a Reference Point O (continue – 1)

Since for any particles i and j the internal forces are of

equal magnitude but of opposite directions

we have

jiij fdfd intint

−=

( ) ( )( ) ( )

( ) collinearfandrfdrfdRR

fdRRfdRR

fdRRfdRR

jitijjitijjitij

jitOjjitOi

jitOjijtOi

intintint

intint

intint

0

←=×=×−=

=×−+×−−=

=×−+×−


We assumed that the equal but opposite forces between i and j act along the line joining them; i.e.

Note

collineararefandr jitij int

This is not always true (see H. Goldstein “Classical Mechanics”, 2nd Edition, pg.8, R. Aris “Vectors, Tensors and the Basic Equations of Fluid Mechanics”, pp.102-104, Michalas & Michalas “Radiation Hydrodynamics”, pg.72, Jaunzemis “Continuous Mechanics” Sec. 11, pg.223)

End Note

24

SOLO

( )( )

( ) ( )∑ −×−+×+∫ ×−=openings

iflowOiflowOiopenO

I

O

tmI

O

I

O mVVRRVtd

cddm

td

RdRR

td

Hd

,

2

2


We have:

( ) ( )∑ ×−=∑ ×−=∑==

N

ii

I

iOi

N

ii

I

iOiOext dm

td

RdRRdm

td

VdRRM

12

2

1,

∞→↓ N

( )( )

( )( )∫ ×−=∫ ×−=∑tv

I

Otm

I

OOext dvtd

VdRRdm

td

VdRRM ρ

,

to obtain

( ) ( )∑ −×−+×+∑=openings

iflowOiflowOiopenO

I

OOext

I

O mVVRRVtd

cdM

td

Hd

,,


Substitute the previous equation with inII

td

Rd

td

Vd2

2

=

25

SOLO

( ) ( ) ∑∑ −+=−−+=openings

OiopeniflowO

I

O


I

O rmVmtd

cdRRmVm

td

cdtP ,

,,


Let substitute in this equation the following

to obtain

( ) ( )∑ −×−+×+∑=openings

iflowOiflowOiopenO

I

OOext

I

O mVVRRVtd

cdM

td

Hd

,,

( ) ( ) O

I

O

openingsiflowOiflowOiopenOext

I

O Vtd

cdmVVRRMH

td

d

×+∑ −×−+∑= ,,,

( ) ( ) ( ) Oopenings

iflowOiopenOopenings

iflowOiflowOiopenOext VmRRmVPmVVRRM

×

∑ −+−+∑ −×−+∑= ,

( )∑ ×−+∑ ×+=openings

iflowiflowOiopenOOext mVRRVPM

,

or

∑ ×+∑ ×+=openings

iflowiflowOiopenOOext

I

O mVrVPMHtd

d

,,,

( )OCOCO RRmrmc −== ,,


Table of Content

26

SOLO

External Forces and Moments Applied on the System

We have a system of particles enclosed at the time t by a surface S(t) that bounds the volume v(t). There are no sources or sinks in the volume v(t). The change in the mass of the system is due only to the flow through the surface openings Sopen i (i=1,2,…). The surface S(t) can be divided in:

• Sw(t) the impermeable wall through which the flow can not escape .( )0,

=sV

• Sopen i(t) the openings (i=1,2,…) through which the flow enters or exits .( )0>m ( )0<m


27

SOLO

External Forces and Moments on the System (continue -1 )

The external forces acting on the system are:

v(t)

I

ds

R

CR

dm

C

( )tS

2openS

1openSg

σn1

t1

OR

O

Or,

OCr ,

jR

jF

kM

• Gravitation acceleration (E center of Earth).E

E

RR

MGg

3

=

• Force per unit surface applied by the surroundings on the surface of the system.( )2/mNσ

( ) dstfnpsdTsdnsd111 +−==⋅=⋅ σσ

where:

( ) ndsnnsdsd111 =⋅= - vector of surface differential

( )2/mN p - pressure on (normal to) the surface .

( ) ( )∑∫∫∑∑∑ +⋅+=→=

jj

tStv

exti

iextext FsddvgFfdF

σρ

( )( )

( )( )

( )( ) ∑+∑ ×−+∫ ⋅×−+∫ ×−=∑→

∑ ×−=∑

kk

jjOj

tSO

tvOOext

iiextOiOext

MFRRsdRRdvgRRM

fdRRM

σρ,

,

The moment of the external forces, relative to a point O, is:

f - friction force per (parallel to) unit surface .( )2/mN

• Discrete force exerting by the surrounding on the point , and discrete moments . ∑j

jF

jR ∑

kkM


28

SOLO

( ) ( ) ( )∑∑∑∑ −+−+−+=openings

Ciopeniflowopenings

Ciopeniflowopenings

iopeniflowiflowext

I

C RRmVVmVVmFtd

Vdm

2

External Forces Equations (continue -2)

( ) ( ) ( )( )

( )∑∫∫∑∫∫∑ ++−+=+⋅+=

jj

tStvjj

tStv

ext FdstfnpdvgFsddvgF

11ρσρ

( ) ( ) ( )0111

0

=⋅∇== ∫∫∫ ∞∞∞tv

Gauss

tStS

dvnpdsnpdsnp

Since the pressure far away from the body is constant ∞p

Let add this equation to the previous one

( ) ( )( ) ( )[ ]

( )∑∫∑∫∫∑ ++−+=+⋅+= ∞

jj

tSjj

tStv

ext FdstfnpptmgFsddvgF

11σρ

( ) ( )[ ] ( )[ ] ∑∫∫ ∑ ∫∫ ++−++−+= ∞∞j

j

S openings S

FdstfnppdstfnpptmgW iopen

1111

Finally since on Sopen i (t) the openings (i=1,2,…) the friction force f = 0

( ) ( )[ ] ( ) ∑∫∫ ∑ ∫∫∑ +−++−+= ∞∞j

j

S openings S

ext FdsnppdstfnpptmgFW iopen

111

Substitute this equation in


29

SOLO

External Forces Equations (continue – 3)

or

( ) ( )[ ] ( ) ( )

( ) ( )∑ −+∑ −+

∑+∑

∫∫ −+−+∫∫ +−+= ∞∞

openingsiflowCiopen

openingsiflowCiopen

jj

openings Siflowopeniflow

SIC

mRRmVV

FdsnppmVVdstfnpptmgmVdt

d

iopenW

2

111 1

( ) ( ) ( )∑∑∑∑∑ −+−++++=openings

iflowCiopenopenings

iflowCiopenj

ji

TiAI

C mRRmVVFFFtmgmVdt

d

2

where


v(t)

I

ds

R

CR

dm

C

( )tS

2openS

1openSg

σn1

t1

OR

O

Or,

OCr ,

jR

jF

kM

( ) ( )∫∫ −+−= ∞

iopenS

iflowiopeniflowTi dsnppmVVF

1:Thrust Forces

( )[ ]∫∫∑ +−= ∞

WS

A dstfnppF11: Aerodynamic Forces

30

SOLO

External Forces Equations (continue – 4)

Let substitute

( ) ( ) ( )∑ −+∑ −+∑+∑+∑+=openings

iflowCiopenopenings

iflowCiopenj

ji

TiAI

C mRRmVVFFFtmgmVdt

d

2

in

CIO

O

C

I

C Vtd

Vd

td

Vda

×+== ←ω

to obtain

RIGID-BODY TERMSmVtd

VdCIO

O

C

×+ ←

ω

∑−∑

×+− ←

openingsiflowCiopen

openingsiflowCiopenIO

O

Ciopen mrmrtd

rd

,,,2 ω FLUID-FLOW TERMS

AERODYNAMIC & PROPULSIVE ∑+∑=

iTiA FF

v(t)

I

ds

R

CR

dm

C

( )tS

2openS

1openSg

σn1

t1

OR

O

Or,

OCr ,

jR

jF

kM

∑++j

jFmg

GRAVITATIONAL & DISCRETE TERMS


31

SOLO

External Moments Equations (continue – 5)

The moments of the external forces relative to the point O are given by

( )( )

( )( )

( ) ∑+∑ ×−+∫ ⋅×−+∫ ×−=∑k

kj

jOjtS

OStv

OOext MFRRsdRRdvgRRM

σρ ~,

( )( )

( ) ( )( )

( ) ∑+∑ ×−+∫ +−×−+×

∫ −=k

kj

jOjtS

OStv

O MFRRdstfnpRRgdvRR

11ρ

Let add to this equation the following

( )( )

( ) ( ) 010

5

=−×∇=×− ∫∫∫∫ ∞∞V

OS

GGauss

tS

OS dvRRpdsnpRR

to obtain

( )( )

( ) ( )[ ]( )

( ) ∑+∑ ×−+∫ +−×−+×

∫ −=∑ ∞k

kj

jOjtS

OStv

OOext MFRRdstfnppRRgdvRRM

11, ρ

( ) ( ) ( )[ ] ( ) ( )

( ) ∑+∑ ×−+

∫∫ ∑ ∫∫

+−×−++−×−+×−= ∞∞

kk

jjOj

S openings SSon

OOOC

MFRR

dstfnppRRdstfnppRRgmRRW iopen

W

1111

0

v(t)

I

ds

R

CR

dm

C

( )tS

2openS

1openSg

σn

1

t

1

OR

O

Or,

OCr ,

jR

jF

kM


32

SOLO

( ) ( ) ( )[ ] ( ) ( )[ ]

( ) ∑+∑ ×−+

∫∫ ∑ ∫∫ −×−++−×−+×−=∑ ∞∞

kk

jjOj

S openings SOOOCOext

MFRR

dsnppRRdstfnppRRgmRRMW iopen

111,

( ) ( )∑ −×−+×+∑=openings

iflowOiflowOiopenO

I

OOext

I

O mVVRRVtd

cdM

td

Hd

,,

External Moments Equations (continue -6)

Using

together with

we obtain

( ) ( ) ( ) ∑+∑ ×−+∑ −×−+

×+∑+∑+×=

kk

jjCj

openingsiflowOiopenOiopen

O

I

O

openingsOTiOAO

I

O

MFRRmVVRR

Vtd

cdMMgc

td

Hd

,,,,

,

( ) ( ) ( ) ( )∑ −×−+∑ −×−+

×+∑=


openingsiflowiopeniflowOiopen

O

I

OOext

mVVRRmVVRR

Vtd

cdM

,


33

SOLO


where

( ) ( )[ ]∫∫∑ +−×−= ∞

WS

OOAero dstfnppRRM

11:, Aerodynamic Moments

( ) ( ) ( ) ( )∫∫ −×−+−×−= ∞iopenS

OiflowiopeniflowOiopenOTi dsnppRRmVVRRM

1:, Thrust Moments on the opening i

discrete forces exerting by the surrounding at point∑j

jF

∑k

kM

jR

discrete moments exerting by the surrounding on the system

v(t)

I

ds

R

CR

dm

C

( )tS

2openS

1openSg

σn

1

t

1

OR

O

Or,

OCr ,

jR

jF

kM


34

SOLO

( )( )

Itm O

OO

Ij

ORjCjrotor

I

IOOIO

I

O

I

OOO

I

O

I

O dmtd

rdr

td

dI

td

d

td

dI

td

Id

td

VdcV

td

cd

td

HdRj

∫

×+∑+⋅+⋅+×+×= ←

←←

,,,,

,,

,,

ωωω


Using

together with

we obtain

( )( )∑+∫

×+⋅×+⋅+⋅ ←←←←

←

jI

ORjCjrotor

I

tmFrozenRotors

O

OOIOOIOIO

O

O

O

IOO Rj

Itd

ddm

td

rdr

td

dI

td

Id

td

dI ωωωωω

,

,,,

,,

( )( ) ( ) ( ) ∑+∑ ×−+∑ −×−+

×+∑+∑+×−=

kk

jjCj


O

I

O

openingsOTiOAOC

I

O

MFRRmVVRR

Vtd

cdMMgmRR

td

Hd

,,,

( ) ( ) ( ) ∑+∑ ×−+∑ −×−+

∑+∑+

−×=

kk

jjCj


openingsOTiOA

I

OO

MFRRmVVRR

MMtd

Vdgc

,,,


Table of Content

35

SOLOSUMMARY OF THE EQUATIONS OF MOTION OF

A VARIABLE MASS SYSTEM

FIRST MOMENT OF INERTIA

SECOND MOMENT OF INERTIA DYADIC

EQUATIONS OF MOTION OF A VARIABLE MASS SYSTEM

( )[ ]( )∫ −⋅=tm

OOOOO dmrrrrI ,,,,, 1:

2nd Moment of Inertia Dyadic of all the mass m(t) relative to O

The First Moment of Inertia of the System, relative to the point O, is defined as:Oc,

( )( ) ( )

( ) OCOC

tm

O

tm

OO rmRRmmdrmdRRc ,,, : =−==−= ∫∫

36

SOLO

( )( )

∑∑ ∫∫∫

===openings iopenopenings S

i

tm td

mdmdmd

td

dtm

iopen

MASS EQUATION

FORCE EQUATION


RIGID-BODY TERMSmVtd

VdCIO

O

C

×+ ←

ω

∑−∑

×+− ←

openings

iiflowiopen

openings

iiflowiopenIO

B

iopen mrmrtd

rd

ˆˆˆ

2 ωFLUID-FLOW TERMS

GRAVITATIONAL, AERODYNAMIC, PROPULSIVE &

∑+∑+=i

TiA FFmg

∑+j

jF

DISCRETE TERMS

SUMMARY OF THE EQUATIONS OF MOTION OF A VARIABLE MASS SYSTEM (CONTINUE – 1)

37

SOLO


MOMENT EQUATIONS RELATIVE TO A REFERENCE POINT O

RIGID-BODY TERMSIOOIOOIOIOO III ←←←← ⋅+⋅×+⋅ ωωωω

,,,

∑ ⋅×+∑ ⋅+ ←←←j

OjrotorCrotorjIOj

OjrotorCrotorj RjRjII ωωω

,, ROTORS TERMS

( )

( )

∫

××+

∫

×+

←tm

FrozenRotorO

OOIO

O

tmFrozenRotor

O

OO

dmtd

rdr

dmtd

rdr

td

d

,,

,,

ω

BODY FLUIDS, MOVING PARTS, ELASTICITY,… TERMS

FLUID CROSSING OPENINGS TERMS∑

×+×− ←

openingsiflowOiopenIO

O

OiopenOiopen mr

td

rdr

,

,, ω

AERODYNAMIC & PROPULSIVE

∑+∑=i

OTiOA MM ,,


( ) ∑+∑ ×−+k

kj

jOj MFRR DISCRETE FORCES

& MOMENTS TERMS

−×+

I

OO td

Vdgc

, NON-CENTROIDAL MOMENTS TERMS

38

SOLO


( )[ ]∫∫∑ +−= ∞

WS

A dstfnppF

11: AERODYNAMIC FORCES

( ) ( )∫∫ −+−= ∞iopenS

iflowiopeniflowTi dsnppmVVF

1: THRUST FORCES

( ) ( )[ ]∫∫ +−×−=∑ ∞WS

OOA dstfnppRRM

11:,AERODYNAMIC MOMENTS

RELATIVE TO O

( ) ( ) ( ) ( )[ ]∫∫ −×−+−×−= ∞iopenS

OiflowiopeniflowOiopenOTi dsnppRRmVVRRM

1:,THRUST MOMENTS

RELATIVE TO O


Table of Content

39

SOLO

References

1. Meriam, J.L., “Dynamics”, John Wiley & Sons, 1966


2. Greensite, A.L., “Elements of Modern Control Theory”,Vol. 2,

Spartan Books, 1970

3. Greenwood, D.T., “Principles of Dynamics”, Prentice-Hall Inc., 1965

Table of Content

January 5, 2015 40

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

equation of motion of a variable mass system1

Science