Seely F.B. Analytical Mechanics for Engineers

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BALANCING

ROTATING

MASSES

371

exert

forces

on the

bearings

which

turn

are transmitted

the

machine

frame.

The

reactions

the

bearings may

be elim-

inated

balancing

the

rotating

mass.

This

may

be done

the

addition

single

mass

diametrically opposite

(Fig.

3846),

the center

gravity

being

distance

from

the

axis of

rotation,

such

that,

TFi._ ,

But,

since

a common

factor,

the

conditions

for

running

dynamic

balance

may

expressed by

the

equation

nf our o

1 o

Mirior

rior=

-If

shaft

carries

Now,

is evident

from

Fig.

384(6),

this

equation

expresses

the

condition

for

standing

balance

also.

Thus,

shop

method of

obtaining

approximate

running

balance

with

rotating

member,

which

the

material is

substantially

plane

of rotation

such as

disc,

pulley,

flywheel,

consists

drilling

out

material on the

heavy

side or

adding

material on the

light

side until

standing

bal-

ance

is obtained.

Two

Rotating

Masses.

Centrifugal

Couple.-

two

rotating

masses

and

different

planes

rotation but

the

same axial

plane (Fig.

385)

and, further,

if the

centrifugal

forces

Miriu

and

of the

two

masses are

equal,

then

the

shaft

subjected

to an

un-

balanced

centrifugal couple

which

resisted

equal

couple

exerted

the

bearings.

Or,

the

two

rotating

masses

are to be

balanced,

two

additional

masses,

and

must be

introduced

the

same

axial

plane (Fig.

385)

such that

the

centrifugal couple

Msraco

^ or

M^r^b

which

they

set

equals

the

centrifugal couple

and

Hence,

Or,

FIG.

385.

372

FORCE,

MASS,

AND

ACCELERATION

And,

before, omitting

the

common factor

may

write,

Wiria

Wznb,

etc.

It will

noted

that the

shaft

when

carrying

only

the two masses

and

M^.

standing

balance

but not

running

balance.

163. Several

Masses

Single

Plane

of Rotation.

If sev-

eral masses

MI,

M%,

Afa,

etc.,

lie

the same

transverse

plane

(Fig.

386a),

the

shaft

subjected

the

centrifugal

forces,

Mirico

.M^co

Tkfsrsco

etc.,

which

form a

concurrent

system

of forces.

The

condition

that

such a

force

system

shall balance is

that the

force

polygon

shall

close.

That

is,

the vectors

representing

the

, etc.,

if laid off in

succession,

each

its

proper

direction,

shall

form a

closed

polygon.

Or,

since

Mru

may

be written

rco

and

since is a factor common

to the

expres-

sion

for

each

force,

the

products TFiri,

W2f2,

etc., may

used

instead

the

actual forces.

Thus,

let

the

four masses

as shown

Fig.

386

(a)

be a

system

of masses which rotate

a transverse

plane.

Suppose

that the

products

Win, Wir<y,,

Wsrs,

and Wr

when

laid off

vectors

(Fig.

3866)

do not

form a

closed

polygon.

is evident

then

that the four masses are not

running

balance.

In order to balance the

system

masses,

mass,

MQ,

must be added

distance, ro,

such

that the

product

Woro

represented

both

MASSES

IN DIFFERENT TRANSVERSE PLANES 373

magnitude

and

direction

the

closing side, EA,

the vector

polygon. By assuming

convenient value

for

TO,

a value of

may

be found.

The

gap,

EA,

may

closed,

however,

two

vectors

from

which

two

or more masses

may

found that

will balance the

system.

164.

Masses

in Different Transverse

Planes. In

Fig. 387,

let

the

masses

and

M<z

be connected

with

the

shaft at A

and

respectively,

and

through

some

point,

the

shaft let a

FIG.

387.

transverse

plane,

called

reference

plane,

be chosen.

The mass

exerts a

kinetic

load

centrifugal force, F\,

the

shaft

such that

Now at

introduce

two

equal

and

opposite

forces each

equal

and

parallel

FI.

The

force

and

the

equal opposite

force

form

couple, Ci,

the

moment of

which

Thus the

single

force

replaced

equal parallel

force

at and

the

couple

. In

like manner

the

single

force

M^r^^

may

replaced

the

equal

parallel

force

the reference

plane

and a

couple

^r^a^.

Thus,

the

kinetic loads

shaft

exerted

system

rotating

masses

may

be reduced to

system

of concurrent

forces

374

FORCE,

MASS,

AND

ACCELERATION

arbitrarily

chosen

reference

plane

and a

system

couples

which he

different axial

planes.

The

resultant of

the

system

concurrent

forces,

not

balanced,

is a

single

force

the refer-

ence

plane,

and the resultant of the

system

couples,

not bal-

anced,

is a

single couple

some

axial

plane.

Hence,

general,

the

system

of kinetic

loads,

not

balanced,

may

be reduced

to a

single

force and a

couple.

The

magnitude

of the

couple

will,

course, depend

on the

position

chosen

for

the reference

plane.

The

conditions then

which

must be

fulfilled to have a

system

rotating

masses

equilibrium are,

(1)

The resultant

the

system

concurrent

forces must

be zero.

That

is,

the force

polygon

for

the forces

the reference

plane

must close.

(2)

The resultant

the

system

couples

must be zero.

That

is,

the

couple

polygon

must close.

These conditions

may

be satisfied

the addition

two

rotating

masses in

different

transverse

planes. Thus,

let the shaft

(Fig.

388)

carry

unbalanced

system

rotating

masses,

MI,

and

:M 2

f)M

FIG.

388.

MS,

and let the

two

balancing

masses be denoted

and

M'Q.

Let the

transverse

planes

which these

balancing

masses

are to

lie

be chosen

arbitrarily

but let the

plane

of one of the

masses be

chosen

MASSES

DIFFERENT

TRANSVERSE

PLANES

the

reference

plane.

The

plane

mass

will

here be

selected

the reference

plane.

Let

OQ, i, &2,

and

denote the

respective

distances

the

masses

MQ,

MI,

and

from

the

reference

plane.

The

couples

W\r\ai

y Wtfiai,

etc.,

may

now

calcu-

lated

(the

common

factor

is omitted

each

term for

simplicity).

The

only

unknown

couple

TFoWo,

since the

couple

is zero

due

to the

fact that the reference

plane

was

chosen as the

plane

M'o,

which

makes

a'o

equal

to zero.

laying

off

the

vectors

that

represent

the known

couples

the sides

polygon,

the

closing

side

determines both

the

magnitude

the

unknown

couple

Woroao

and the

axial

plane

which it

lies.

By assigning

any

convenient

value

to the moment

arm

OQ,

the

product

Woro

may

found,

and

assuming

convenient value for

ro,

may

be determined

and

placed

the

plane

indicated

the

closing

vector of the

couple

polygon.

Thus,

the addition of

the

couple

WoToOo

condition

(2)

is satisfied.

Condition

(1)

may

now be satisfied as

follows: The kinetic

load due to

the added mass

replaced

force

the refer-

ence

plane

and the

couple

Moroao

was done

for the

kinetic loads

due to the

original

masses.

Now,

the

products

W\ri,

Wtfi,

etc.

(including

Woro)

are

laid off as

the sides of a

polygon,

the

closing

side

gives

the

magnitude

and

the direction

of a

product

V'o

for a

body

W'o

which must be

added

the

reference

plane

to balance the

system

of concurrent forces in

the

reference

plane

and

thereby satisfy

condition

(1).

choosing

convenient

value

for

r'o,

the mass

M'o

may

determined

and

the

direction

of the

closing

side of the

polygon gives

the

direction

M'Q

from

the

axis of the shaft.

the

couple polygon

formed

drawing

the

couple

vectors

perpendicular

the

planes

of the

couples,

explained

Art.

17,

and then

turned

through

90,

it will

the

same

the

polygon

formed

by drawing

the

couple

vectors

according

the

following

rule:

Draw the

couple

vectors

parallel

the

respective

crank

directions;

outwards for masses on

one

side of

the

reference

plane;

inwards

towards the axis for

masses

the

opposite

side of the

reference

plane.

The vectors

the force

polygon

are

drawn,

course,

from

the

axis

outwards

parallel

to the radii

the

masses,

376

FORCE,

MASS,

AND

ACCELERATION

ILLUSTRATIVE

PROBLEM

419.

Three

weights

and

(Fig.

389)

which

revolve in

the

planes

and

are

to be

balanced

the addition of

two

weights.

Plane 1

chosen

as the

plane

one of

the

weights (W

)

and

the

plane

the

other

weight

(W'o)

will

arbitrarily

be taken

1.4

ft. to the

right

plane

3 and will

selected

as the

reference

plane.

It is

required

determine values

of W

and

and the

lengths

and

directions of

the

corresponding

radii for

kinetic

balance. The

accompanying

table

gives

the

values of the

weights,

the

lengths

and

directions

the

radii,

and the

distances

from the

reference

plane.

Plane.

MASSES

DIFFERENT

TRANSVERSE

PLANES 377

planes

and

radial

distances

8 in.

and 5

in.,

respectively,

as indicated

Fig.

389,

the

system

will be

in kinetic

balance.

378

FORCE,

MASS,

AND

ACCELERATION

Determine

sufficient

data for

kinetic

balance of

the

system

bodies:

(a)

the

addition of

single

weight

placed

ft.

from

the

axis of

rotation.

(6) By

the addition

of two

weights,

one of 2

Ib.

placed

the

?/-axis,

and

the other of

2.25 Ib.

placed

ft.

from

the

axis

rotation.

421.

Four bodies

are

attached

revolving

shaft

different

transverse

planes.

The

weights

and

positions

the

bodies

are

indicated in

the

following

table,

the

reference

plane being

the

transverse

plane

which

the

mass-center

the 6-lb.

body

lies.

FORCES IN

THE GYROSCOPE

379

The

gyroscope

here considered

is a

body

symmetrical

with

respect

each of

three

rectangular

axes.

The

body

rotates or

spins

with

constant

angular

velocity

about

one

of the

axes

and

the same time

turns about

one of

the other

axes with

constant

angular velocity.

Thus,

Fig.

390,

let

BCD

represent

a circular

disc

which

symmetrical

with

respect

to the three

axes

x, y,

and

Let

the disc rotate

spin

with

high

constant

angular

velocity

FIG.

390.

about

the

z-axis

and at the

same time let

the disc turn

about the

y-axis

with

the

constant

angular

velocity

12.

The

problem

which

arises

is that

determining

the forces

which

must act on

the disc

its

axles

(axle

reactions)

order

to maintain

this

gyroscopic

motion.

166.

Analysis

of Forces

in the

Gyroscope.

Due to

the

two

rotations

imposed

the

disc,

particle, m,

at the distance

(assumed

the

circumference

the

disc for

convenience)

has

any

instant,

two

velocities:

(1)

constant

velocity

the

plane

the

disc

due

the rotation about

the

z-axis with

angular

380

FORCE, MASS,

AND

ACCELERATION

velocity co,

and

(2)

the

velocity

ftp

cos

perpendicular

the

disc

(parallel

to the

z-axis)

due to the

rotation

about the

2/-axis

with

angular

velocity

ft.

The

changes

that occur

each

these two

velocities of

any

particle

due to each of

the rotations

which

are

given

the disc

will

first

investigated,

since the

effective force for

any particle

must

have

components

corresponding

to the

change

the

mag-

nitude

and to the

change

the

direction of each of

these

two

velocities

(Theorems

and

II,

Art.

118,

and Newton's

second

law,

Art.

140).

And,

the

effective forces for

the

particles

must

found in

order to

determine

the external

forces

required

to main-

tain

the

gyroscopic

motion.

(Read

Art.

144

for the

general

method

procedure,

keeping

mind that

gyroscopic

motion

not

uni-

planar

motion.)

will

found,

however,

that

external

forces

come into action

as the

result of

only

two

the

changes

which

occur in

the two

velocities;

namely,

the

change

ftp

cos 6

due

the rotation

about the

z-axis

and the

change

cop

due to

the

rotation

about the

i/-axis.

CHANGES IN

THE

VELOCITIES DUE TO

ROTATION

ABOUT

Z-AXIS

(1)

Change

cop.

Since co is

constant,

cop

changes

direction

only.

The

resulting

acceleration

(co

and the

corresponding

effective

force

for

the

particle,

directed

towards

the

center of

rotation

(towards

the

z-axis).

And,

since the

body

symmetrical,

these

effective forces for

all the

particles

form a

balanced

system

and

hence,

according

D'Alemberfc's

principle,

external

forces

act on

the

body by

reason of

this

change

the

velocity

of the

particles.

(2)

Change

ftp

cos

6. In

one

revolution of the

disc

about

the

z-axis,

this

component

the

velocity

changes,

follows: It

gradually

increases

magnitude

downwards from

zero

to a

maximum

it then

decreases

gradually

to zero at

then

increases

the

opposite

direction to a

maximum

;

and

finally

decreases to

zero

again

The

acceleration,

any

instant

then

(Theorem

Art.

118),

d(ftp

cos

-dB

ai=

-r

ftp

sin

Q-j-

ftp

sin 0co

cofti/,

which

is the

distance of the

particle

from the

z-axis,

and

the