44

3. Kinetics

Here, we investigate the interaction of motion of a body or system of bodies and the forcesand moments causing this motion.

3.1 Kinetics of a Single Particle

The idealization of a „real world body“ as a particle is admissible, if only the translation ofthe body is of interest. The mass of the body is concentrated to the centre of gravity (CG).

3.1.1 Momentum and Angular Momentum, Newton’s Law

Newton’s Axioms (Philosophiae Naturalis Prinzipia Mathematica, 1687) belong to the mostimportant fundamentals in mechanics and had a great impact on the further scientificdevelopment at that time. In applied mechanics (where we are far away from the speed oflight) these basic laws are still valid today. Especially, the second axiom is one of the mostimportant foundations of kinetics. It says that: „the temporal change of the motion (today wewould say „momentum“) is proportional to the impressed force …” . This implies that, if noimpressed force is present, the motion remains unchanged.

yx

z

m

(I)

v

F

r

0

Path

Fig 3.1: Particle under the influence of an impressed force

The momentum p is defined by

vmp = (3.1.1)

45

In mathematical terms Newton’s 2. axiom reads as:

( )vmdtd

dtpd

F == (3.1.2)

which is valid in an inertial system. For many applications the mass can be consideredconstant. For constant mass m we get:

amdtvdmF == for m = const. !! (3.1.3)

where a is the absolute acceleration.

In integrated form Newton’s law is

∫=−t

t

dttFpp0

**)(0 (3.1.4a)

The integral of the force over time is call impulse. So, we see that change of momentum =impulse. Especially, for constant m we get

∫=−=−t

tdttFpptvtvm

0

**)())()(( 00 (3.1.4b)

For a particle we can define the angular momentum which can be derived the cross product ofthe position vector and the momentum:

prL ×=0 (3.1.5)

The angular momentum is related to the origin of the reference frame (as we will see later,any other reference point can be chosen).

If we multiply eqn. (3.1.2)

Fvmdtdp == )(! (3.1.6)

by the position vector from the left hand side we get:

Frvmdtdrpr ×=×=× )(! (3.1.7)

On the right hand side of this equation, we see the moment of the impressed force withrespect to the origin of the reference frame. The left hand side leads to

( ) ( ) vmdtrdvm

dtdrvmr

dtd ×+×=× and because 0=×=× vmvvm

dtrd

46

we obtain the relation for the angular momentum and the moment of the impressed forcecalled angular impulse-momentum priciple:

00 MLdtd = with FrM ×=0 (3.1.8)

Note: While we could derive this relation for a particle from Newton’s law the relationbetween angular momentum and moment is an independent axiom for rigid bodies.

The integrated form is:

∫=−t

tdttMtLtL

0

**)()()( 0000 (3.1.9)

Instead of cross product in vector notation, in matrix notation we can write the angularmomentum in the following form:

prL ~0 = (3.1.10)

where the skew-symmetric matrix r~ is built up from the components of the position vector

−−

−=

00

0~

xyxz

yzr (3.1.11)

Note: Compare this to angular velocities in vector and matrix notation in chapter 2.

3.1.2 Rotation of a Body About a Fixed Axis

Although we mainly deal with particles here, the rotational motion of a spinning rigid body ofmass m about a fixed axis can be easily derived from Newton’s law. We can find rotationalmotion about a fixed axis in many machines such as turbines, gears etc.

If we consider a small mass element dm of the rigid body (Fig. 3.2), we can see that it performa circular motion about the rotation axis (with constant radius r). From the previous kinematicconsiderations we know the circular motion causes acceleration (also for constant angularspeed). In cylindrical coordinates we got (please compare to chapter 2.1):

−=

−=

=

000

22

ϕϕ

ωω

ϕ !!

!

! rr

rr

aa

ar

(3.1.12)

The (infinitesimal) moment we need to accelerate a single element dm in circumferentialdirection (ϕ-direction) is

47

ϕdFrdM =0 (3.1.13)

Fig. 3.2: Rotational motion of a rigid body about a fixed axis

According to Newton’s law for an infinitesimal small element dm we can write

ϕϕ admdF = (3.1.14)

dmrdM ϕ!!20 = (3.1.15)

The integration over the whole rigid body yields:

dmrMm∫=

)(

20 ϕ!! (3.1.16)

orϕ!!00 JM = (3.1.17)

with the mass moment of inertia

dmrJm∫=

)(

20 (3.1.18)

with respect to the rotation axis through the origin of the reference frame.

z

r

er

dm

ϕeϕ

x

y

48

For an arbitrary axis through a point A we get (see Fig. 3.3):

ϕ!!AA JM = with dmrJm

A ∫=)(

2 (3.1.19)

dmr

SA

xx’

xx’

rs

Fig. 3.3: On the rotational motion about different fixed axes: explanation of geometricquantities

In many cases the mass moment of inertia for a fixed axis x‘-x‘ through center of gravity S isknown (e.g. from tables). The transformation for a parallel axis through A can be done bymeans of Steiner’s theorem

mrJJ SSA2+= (3.1.20)

As we can see, the mass moment of inertia with respect to an axis through the CG is alwaysthe smallest possible one, because the additional Steiner term is always positive.

3.1.3 Kinetics of a Particle for Relative Motion

As stated earlier, Newton’s law is valid only when applied in an inertial system. For constantmass and indicating the inertial system by the subscript I we can write:

amF II = (3.1.21)

Now we can replace the absolute acceleration in the I-system by the components of themoving R-system by using the rotation matrix:

amaAmFAF RIRIIRIR === (3.1.22)

In a next step we express the acceleration by means of the guidance, the Coriolis and therelative acceleration

49

relCorFü aaaa ++= (3.1.23)

where the subscript R has been omitted for simplicity. Now we introduce this into equation(3.1.22):

)( relCorFü aaamamF ++== (3.1.24)

or after re-arranging the last equation:

corgrel amamamam −−=

or

corgrel amamFam −−= (3.1.25)

What can be seen clearly is that we are not allowed to write Newton’s law in it’s simple formin the moving R-system. However, we have to „correct“ the formula by the inertia forces: the

Coriolis force corcor amF −=and the

guidance force gg amF −=(which consists of three terms and among them the centrifugal force as the most prominentrepresentative of the inertia forces). If we introduce these forces into the last equation, we seethat:

corgrelrel FFFamF ++== (3.1.26)

The appearance of the inertia forces is directly coupled with the shift from the inertial system(where we have no inertia forces) to the moving reference frame R. The inertia forces play animportant role in D‘Alembert’s principle.

Finally we consider the special case that the reference frame is rigidly connected with ourmass particle m. Then per definition of the R-frame we cannot have any relative motion. Therelative acceleration and the relative velocity both are zero. The latter implies that the Coriolisforce is zero. So we end up formally in an dynamic equilibrium where the sum of all forcesincluding the inertia forces is zero:

gFF +=0 (3.1.27)

50

3.1.4 Work and Work-Energy Principles

x

z

m dr

F

r

Bahn

αs

Fig. 3.4: Quantities to explain the concept of w

3.1.4.1 Translational Motion of a Particle

The force F moves the particle m along the path. If we considerget an infinitesimal contribution of work

.cos dsFFdrrdFdW s=== α

The work is calculated by means of the scalar product of forangle between the vectors F and dr . The magnitude of the infds. Fs is the component of the force in direction of the tangcomponent which is orthogonal to the path tangent cannot contr

Integration along the path from s0 to s1 yields the work

∫=1

0

))(cos()(s

sdsssFW α

Next we set dtvrd = in eqn.(3.1.28) and F is replaced by ma

dtvdmamF == (m = const.),

so that

vdvmrdFdW ==

The integration now yields the very important relation

( )∫ −==1

0

20

212

v

vvvmvdvmW

h

Pat

y

(I)

0

ork of a force F

a small displacement dr we

(3.1.28)

ce and displacement, α is theinitesimal displacement is dr =ent to the path. Obviously, theibute to the work of the force.

(3.1.29)

(Newton’s law)

(3.1.30)

(3.1.31)

51

where the right hand side can be identified as change of the kinetic energy of the particle(mass m). The kinetic energy is

221 mvEkin = (3.1.32)

Note: In the anglo-american literature, frequently the symbol T is used for the kinetic energy.

From eqn.(3.1.31) follows the work-energy theorem:

0,1, kinkin EEW −= (3.1.33)

The work of the external force causes a change of the kinetic energy (increase or decreasedepending on the angle between the force and the path).

Note: the left hand side is a path integral which considers what happens along the path, whilethe energies only take the initial and final state into account.

3.1.4.2 Conservative and Non-Conservative Forces, Potential, Energy-Theorem

We can distinguish between conservative and non-conservative forces. For conservativeforces the value of the path integral (3.1.29) for a fixed starting and final point is independentof the special shape of the path between these points. This is a very important property ofconservative forces.

For conservative forces a scalar function Π, the potential, exist which allows the calculationof the force vector from the negative gradient of the potential:

Π−= gradF (3.1.34)

(note: negative sign by definition). For example, in cartesian coordinates we get

xFx ∂

Π∂−= ;y

Fy ∂Π∂−= ;

zFz ∂

Π∂−= (3.1.35)

Example:The potential of the gravitational field of the earth (close to the surface) is

mgzzyx =Π ),,(

where the z-axis is orthogonal to the surface and points away from the center of the earth. Thecalculation of the gradient immediately yields: 0=xF ; 0=yF ; mgFz −= .

The total differential of Π is

52

dzz

dyy

dxx

d∂Π∂+

∂Π∂+

∂Π∂=Π (3.1.36)

and on the other hand from eqn.(3.1.28) and (3.1.29) (in cartesian coordinates) we get

( )∫ ++=1

0dzFdyFdxFW zyx (3.1.37)

If we replace the force components by the derivatives of the gradient an using eqn.(3.1.36),we obtain

( )011

0

1

0Π−Π−=Π−=

∂Π∂+

∂Π∂+

∂Π∂−= ∫∫ ddz

zdy

ydx

xW (3.1.38)

which shows us that the work W does not depend on the path but only on the values of thepotential Π0 and Π1 at the initial and final point of the path which proves the independence ofthe path. The condition that a force field is conservative is the existence of a potential!

In the special case that initial and final point coincide, it follows from eqn.(3.1.38), that W =0: if we walk along a closed loop in a conservative force the work resulting from this processis zero!

For non-conservative forces such a potential does not exist. The work integral is path-dependent and the integration along a closed loop yields (in general) a value W ≠ 0.

Examples for conservative forces are- gravitational forces,- elastic forces (e.g. of elastic springs, bending of beams, etc.),- magnetic forces

Non-conservative forces are- friction forces or- forces and moments from external sources

Finally, for forces having a potential, from eqn. (3.1.33) and (3.1.38) we can derive theimportant energy theorem:

( )010,1, Π−Π−=−= kinkin EEW (3.1.39)

from which follows:

00,11, Π+=Π+ kinkin EE (3.1.40)

With the potential energy Epot which is corresponding to the value of the Π we get

0,0,1,1, potkinpotkin EEEE +=+ (3.1.41)

53

This describes the invariance of the total energy (which is the sum of kinetic and potentialenergy) for a process in conservative fields.

3.1.4.3 Rotation About a Fixed Axis

In analogy to the translational motion of a particle we obtain for the rotation of a rigid bodyabout a fixed axis:

ϕdMdW = (3.1.42)

∫=1

0

)(ϕ

ϕϕϕ dMW (3.1.43)

Again, the work changes the kinetic energy. For the rotation about a fixed axis we get:

221 ωJEkin = (3.1.44)

where ω is the angular velocity and J is the moment of inertia with respect to the spinningaxis. We obtain (according to eqn. (3.1.31) and (3.1.33)):

( )20

210,1, 2

1 ωω −=−= JEEW kinkin (3.1.45)

A more general analysis of the kinetic energy for the rotation about a free axis will be givenlater.

3.1.5 Power

Relating the work to time, we come to the concept of power which is very important in alltechnical disciplines.

The power P is mathematically defined by the gradient:

dtdWP = (3.1.46)

For a translational motion we get

vFvFdt

rdFP s=== αcos (3.1.47)

where Fs is again the tangential component of the force F.

54

For the rotational motion with a fixed axis, the power is

ωϕ MdtdMP == (3.1.48)

The unit of the power is 1 W (Watt): s

NmW 11 = .

For the old unit HP (horse power) the conversion is: kWHP 7355.01 = . This unit is still inpractical use e.g. in the automotive industry.

3.2 Kinetics of Rigid Bodies

3.2.1 Momentum of a Rigid body and the Momentum Theorem

The total mass of a body results from integration over the whole body

∫∫ ==)()( Vm

dVdmm ρ (3.2.1)

where ρ is the mass density and dV is a small volume element. In fig. 3.1, P is an arbitrarypoint of the body, while S is the center of gravity (German: “Schwerpunkt” or“Massenschwerpunkt”)

Fig. 3.5: On the kinetics of a rigid body

In the same way, the resultant of the external forces dF acting on a small mass element dmcan be obtained by integration:

∫=)(K

FdF (3.2.2)

(I)

S

Pdm

PSP

S

dF

rr

r

x

z

y

55

If we consider the contribution of the momentum of a single mass element dm (at point P):

dmvpd = (3.2.3)

As the rigid body performs translational and rotational motion as well, we apply Euler’sformula and use the center of gravity S as reference point

SPSP rvvv ×+== ω (3.2.4)

If we put this into eqn.(3.2.3) we get

dmrdmvdmrvpd SPSSPS )()( ×+=×+= ωω (3.2.5)

Again the total momentum of the rigid body is obtained by integration:

∫∫∫ ×+=×+=)()()(

)(m

SPSm

SPm

S dmrmvdmrdmvp ωω (3.2.6)

The second integral vanished because we have chosen the reference point as the mass centerof gravity (good choice!).

Note: Recall that the formula ∫=)(

1

mRPRS dmr

mr yields the position of the center of gravity S

as vector pointing from the reference point R to S. If the reference point R is already identicalwith S it follows immediately that 0== SSRS rr

The momentum then is

Svmp = (3.2.7)

It is the product of the mass m which is concentrated in the center of gravity multiplied by thevelocity vS of point S. From Newton’s law formulated for a small element dm

Fddmr =!! (3.2.8)

we obtain after integration

∫∫ ==)()( KK

FFddmr!! (3.2.9)

In chapter 2 we derived an expression for the acceleration of a point P

)( SPSPSPP rraarr ××+×+=== ωωω!!!!! (3.2.10)

56

putting eqn.(3.2.10) into (3.2.9) we get (the integrals ∫)(m

RP dmr are zero):

madmadmr Sm

Sm

∫∫ ==)()(

!! (3.2.11)

The yields the momentum theorem:

pvdtdmamF SS !=== (3.2.12)

where m =const. was assumed.

We see that the center of gravity (CG) moves as if the force resultant F would act directly onthe CG where the total mass m is concentrated. The resulting force F determines theacceleration as of the CG. In general there is an additional rotation of the rigid body, becauseF does not act directly on the CG which causes a moment about S.While we could derive the angular momentum – moment relation from the momentumprinciple for a mass particle this is not possible for a rigid body. In the latter case we needadditional assumptions about the internal forces. Instead the angular momentum theorem wasformulated as an independent law for the rigid body by Euler.

3.2.2 Angular Momentum and Moments of Inertia

A small element dm delivers a contribution

dmvrpdrLd ×=×=0 (3.2.13)

to the angular momentum (compare to our considerations with respect to a particle) whichafter integration leads to

dmrvrrdmvrL SPSSPS )()(0 ×+×+=×= ∫∫ ω

and multiplying all the terms in brackets of the cross product:

∫∫∫∫ ××+××+×+×= dmrrdmrrdmvrdmvrL SPSPSPSSSPSS )()(0 ωω

The vectors rS , vs and ω are constant with respect to the integration over the body and can bemoved outside the integral Now, the second and third term vanish because the integral

0=∫ dmr SP as we showed before so that

∫ ××+×= dmrrmvrL SPSPSS )(0 ω (3.2.14)

remains

∫ ××+×= dmrrprL SPSPS )(0 ω (3.2.15a)

57

The first term is the translational part, the second part is the rotational part of the angularmomentum:

prL Strans ×=,0 (3.2.15b)

∫ ××= dmrrL SPSProt )(,0 ω (3.2.15c)

rottrans LLL ,0,00 += (3.2.15d)

We see that also if the body does not rotate we have a contribution to the angular momentumfrom the translation.

The rotational part can be further investigated and expressed as :

ω

ωωω

ωωω

ωωω

ω S

zyx

zyx

zyx

mSPSProt J

dmzzyzx

dmyzyyx

dmxzxyx

dmrrL =

−−

−−

−−

=××=

∫∫∫

∫)(

)(

)(

)(2

2

2

)(,0 (3.2.16)

Splitting the integral terms and the angular velocities, the integral yields the symmetric tensorof the moments of inertia:

=

zzyzxz

yzyyxy

xzxyxx

SJJJJJJJJJ

J (3.2.17)

The subscript ‚S‘ indicates that the inertia tensor is related to center of gravity. The matrix iscalled moment of inertia matrix or simply inertia matrix. The diagonal elements are called themoments of inertia (with respect to the x-, y– or z-axis) and the off-diagonal terms are theproducts of inertia or cross products of inertia:

S

dmz

xy

r

Fig. 3.6: Rigid body with reference frame in the center of gravity

58

Moments of Inertia:

∫ += dmzyJ xx )( 22 (3.2.18a)

∫ += dmzxJ yy )( 22 (3.2.18b)

∫ += dmyxJ zz )( 22 (3.2.18c)

Products of inertia:

∫−== dmxyJJ yxxy (3.2.18d)

∫−== dmzxJJ xzzx (3.2.18e)

∫−== dmyzJJ yzyz (3.2.18f)

where the coordinates SPxx = , SPyy = and SPzz = . For a change of the reference frame toanother reference frame which has axes parallel to the first one Steiner’s theorem is valid.While the coordinates always appear in quadratic terms in the moments of inertia, the sign ofthe coordinates are not important, however, in the products of inertia the sign of thecoordinates have to be considered.

Principle Axes:

If the x-, y- and z-axes are principle axes, the products of inertia (off-diagonal terms) vanish

=

3

2

1

000000

JJ

JJ S (3.2.19)

and J1 and J3 are the maximum and minimum values of the moments of inertia, respectively.The three diagonal moments of inertia are the eigenvalues of matrix in eqn.(3.2.17) and arecalled principal inertias. In a basis with symmetric planes of a body, the off-diagonal termsare zero.

Due to the fact that the integration is a linear operator, the inertia-matrix calculation is anadditive operation. The inertia matrix of a body with a volume V1 + V2 is the sum of theinertia matrix of volume V1 and the inertia matrix of V2.

59

Steiner’s theorem or parallel axes theorem:

S

dmz

xy

y’

z’

x’

r

Fig. 3.7: Rigid body with reference frame in the center of gravity

If we know the inertia matrix with respect to the center of gravity S we can easily calculatealso the inertia matrix for any point, say A, which is the origin of a reference frame which hasaxes (x’-y’-z’-axes) which are parallel to the x-y-z-axes through S. Steiner’s theorem says that

mzyJJ SSxxxx )( 22'' ′+′+= (3.2.20a)

mzxJJ SSyyyy )( 22'' ′+′+= (3.2.20b)

myxJJ SSzzzz )( 22'' ′+′+= (3.2.20c)

and

myxJJ SSxyyx ′′−='' (3.2.20d)

mzxJJ SSxzzx ′′−='' (3.2.20e)mzyJJ SSyzzy ′′−='' (3.2.20f)

where the coordinates x’S, y’S and z’S define the position of S in the x’-y’-z’ coordinate systemin A.

Note: For completeness we note that we can also find the definition of the products of inertiawithout the negative sign (eqns.(32.18d-f)) in the literature. But in this case the negative signis included directly in the sJ ,eqn.(3.2.17), where all Jkl , k,l = x,y,z, k ≠ l having negative

signs.

The rotational part of the angular momentum is

ωSrot JL =,0 (3.2.21)

60

The components of this vector in Cartesian coordinates are

zxzyxyxxxxrot JJJL ωωω ++=,,0

zyzyyyxyxyrot JJJL ωωω ++=,,0

zzzyzyxzxzrot JJJL ωωω ++=,,0

or for the principle axes:

111,,0 ωJL rot =

222,,0 ωJL rot =

333,,0 ωJL rot =

The total angular momentum finally is

ωSS JprL +×=0 (3.2.22)

In matrix notation the angular momentum is

ωISISII JprL += ~0 (3.2.23)

3.2.3 Angular Momentum Theorem

The theorem of angular momentum introduced by Euler is –as already mentioned- anindependent axiom, that cannot be derived from the momentum theorem without furtherassumptions about the internal forces.

The angular momentum theorem is

00 ML =! (3.2.24)

The external moment changes the angular momentum. For a rigid body the internal forces donot have any influence on the momentum and the angular momentum.

The resultant of all external moments follows from the integration of all contributions r × dFover the whole body

∫ ×= FdrM 0 (3.2.25)

It is important to note that the reference point 0 of the coordinate frame is arbitrary, howeverit must be a fixed point. Angular momentum L0 and moment M0 must be related to the samepoint (0). If the reference point 0 is not fixed but moving, additional terms describing themotion of the reference frame have to be added.

61

00 )()( MJvmrL SSS =+×= !!! ω (3.2.26)

The product rule has to be applied. The term

( ) ( )SSSSSSS amrvmvvmrpr ×+×=×=× )()( !!

and because 0=× SS vv only the second term remains

( ) 00 )( MJamrL SSS =+×= !! ω (3.2.27)

3.2.4 Angular Momentum Theorem for Pure Rotation

For a fix-point A only rotation of the body about this point is possible. Euler’s equation for thevelocities was

APAP rrr ×+= ω!!

If A is a fix-point, then 0=Ar! so that APP rr ×= ω! . In the case that point P is the CG S:

SSS rrv ×== ω!

Putting this into eqn.(3.2.26) with 0 = A:

ASSSA MJrmrL =+××= )()( !!! ωω

The double cross product is

( )( )

( )( )

( )( )

++−−−++−−−+

=

++−−

−++−

−−+

=××

z

y

x

SSSSSS

SSSSSS

SSSSSS

zSSySSxSS

zSSySSxSS

zSSySSxSS

SS

mxymyzmxzmyzmzxmxymxzmxymzy

mxymyzmxz

myzmzxmxy

mxzmxymzy

rmr

ωωω

ωωω

ωωω

ωωω

ω

22

22

22

22

22

22

)(

(3.2.28)

62

The matrix can be identified from Steiner’s theorem, see eqn.(3.2.20) so that we can write

( )( )

( ) ASSSSSS

SSSSSS

SSSSSS

S Jmxymyzmxz

myzmzxmxymxzmxymzy

J =

++−−−++−−−+

+22

22

22

(3.2.29)

which is the inertia matrix with respect to fix-point A. The angular momentum theorem then is

AAA MJL == )( !! ω (3.2.30)

As we can see the translational part of the angular momentum is eliminated.

3.2.5 Angular Momentum Theorem with Respect to the Center of Gravity

The center of gravity is fixed on the body, but moving in space. Until now, we considered theangular momentum theorem only for an inertial system. Starting our considerations witheqn.(3.2.27):

( ) 00 )( MJamrL SSS =+×= !! ω

we express the moment according to eqn.(3.2.25)

( ) SSS MFrFdrrFdrM +×=×+=×= ∫∫ '0 (3.2.31)

where r’ is the position vector from point S to the actual mass element. As can be seen, M0 canbe split into a part of the resultant force (1.part) and a moment about the center of gravity MS.

( ) SSSSS MFrJamrL +×=+×= )(0!! ω (3.2.32)

Now we use the momentum theorem (Newton’s law for a rigid body):Fam S =

and we see that eqn.(3.2.32) simplifies to

SSS MJL == )( !! ω (3.2.33)

As can bee seen the choice of the center of gravity as reference point turns out to be a goodchoice because only the rotational part of the angular momentum remains.

63

S

z

xy

MS

F

ω

Fig. 3.8: Rigid body(with reference frame in the center of gravity) under external force andmoment loading

We should keep in mind that although the reference frames moves with S, the directions of x-y-z-axes are always constant. This means also that if the body is rotating the moments andproducts of inertia change with time during the rotational motion. This is a seriousdisadvantage of this formulation and we have to think about the possibility of a referenceframe which is fixed on the body (in S).