Teodorescu P.P. Mechanical Systems, Classical Models Volume II: Mechanics of Discrete and Continuous Systems

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wherefrom

()

sin cos secψω ϕω ϕ θ=+



cos sinθω ϕω ϕ=−



()

12 3

sin cos tanϕω ω ϕω ϕ θ=+ +



(14.2.36')

The angles

ψ, θ and ϕ are measured by the gyrocompass, the angular velocity being

then given by (14.2.36). With the aircraft instruments one can also measure the

components

′

of the velocity

′

v in the frame

′

, along the axes of the frame R,

the formula

Cj Ci

Vvβ

′′

, 1, 2, 3j = , allowing then to pass to the axes of the frame

′

R ; integrating the system of equations d/d

tVρ

′′

= , 1, 2, 3j = , we determine the

motion of the mass centre C in the latter frame of reference.

The equations (14.1.48), (14.1.52) take the form (

23 12

0II==, because the plane

Cx x is a plane of symmetry)

()

23 1

132

sin

CCC

Mv v v Mg Rωω θ

′′′

+− =− +



()

31 2

213

cos sin

CCC

Mv v v Mg Rωω θϕ

′′′

+− = +



()

12 3

321

cos cos

CCC

Mv v v Mg Rωω θϕ

′′′

+− = +



(14.2.37)

()

11 1 31 3 33 22 2 3 31 1 2

II II I Mωω ωωωω++− + =



()

22 2 11 33 3 1 31 3 1

III I Mωωωωω+− + −=



()

33 3 31 1 22 11 1 2 31 2 3

II II I Mωω ωωωω++− − =



(14.2.37')

263

14 Dynamics of the Rigid Solid

where

M=Gg is the weight of the airplane, while

()

′

RR vω and

()

CCC

′

MM vω are the components of the torsor of the aerodynamic actions at C.

We notice that components of a pseudomoment of momentum



K can intervene too,

leading to supplementary terms of the form



ijk

KKω+∈





, 1, 2, 3i = , in the left

member of the equations (14.2.37'), due to the rotation of some parts of the airplane

(e.g., of a wing of it); thus, in the right member of the same equations appear

supplementary moments, called moments of gyration. In any case, in a leeway (drift

angle) navigation we have



=K0.

The airplane can change its direction of advance modifying the angle between the

fixed and the movable surfaces by which its wings are fitted out; the movable surfaces

(called ailerons) corresponding to the frontal wings are coupled so that if one of them is

moving upwards the other one is moving downwards, acting thus upon the rolling axis

Cx . The horizontal empennage (the movable part of which is called depth rudder) and

the vertical empennage (the movable part of which is called direction) are at the back

part of the airplane, helping – as well – to change its direction. We denote by

′

the

angle between the horizontal empennage and the depth rudder, by

′

the angle between

the vertical empennage and the direction and by

′

the angle between the wing and the

aileron; one can establish the driving (manoeuvre) equations

()

11 23 13121

132 1

CCC d

IMev v v J RRψωωωωω

′′′′ ′

++−+−=+



()

22 31 2123

213CCC

IMev v v Jθωωωωω

′′′′

−+−−+



()

23 1 2

JRRωω ω

′

−−=+



()

332313

IJ RRϕωωω

′′

++=+





(14.2.38)

where

IJ, 1, 2, 3k = , and J are axial and centrifugal moments of inertia,

respectively, of some movable parts (e.g.,

I is the moment of inertia of the depth

rudder with respect to its axis of rotation),

M and

M ,

e and

e are the masses and

the eccentricities of the depth rudder and of the direction, respectively,

′

are the

aerodynamic forces which appear due to these rotations, while

R ,

1, 2, 3k =

, are the

corresponding driving actions. To determine the 12 unknown functions

()

Ck Ck

vvt

′′

()

tωω= ,

1, 2, 3k =

, ()tψψ= , ()tθθ= , ()tϕϕ= , ()tψψ

′′

= , ()tθθ

′′

and ()tϕϕ

′′

= we have thus at our disposal the system of 12 differential equations

(14.2.36')-(14.2.38), corresponding to a given command. In particular problems, these

equations can be simplified (e.g., in case of dynamic equilibrated commands we have

123

0JJJJ====,

0ee==, while in problems of stability with free wings

we put

123

ddd

RRR===).

Besides the aerodynamic and manoeuvre loads, one can take into consideration the

storm loads (of aerodynamical nature too), the loads which arise at take-off and

landing, various types of loads with a local character etc.

264

MECHANICAL SYSTEMS, CLASSICAL MODELS

In case of the plane-parallel motion, the problem in the preceding subsection is

considerably simplified; we put thus in evidence the motion in the symmetry plane of

the airplane, which is a vertical plane. Unlike the general case, we report the motion to

the inertial frame of reference

Oxx

′′′

(the

′′

-axis being horizontal) and to a non-

inertial frame

Cx x , with the axes parallel to those of the inertial one. At the mass

centre C, which moves with the velocity

′

v , which makes the angle α with the

longitudinal axis

Δ of the airplane, acts the own weight

M=Gg

, the propelling force

F and the torsor

{

}

RM of the aerodynamic forces exerted upon the aircraft’s

surface. It is convenient to decompose the resultant of the aerodynamic forces in the

form

=+RWN, where W (the so-called resistance) is along the velocity

′

v , while

N (force of uplift) is normal to W (Fig. 14.19a). The magnitudes of these components

are obtained by studies of aerodynamical nature, in the form

()

WC Av

′

()

NC Av

′

(14.2.39)

14.2.2.5 Plane-Parallel Motion of the Airplane

where A is the area of the lifting surface,

γ is the unit weight of the air, while the

coefficients of resistance

()

C α and ()

C α , characteristic for each airplane, are two

non-dimensional functions of angle

α; we assume that for 0α = we have

()

C = , hence 0N = , the velocity

′

v being along the longitudinal axis of the

aircraft.

The motion of the aircraft’s centre of mass will be specified by the equation

′

=+++

FNW

(14.2.40)

written with respect to the inertial frame of reference

′

, and its rotation by the

equation

265

Fig. 14.19 Plane-parallel motion of an airplane

14 Dynamics of the Rigid Solid

IMθ =



(14.2.40')

where

()tθθ= is the angle made by the longitudinal axis Δ with the

′′

-axis, while

I and

M correspond to the

Cx -axis, normal to the considered vertical plane. The

magnitude of the moment

M is obtained, by aerodynamical research too, in the form

(,)

Mmvnvαδ θ

′

=− −



(14.2.39')

where the function

(,)m αδ , of the nature of a mass, depends on α and on the angle δ

made by the altitude rudder with its normal position, while the coefficient n, of the

nature of a product of a mass by a length, is due to the motion of the rudder surfaces

(the damping action of the moment

M ); here too, m and n are characteristics of each

aircraft. We obtain thus three scalar equations (14.2.40), (14.2.40') for the unknown

functions

()tρρ

′′

= ,

()tρρ

′′

= and ()tθθ= .

If the airplane advances with switched off motor (

=F0) and constant velocity

const

′





(case of a gliding flight), then the equation (14.2.40) leads to

M ++ =gNW0

. (14.2.41)

Hence, the resultant

R of the aerodynamic forces equilibrates the weight of the aircraft

(

M=−Rg). The soaring angle ϕ, made by the velocity

′

v with the horizontal axis

Cx (Fig. 14.19b), is given by tan / /

WN C Cϕ =− =− , where we took into

account (14.2.39); analogously, starting from the relation

22222

GMgNW==+,

the magnitude of the soaring velocity

′

v is given by

AC Cγ

′

(14.2.41')

In case of a normal flight (horizontal flight with constant velocity) we have

0ϕ =

and const

′

; the equation (14.2.40) takes the form

M +++ =gFNW0. (14.2.42)

266

MECHANICAL SYSTEMS, CLASSICAL MODELS

In projection on the axes

Cx and

Cx , we get (Fig. 14.20a)

cos 0WF β−+ =, sin 0NGF β−+ =, (14.2.42')

where

β is the angle made by the force F with the velocity

′

v . We obtain thus

tan

−

= ,

()

FGNW=−+.

(14.2.42'')

We can easily calculate the angle

αββ

′

, because the angle

′

between the force

F and the longitudinal axis Δ is known. Observing that constθ = , from the equations

(14.2.39'), (14.2.40') it results

(,) 0m αδ = , determining thus δ, that is the necessary

deviation angle of the altitude rudder.

Fig. 14.20 Case of a horizontal flight with constant velocity

Lanchester studied in 1909 the motion of an aircraft with switched of motor (

=F0)

and with a nearly vanishing resultant (

≅W0), the velocity

′

v of which makes a

constant angle with the longitudinal axis (

constα = ). It results the equation of motion

of the mass centre

′

(14.2.43)

Projecting on the direction of the velocity

′

v and on a direction normal to it, we obtain

(using the expressions (5.1.14) of the velocity in polar co-ordinates (Fig. 14.20b)

sin

′

=− ,

cos

vgKv

′

=− +

(14.2.43')

where

KCAMgγ= is a coefficient of the nature of the inverse of a length. We

have thus

d/d sin

vv s g ϕ

′′

=− for an element of arc along the trajectory of the

centre C; observing that

ddsinxsϕ

′

= , it results

267

14 Dynamics of the Rigid Solid

vgx

′′

=− +

(14.2.44)

where h is the energy constant (corresponding to the first integral of the kinetic energy).

Observing that

d/d sind/d

vtv xϕϕϕ

′′ ′

= and taking into account (14.2.44), we

can write the second equation (14.2.43') in the form

()

dcos cos

xhMgx

ϕϕ

−+=

′′

−

The change of variable

′

/yhMgx

′

=−

(one passes to a left-handed inertial

frame of reference Oxy , with the Oy -axis along the ascendent vertical and with the

Ox -axis at a level at which, in conformity to the formula (14.2.44), the velocity of the

mass centre vanishes) leads to

dcos cos

ϕϕ

wherefrom

cos

ϕ =+,

bK= , ,constab= .

(14.2.45)

Observing that

()

1/2 1/2

22 2

dd 1

cos

1d/d

ϕ == =

⎡

+⎤

⎣

⎦

we obtain the differential equation of the trajectory of the mass centre in the form

()

+=+

Differentiating with respect to the variable x, we obtain

()

3/2

d/d

1d/d

Ryy

== −

⎡+ ⎤

⎣⎦

(14.2.45')

where R is the curvature radius of the trajectory at the point C. Using the relations

(14.2.45), (14.2.45'), one obtains the trajectory by a graphical integration; the curve

thus obtained is called a figoid, the corresponding motion being a figoidal motion.

268

MECHANICAL SYSTEMS, CLASSICAL MODELS

Let be a homogeneous circular cylinder of radius R and weight

M=Gg, situated on

a plane inclined with the angle

α with respect to the horizontal; considering the

plane-parallel motion of the cylinder, it is sufficient to make a study in a vertical plane,

which passes through the mass centre of the cylinder (Fig. 14.21). Assuming at the

beginning that sliding friction does not exist, the circular disc of radius R (to which is

reduced the cylinder in our study) glides downwards in a uniform accelerated motion,

under the action of the own weight

G and of the constraint force N. Writing the

equations of motion of the mass centre in the inertial frame of reference

Oxx

′′′

, we

find the magnitude of the normal constraint force

cos cosNG Mgαα== , (14.2.46)

as well as

d/d sin

Mv t G α

′

= , wherefrom d/d sin

vtgα

′

= . Hence, the centre C

moves frictionless as a heavy particle of mass M on a plane inclined by the angle

with respect to the horizontal; its trajectory is, obviously, the

Cx -axis, the disc having

a motion of translation, while the component of

′

along the

′′

-axis is given by

()

/2 singtραρ

′′

, where

′

corresponds to the initial moment 0t = .

If a tangential constraint force

T, applied at the point I, intervenes too, the inclined

plane being rough, we obtain the equations of motion (

I is the moment of inertia with

respect to an axis normal at C to the fixed plane in which the motion takes place; we

use a left-handed frame of reference)

14.2.2.6 Rolling on an Inclined Plane

sin

MMgT

′

=−

ITRω =



(14.2.47)

the ideal constraint force being given by the relation (14.2.46) too; here is the angular

velocity by which the disc is rolling without sliding on the inclined plane and we have

vRω

′

= , I being the instantaneous centre of rotation. We notice also that

()

11 0

Rρρ θθ

′′

=+ −

, where

θ corresponds to the initial moment 0t = , being the

angle made by the

Cx -axis with the

Cx -axis. Hence,

()

/d/d

TIRvt

′

= ;

replacing in the first equation (14.2.47), it results

269

14 Dynamics of the Rigid Solid

sin

′

λ =

(14.2.48)

for the rolling without sliding, where m is the peripheral mass of the disc (the mass

which, situated at the distance R from the centre C, leads to the axial moment of inertia

). Observing that

IMR= , it results 1/2λ = . If the disc is reduced to a

peripheral circle of mass M (corresponds to a hollow cylinder, with a very thin wall),

then we have

1λ = . We can thus state that a homogeneous full cylinder is rolling

without sliding on the inclined plane with an acceleration greater (hence, quicker) than

that of the hollow cylinder, assuming that both cylinders are rolling without initial

velocity.

Studying, analogously, the rolling without sliding of a homogeneous sphere in a

vertical plane, which contains its centre of mass, we notice that

2/5

IMR=

; we

find thus

2/5λ = , the corresponding acceleration being greater than the acceleration

of the two cylinders. Observing that

()

[]

/2 1 singtρλαρ

′′

=+ + and denoting

Fig. 14.21 Rolling on an inclined plane

t ,

t and

t the times in which a full and a hollow cylinder, a sphere and a

particle, respectively, are rolling without sliding on the same inclined plane, without

initial velocity, travelling through the same space, we obtain

cf ch

7/5 3/2

t ===

(14.2.49)

The tangential constraint force is given by

sin

TMg

(14.2.50)

270

MECHANICAL SYSTEMS, CLASSICAL MODELS

The above results hold also for a rigid solid with an axis of geometric and mechanical

symmetry and with a plane of geometric and mechanical symmetry, normal to that axis

(including thus also the case of a non-homogeneous rigid solid).

To have a rolling without sliding, it is necessary that

TfN≤

, where tanf ϕ= is

the coefficient of sliding friction, while

ϕ is the angle of sliding friction between the

cylinder and the inclined plane; hence, the condition of rolling without sliding of the

cylinder is

tan tanf

λλ

αϕ

λλ

≤=

(14.2.51)

αϕ

′

≤ ,

tan tan

ϕϕ

′

, ϕϕ

′

> .

(14.2.51')

Corresponding to the results in Chap. 4, Subsecs 1.1.8 and 2.1.6 too, a particle would

slide on the inclined plane if

ϕα< ; if αϕ

′

> , the cylinder is rolling with sliding.

Experimentally, one sees firstly that, for a sufficiently small angle

α, the cylinder

remains at rest, due to the apparition of a moment of rolling friction

M , which equates

to zero the couple TR ; observing that, in this case,

′

= , the first equation (14.2.47)

gives

sinTMg α= , so that the respective couple will be sin tanMgR NRαα= . For

an angle

αα> ,

tan

MNRα=

, the cylinder recommences to roll; introducing the

coefficient of rolling friction

tansR α=

, it results

MsN= , corresponding to the

considerations in Chap. 3, Subsec. 2.2.12 and Chap. 4, Subsec. 2.1.4. The experiments

show that, in general, the influence of the rolling friction is smaller than the influence of

the sliding friction and we have

αϕ< , hence /fsR< .

If the rolling friction appears too, then the second equation (14.2.47) is completed in

the form

ITRsNω =−



, (14.2.47')

so that, analogously, we obtain

(

)

()

sin

sin cos

d1 1 cos

αα

λλα

′

−

=−=

(14.2.48')

Comparing the relations (14.2.48) and (14.2.48') and observing that

()

sin sin cosαα α α−< , it results that the intervention of the rolling friction

diminishes the acceleration of the mass centre of the disc. The tangential constraint

force will be given, in the same way, by

(

)

sin cos

Tm N

tR R

λα α

′

=+= +

(14.2.50')

271

14 Dynamics of the Rigid Solid

The condition of rolling without sliding (

TfN≤

) becomes

(

)

()

tan tan tan tan

αϕϕα

λλ

≤+ − = + −

(14.2.52)

αϕ

′′

≤ ,

()

tan tan tan tanϕϕ ϕα

′′

=+ −

(14.2.52')

where we suppose that

ϕα> . Finally, for

αα≤

the disc is at rest, for

ααϕ

′′

<≤ the disc is rolling without sliding, while for αϕ

′′

> the rolling takes

place with sliding, assuming a null velocity at the initial moment.

The above results can be put in connection also with the study of the equilibrium

problems of the drawn and of the motive wheels (see Chap. 4, Subsec. 2.1.4).

If the cylinder considered at the preceding subsection lies on a horizontal plane (the

case

0α = ), then the problem has a different character; indeed, if upon the cylinder (a

disc in a vertical cross section) acts only its own weight

M=Gg as a given force, the

initial velocity being equal to zero, this one remains at rest. To roll on the horizontal

plane, the disc must be acted upon also by other given forces (which will be considered

constant in time), of torsor

{

}

FM at the mass centre C (Fig. 14.22); as well, the

initial velocity can be non-zero. In the case of rolling without sliding, the contact point

I of the disc with the horizontal plane is the instantaneous centre of rotation, while the

velocity of the mass centre C (which moves along a horizontal) is given by

′

=v ω ,

(14.2.53)

where is the rotation angular velocity about the centre C; the relation takes place at

the initial moment

0t = too, in the form

′

=v ω . If the point I slides along the

14.2.2.7 Rolling on a Horizontal Plane

′′

-axis with the velocity

′

v (the disc has a motion of translation too), then – by

composition of velocities – we can write

′′

=+vv ω.

(14.2.53')

At the initial moment we have, obviously,

′′

=+vv ω . In general, arises a force

of sliding friction

T, opposite to the motion (to the velocity

′

v ) and a moment of

rolling friction of magnitude

MsN= ,

tansR α= , which, as well, is opposite to

the motion or to the tendency of motion.

272

MECHANICAL SYSTEMS, CLASSICAL MODELS

From the theorem of motion of the mass centre, it results

MTF

′

=− +

0NGF−+ =

(14.2.54)

while the theorem of moment of momentum with respect to the same centre gives (we

use a left-handed frame of reference)

IMMTR TRω =−+=+



M ,

MM=−M .

(14.2.54')

We find the normal constraint force

NGF=−

, (14.2.55)

where we assume that

GF> (otherwise, the cylinder would be detached from the

plane). Eliminating the constraint force T, it results

IMR FR

′

+=+



M .

(14.2.56)

In case of rolling without sliding (pure rolling) we can assume that at a moment t (as

well, at the initial moment

0t = ) takes place the relation (14.2.53). The equation

(14.2.56) becomes

Fig. 14.22 Rolling on a horizontal plane