DYNAMICS OF ROTATING SYSTEMS questions Mechanical Principles

MODULE TITLE : MECHANICAL PRINCIPLES

TOPIC TITLE : DYNAMICS OF ROTATING SYSTEMS

LESSON 5 : EFFECTS OF COUPLING

MP - 4 - 5

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

In this lesson, we shall consider coupled mechanical systems, in which the

components may rotate in different directions. In such systems, the component

motions are coupled by transducers of various forms. A transducer is defined

as a device which transforms energy or power from one form into another

form, so it also called a mechanism. A pure transducer performs this function

without energy loss or storage, so that the net power flow in a pure transducer

is zero.

The applications of such mechanical transducers are:

• to change the speed of a motor or other power source to meet the

need for a lower or higher output speed (as in automobile

transmission which couples the engine to the driving wheels)

• to obtain a non-uniform motion from a uniform one or vice versa.

This is a common application in automatic machinery

• to achieve a mechanical advantage, i.e. an increase in torque level.

Common examples of transducers coupling bodies in different translational

motions are:

• translatory-to-translatory mechanisms; levers and pulleys

• translatory-to-rotary mechanisms; rack-and-pinion, pulley-and-

sprocket, lead-screw, slider-crank, or cam assemblies

• rotary-to-rotary mechanisms; gear trains, belts and pulleys, sprockets

and chains, or harmonic drives.

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The first two mechanisms have been discussed in the previous lesson. Now,

we will deal with the last one.

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YOUR AIMS ________________________________________________________________________________________

After studying this lesson, you should be able to:

• understand the principle of conservation of angular momentum

• determine the effects of coupling freely rotating systems

• calculate the energy loss due to coupling and the final common

rotational speed.

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COUPLED ROTATIONAL SYSTEMS ________________________________________________________________________________________

Rotary-to-rotary coupling is the coupling between one rotational motion about

a fixed axis and another rotational motion about the same or a different axis.

The simplest and most familiar example of a rotary-to-rotary coupling is the

gearbox assembly with rotors as shown in FIGURE 1.

FIG. 1

Gearbox

T i

ω i

Input Output

Rotors

T o

ω o

Hold-down torque T h

Rotors

Hold-down torque T h

Input Output

Gearbox

T i

ω i

T o

ω o

(a)

(b)

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Because we are already armed with the necessary mechanical knowledge from

our previous study on the dynamics of rotating systems, we are now able to

examine the torque, angular velocities, momentum and energy change

involved in the coupled systems. The following examples will show you how

to draw together and consolidate what we have learnt in the earlier lessons.

GEARBOX ASSEMBLY WITH ROTORS

Consider the gear system shown in FIGURE 1, in which the gearbox has an

input torque of Ti at an angular velocity of ωi and an output, i.e. the resisting torque on the shaft, of To at an angular velocity of ωo. The input torque is in the same direction as the angular velocity, the output resistive torque is,

however, in the opposite direction to the output angular rotation. FIGURE 1(a)

shows the case when the input and output shafts rotate in the same direction,

whereas FIGURE 1(b) shows the situation when they rotate in opposite

directions.

For the gearbox, the transmission efficiency η is defined by

or

The negative sign means that the output torque To acts in the opposite sense to

the output angular velocity ωo.

Note that if the input and output shafts rotate in opposite directions, ωi and ωo will have opposite signs.

η ω

ω

ω η

= =

+

power output power input

o o

i i

o o i

–T

T

T T ωω i = 0

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FIG. 2 Free-body diagrams

FIGURE 2 shows the free-body diagrams for the system in FIGURE 1. For

the input rotor or flywheel in FIGURE 2(a), the angular acceleration depends

on the resultant torque acting on it, so

The output rotor or flywheel rotates freely as shown in FIGURE 2(c), then we

have

Referring to FIGURE 2(b), the net torque acting on the gearbox must be zero

as it should be stationary. Thus, we must have a hold-down torque Th that

prevents the gearbox from rotating. Hence

The direction of Th depends on the sizes and directions of Ti and To.

T T Ti o h+ + = 0

T Io o o= α

T T I– i i i= α

(b) Gearbox

T i

ω i

T o

ω o

T h

(a) Input rotor (c) Output rotor

T o

ω o

T i

ω i

T

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In the above equations,

I = the moment of inertia (kg m2)

T = the torque acting on the shaft (N m)

α = the angular acceleration (rad s–2) η = the efficiency of the gearbox ω = the angular velocity (rad s–1).

Example 1

Determine the holding torque required for a gearbox, when the input power is

1 kW at 40 rev s–1, and the angular velocity at the output is 20 rev s–1, in the

same direction as the input. The transmission efficiency of the gearbox is

95%.

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Solution

The input power is Tiωi, hence

thus

The output power can be determined by

where ωi and ωo have the same sign in this case as the input and the output shafts rotate in the same direction. Then, we have:

The direction of the output torque is opposite to the input.

Since the net torque on the gearbox must be zero, the holding torque is thus

given by

then

The direction of the holding torque is the same as the input torque.

T T T

T T T

i o h

h i o

+ + =

= +( )

= ( )

=

0

3 979 7 560

3 581

–

– . – .

. NNm

T T

o i i

o

Nm= = × ×

= –

– .

– . η ω

ω 0 95 10

2 20 7 560

3

π

T To o i iω η ω+ = 0

T T

T

i i i

i Nm

ω = × × =

=

2 40 10

3 979

3π

.

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FLYWHEEL AND CLUTCH SYSTEM

Consider two flywheels, initially rotating independently, that are connected by

engaging a clutch, as shown in FIGURE 3.

FIG. 3

Before the engagement, the two flywheels are rotating at the angular velocities

ω1 and ω2, respectively. Thus, the total angular momentum is I1ω1 + I2ω2.

After connection, there will only be a single moment of inertia I3 at a single

angular velocity ω3 in the system. Then, the final angular momentum will be I3ω3.

Applying the principle of conservation of angular momentum, we have

Since

then

I I I

I I I

I I I I I

1 1 2 2 3 3

3 1 2

1 1 2 2 3 3 1 2

ω ω ω

ω ω ω

+ =

= +

+ = = +( )) ω 3

Clutch

Flywheels

I 1

I 2

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Although angular momentum is conserved, energy is not. There is a loss of

energy, in the form of angular kinetic energy, during coupling due to friction

between the plates in the clutch, which is

In the special case that two identical flywheels are rotating at the same speed

but in opposite directions, when the clutch is engaged both flywheels will be

brought to a stop. The clutch will effectively act as a brake that dissipates all

the kinetic energy as heat energy because of friction.

From preceding work we have:

∴ change in angular momentum due to bearing friction = Tf t

where Tf = friction torque

t = time.

Example 2

A flywheel with the moment of inertia of 10 kg m2 rotates at 100 rpm. A

second flywheel has a moment of inertia of 12 kg m2 and rotates at 200 rpm

but in the opposite direction. Determine the final rotational speed when they

are coupled by engaging a clutch. Also, calculate the energy loss during

coupling.

torque change in angular momentum

time

f

=

∴ rriction torque change in angular momentuTf = mm due to friction

time

∆KE I I I

I I I

= +( )

= +

1 2

1 2

1 2

1 2

2 2

3 2

1 2

2 2

1 2 3

1 2

ω ω ω

ω ω

–

– 33ω 32( )

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Solution

According to conservation of angular momentum, we have

From the question, it is known that

Then, the final rotational speed is

The rotating direction after the clutch engagement is same as the second

flywheel.

ω ω ω

3 1 1 2 2

1 2

= + +

= × × +

=

I I

I I

10 10 47 12 20 94 10 12

6

. – .

– ..

– .

66

63 6

rad s

rpm

–1

=

I I1 210 12

100 2 60

10 4

= =

= × =

kg m kg m2 2

1

,

.ω π 77

200 2 60

20 94

rad s

and rad s

–1

2 –ω = × =– – .π 11

I I I I I1 1 2 2 3 3 1 2 3ω ω ω ω+ = = +( )

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The energy loss during coupling is calculated by

Note that I3 = 10 + 12 = 22 kg m 2, thus,

Example 3

Clutch plate A has a mass of 30 kg, a radius of gyration of 0.12 m and rotates

at 900 rev min–1. Clutch plate B has a mass of 36 kg, a radius of gyration of

0.15 m and is stationary. The plates become engaged and slipping ceases after

1.2 seconds. Assuming that there is a total constant frictional torque, during

the 1.2 seconds, of 1.5 N m at the bearings, determine:

(a) the common angular velocity after engagement

(b) the loss of kinetic energy during engagement.

∆KE = × + × ( ) × ( )⎡⎣ ⎤ 1 2

10 10 47 12 20 94 22 6 662 2 2. – . – – . ⎦⎦

= 2691 1. J

∆KE I I I= +( )1 2 1 1

2 2 2

2 3 3

2ω ω ω–

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Solution

Moment of inertia plate A

Initial angular velocity of plate A

Moment of inertia of plate B

Initial angular velocity of plate B

Loss of angular momentum due to friction

Now, angular momentum before engagement

where ω = common angular velocity after engagement.

∴ + = +( ) + a a b b a bI I I Iω ω ω 1 8.

= +angular momentum after engagement angullar momentum lost due to friction during enggagement

=

= ×

=

T tf

1 5 1 2

1 8

. .

. –kg m s2 1

ω b = 0

I m kb b b

2kg m

=

= ×

=

2

236 0 15

0 81

.

.

ω a

1rad s

= ×

=

900 2 60

30

π

π –

I m ka a a

2

2

30 0.12

kg m

=

= ×

=

2

0 432.

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Loss of kinetic energy during engagement

= kinetic energy before engagement – kinetic energy after engagement

Loss of kinetic energy during engagement = 1309 J.

= + +( )

= × × ( )

1 2

1 2

1 2

1 2

0 432 30

2 2 2I I I Ia a b b a bω ω ω–

. π 22 20 1 2

0 432 0 81 31 34

1919 610

1309

+ +( )

=

=

– . . .

–

J

Common angular velocity after engagement 31= ..34 60 2

rev min 1

×

=

π

299 3. –

∴ × + = +( ) +

=

0 432 30 0 0 432 0 81 1 8

40 72 1 2

. . . .

. .

π ω

442 1 8

40 72 1 8 1 242

31 34

ω

ω

+

=

=

.

. – . .

. –rad s 1

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. A shaft has a moment of inertia of 6 kg m2 and rotates clockwise at

100 rad s–1. It is engaged, by means of a clutch, with a shaft of moment

of inertia 10 kg m2 which is rotating anti-clockwise about the same axis at

200 rad s–1. After engagement both shafts rotate together with a common

angular velocity. Determine:

(a) the common angular velocity after engagement

(b) the loss of kinetic energy during engagement.

2. Calculate the holding torque required for a gearbox, when the input power

is 2 kW at 120 rpm, and the output shaft rotates in the opposite direction

at 20 rpm. Assume the transmission efficiency of the gearbox is 98%.

3. A clutch plate A has a mass of 30 kg and a radius of gyration of 0.12 m.

It rotates at 900 rev min–1 and becomes engaged with a stationary clutch

plate B of mass 20 kg and radius of gyration 0.1 m. Slipping of the plates

ceases after 2.2 seconds and the plates then rotate with a common angular

velocity of 560 rev min–1. Determine:

(a) the loss of angular momentum during engagement

(b) the steady frictional torque at the bearings during engagement

assuming that the loss of angular momentum is due entirely to

bearing friction

(c) the average angular acceleration of plate B during engagement.

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________________________________________________________________________________________

NOTES ________________________________________________________________________________________

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ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. (a) Angular momentum before engagement

where ω is the common angular velocity after engagement

Common angular velocity after engagement is 87.5 rad s–1 anti-

clockwise.

(b) Kinetic energy before engagement

= +

= × ×⎛⎝⎜ ⎞ ⎠⎟ + × ×

1 2

1 2

1 2

6 100 1 2

10 200

2 2

2

I Ia a b bω ω

–(( )⎛⎝⎜ ⎞ ⎠⎟

= +

=

2

30 000 200 000

230 000 J

6 100 10 200 6 10

600 2000 16

1

×( ) + × ( )( ) = +( )

=

=

–

–

–

ω

ω

ω 4400 16

87 5= – . –rad s 1

=

+

angular momentum after engagement

a a b bI Iω ω == +( )I Ia b ω

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2. Since the input power is Tiωi and rad s–1, then,

thus

The output power can be determined by

Note that from the question

Then

The output torque has the same direction as the input.

T T

o i i

o

= = × × ×⎛

⎝⎜ ⎞ ⎠⎟

=– – .

– .

η ω ω

0 98 2 10 20 2

60

935 8 3

π 33 Nm

ω o –1 rpm rad s= = ×

– – .20 20 2

60 π

T To o i iω η ω+ = 0

T T

T

i i i

i Nm

ω = × × = ×

=

120 2 60

2 10

159 15

3π

.

ω i = ×120 2

60 π

Kinetic energy after engagement a b= +( )12 2I I ω

== +( ) × ( )

=

1 2

6 10 87 5

61250

2 – .

J

Loss of kinetic ennergy 230 000 61250

168 750 J

168.75 kJ

=

=

=

–

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Since the net torque on the gearbox must be zero, the holding torque is

thus given by

then

The direction of the holding torque is opposite to the input and output

torques.

3. (a) I m ka a a

2

a

kg m

r

=

= ×

=

= ×

=

2

230 0 12

0 432

900 2 60

30

.

.

ω π

π aad s

kg m

1

b b b

2

b

–

.

.

I m k=

= ×

=

=

2

220 0 1

0 2

0ω

T T T

T T T

i o h

h i o

+ + =

= +( ) = +( )

=

0

159 15 935 83

109

– – . .

– 44 98. Nm

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(b) Loss of angular momentum due to friction

= ×

∴ = ×

=

friction torque time

3.65 T

T

f

f

2 2

3 6

.

. 55 2 2.

Friction torque 1.659 N m=

Loss of angular momentum

k

=

=

40 71 37 06

3 65

. – .

. gg m s2 1–

Angular momentum before engagement a a b= +I Iω ω bb

1s

Angular momentum

= × +

=

0 432 30 0

40 71 2

.

. kg m –

π

after engagement a b= +( )

= +( )

I I ω

0 432 0 2 58 6. . . 44

37 06 2= . kg m –s 1

Common angular velocity after engagement ω = 5660 2 60

58 64

×

=

π

ω . –rad s 1

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(c) Average angular acceleration of plate B = ω ω– bb

2rad s

t

=

=

58 64 0 2 2

26 65

. – .

. –

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

The aim of this lesson was to consolidate and intensify our knowledge of

power transmission systems and dynamics of rotating systems.

By using some examples, we have analysed the effects of coupling in

rotational systems, which involves calculations on conservation of angular

momentum, mechanical energy loss and torque balance during coupling.

Now, we have completed our study of this module.

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