DYNAMICS OF ROTATING SYSTEMS questions Mechanical Principles

MODULE TITLE : MECHANICAL PRINCIPLES

TOPIC TITLE : DYNAMICS OF ROTATING SYSTEMS

LESSON 4 : FLYWHEELS

MP - 4 - 4

© Teesside University 2011

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School of Science & Engineering

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

A flywheel is a heavy rotating disk used to store energy in the form of kinetic

energy, so in essence it is a mechanical battery. FIGURE 1 shows an example

of a flywheel attached to the shaft of a motor.

By increasing or decreasing its angular velocity, it absorbs or gives out

mechanical energy. The mechanical energy transmitted by a torque is normally

provided by a motor. The torque increases the flywheel inertial energy, which

can be used to drive a load after the accelerating torque has been removed.

FIG. 1

Reproduced by kind permission of Dirac Delta Consultants Ltd.

(www.diracdelta.co.uk)

Flywheels resist changes in their rotation speed, which helps steady the

rotation of the shaft when an uneven torque is exerted on it by its power

source, such as a piston-based (reciprocating) engine, or when the load placed

on it is intermittent (such as a piston-based pump). In the Industrial

Revolution, James Watt contributed to the development of the flywheel in the

steam engine. Later, the flywheel was combined with a crank to transform

reciprocating motion into rotary motion, as shown in FIGURE 2.

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FIG. 2 A labelled schematic diagram of a typical single cylinder

simple expansion, double-acting high pressure steam engine.

Power takeoff from the engine is by way of a belt.

Flywheels, one of the oldest and most common mechanical devices in

existence, may still prove to be as an important component for tomorrow’s

vehicles and future energy needs.

This lesson is about the use of flywheels to smooth out the rotation of

machines subjected to an erratic torque. In these machines, such as piston

engines, gas compressors and reciprocating pumps, the torques acting on the

shafts go through a cycle change. In the example shown in FIGURE 2, the

flywheel is used to smooth out the motion and keep the variations in the shaft

speed within acceptable limits.

1. 2. 3. 4. 5. 6. 7. 8. 9.

10.

Piston Piston rod Crosshead bearing Connecting rod Crank Eccentric valve motion Flywheel Sliding Valve Centrifugal governor Belt

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________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

After studying this lesson, you should be able to:

• define angular momentum and kinetic energy in rotating systems

• understand the principle of conservation of angular momentum

• determine the energy storage requirements of flywheels

• calculate flywheel mass and dimensions to give required operating

conditions.

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________________________________________________________________________________________

ANGULAR MOMENTUM AND KINETIC ENERGY ________________________________________________________________________________________

ANGULAR MOMENTUM

In dealing with linear motion we have seen that the product of mass and

velocity, mv, is termed ‘momentum’. In rotary motion we have a similar

quantity termed ‘angular momentum’ (sometimes called ‘moment of

momentum’) and this is the product of moment of inertia and angular velocity.

Let us investigate this in greater detail:

Consider a body which rotates about centre O with an angular velocity ω. Consider an elemental mass δm in the body at a distance r from O. The linear velocity (v) of this mass will be rω as shown in FIGURE 3.

FIG. 3

Linear momentum mass velocity

linear mo

= ×

∴ mmentum of elemental mass = ×

= × ×

δ

δ ω

m v

m r

O r ωδm

v = rω

Thus, angular momentum moment of inertia= ××

=

velocity

L Iω

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This 'moment' of momentum is termed ‘angular momentum’.

For the whole body, angular momentum about O:

where: I = moment of inertia (kg m2)

ω = angular velocity (rad s–1) k = radius of gyration (m).

Hence the units for angular momentum are kg m2 s–1.

Note: the radius of gyration is defined as the radius, at which the mass or

masses are situated, from the centre of rotation of the body. For most

engineering components, the mass is distributed and not concentrated at any

particular radius, so k is normally different from the radius r, as shown in

FIGURE 4.

L mr

mr

= ∑

= ∑

δ ω

ω δ ω

2

2 (since is constant for all elements of the body)

Now moment of inertia

A

δ mr I∑ = ( )

∴

2

nngular momentum

and since t

L I

I mk

=

=

ω

2 , hhen L mk= 2ω

i.e. angular momentum L m r= ×δ ω2

Moment of this linear momentum about O = ×δ m r ×× ×

= × ×

ω

δ ω

r

m r 2

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FIG. 4 Definition of the moment of inertia for some engineering components

Rim type flywheel, I = m k2

and r m

= k, so I = m r2

r m

= mean radius

X X

m

X

X

r Point mass at set radius r

I = m k2 and k = r so I = mr2

Solid disc where k = I and I = m k2

so I = m r2

X X r

1

2

√2

Thin plate I = m a2

where k = 0.577 a 2

X

X a

r

X

XX Sphere I = 2

where k = 0.6345 r 5

X

X

l

2l

2

Thin rod I = 1 ml2

where k = 0.289 l 12

mr2

X

m

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KINETIC ENERGY

A flywheel is essentially a device for storing angular kinetic energy KE, which

can be expressed as

where I is the moment of inertia of the mass about the centre of rotation and ω is the angular velocity in radian units. A flywheel is more effective when its

inertia is larger, as when its mass is located further from the centre of rotation,

either due to a more massive rim or due to a larger diameter.

Note the similarity of the above formula to the kinetic energy formula

where linear velocity v is comparable to the rotational velocity ω, and the mass m is comparable to the rotational inertia I. This similarity has been mentioned

in our previous study.

The angular velocity is given by:

where ω = angular velocity (rad s–1) N = rotational speed of the flywheel (revolutions per minute).

ω = 2 60 πN

KE mv= 1 2

2

KE I= 1 2

2ω

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The angular acceleration is calculated as

where α = angular acceleration (rad s–2) ω1, ω2 = angular velocity at time 1 and 2, respectively (rad s–1)

t = time between 1 and 2 (s).

PRINCIPLE OF CONSERVATION OF ANGULAR MOMENTUM

This states that:

The angular momentum of a system, about a given axis, remains constant

provided there is no externally applied torque on the system.

In other words, there may be changes in the angular momentum of the

individual parts of a system, but the total angular momentum of the system

will remain constant unless an external torque is applied.

The following problems will illustrate the application of this principle.

Example 1

A and B are two separate clutch plates. Plate A has a mass of 40 kg, a radius

of gyration of 0.12 m and is stationary. Plate B has a mass of 36 kg, a radius of

gyration of 0.15 m and rotates at 720 rev min–1. The plates become engaged

and then rotate at a common angular velocity. Calculate:

(a) the common angular velocity after engagement

(b) the loss of kinetic energy during engagement.

α ω ω

= 2 1 –

t

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Solution

Assuming there is no externally applied torque then:

angular momentum before engagement = angular momentum after engagement

where ωA, ωB = angular velocity of A and B before engagement, repectively ω = common angular velocity after engagement

(IA + IB) = total moment of inertia of combined plates

40 0 12 0 36 0 15 24 40 0 12 36 0 152 2 2 2× × + × × = × + ×( ). . . .π ωω

ω

ω

ω

0 61 07 0 576 0 81

61 07 1 386

61 07 1

+ = +( )

=

=

. . .

. .

. ..

. –

386

44 06= rad s 1

I m k I m kA A A 2

B B B 2 = =,

∴ + = +( ) A A B B A BI I I Iω ω ω

Initial angular velocity of A

Initial a

A, ω = 0

nngular velocity of B, 720 2

rad s

B

1

ω = ×

=

π 60

π24 –

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Loss of kinetic energy (KE) during engagement:

Example 2

A motor has rotating parts with a moment of inertia value of 0.3 kg m2 and

which rotate at a speed of 420 rev min–1. The motor shaft is suddenly

connected to a flywheel, positioned co-axially. If the flywheel has a mass of

10 kg and a radius of gyration of 0.15 m and is rotating initially at

90 rev min–1 in the opposite direction to the motor shaft determine:

(a) the common speed after engagement, in rev min–1

(b) the loss of kinetic energy during engagement.

= total KE before engagement KE after engage– mment

initial KE of plate A initial KE of p= + llate B KE of combined plates

B B

–

–= +0 1 2

1 2

2I Iω AA B+( )

= × × ( ) × ( ) ×

=

I ω 2

2 21 2

0 81 24 1 2

1 386 44 06. – . .π

9957 1. J

Common angular velocity after engagement = 44..

.

.

–

–

–

06

44 06 2

60

420 8

rad s

rev min

rev min

1

1

1

= ×

=

π

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Solution

We shall consider the direction of rotation of the motor shaft as positive.

The negative sign means the flywheel rotates in the opposite direction to the

shaft.

angular momentum before engagement = angular momentum after engagement

where ω = common angular speed after engagement.

I

I mk

a 2

b 2

kg m

kg m

=

= = ×

0 3

10 0 152 2

.

.

I I I Ia a b b a bω ω ω+ = +( )

Initial angular velocity of motor shaft aω = 4220 2

14

×

=

π 60

π rad s

Initial angular velocity

1–

oof flywheel

rad s

b

1

ω = − ×

=

90 2

3

π 60

π– –

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This is in the same direction as the motor shaft.

=

=

KE before engagement KE after engagement

1

–

22 2 1 2

1 2

0 3 14 1

2 2 2

2

I I I Ia a b b a b 1

ω ω ω+ +( )

= × × ( ) +

–

. π 22

10 0 15 3

1 2

0 3 10 0 15 21 09

2 2

2 2

× × × ( )

× + ×( ) × . –

– . . .

π

== +

=

290 2 10 0 116 8

183 4

. . – .

. J

Loss of KE during engagement

0 3 14 10 0 15 3 0 3 10 0 15

1

2 2. . – . .×( ) + × × ( )( ) = + ×( )π π ω

33 19 2 12 0 525

11 07 0 525

11 07 0 525

2

. – . .

. .

. .

=

=

=

=

ω

ω

ω

11 09. –rad s

Common angular velocity after en

1

ggagement 21.09

rev min 1

= ×

=

2 60

201 4

π

. –

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________________________________________________________________________________________

TORQUE AND ANGULAR MOMENTUM ________________________________________________________________________________________

Consider a shaft or other rotating part subject to a number of external torques

(e.g. applied torque, friction torque). Assuming the resultant torque produces

an angular acceleration, then:

The product of torque and time is termed 'angular impulse' with the units N m s

Thus, change in angular momentum = angular impulse.

i.e. angular impulse = ×T t

Resultant torque

i.e.

T I

I

t

I I

t

=

= ( )

=

α

ω ω

ω ω

2 1

2 1

–

–

change of angular momentum

time

rate

T =

= oof change of angular momentum

Expressed mathhematically, d d

As torque change of an

T t

L= ( )

= ggular momentum time

then, change in angular mmomentum torque time= ×

= ×T t

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It follows then, that the units of angular momentum must be the same as those

for angular impulse, i.e.

Since 1 N = 1 kg m s–2 this may be confirmed as follows

Note that the units for linear momentum are kg m s–1 and clearly linear

momentum values cannot be numerically added to angular momentum values.

When an increase in torque occurs, the flywheel will speed up and absorb

energy. The grater the moment of inertia, the more energy it will absorb.

When the torque decreases, the flywheel will slow down but the inertia of the

system will limit the amount of slowing.

Example 3

A gear wheel has a moment of inertia value of 2 kg m2 and rotates at

360 rev min–1. Its speed is then uniformly increased to 480 rev min–1 in

20 seconds.

Determine:

(a) the change in angular momentum

(b) the torque necessary to produce this change (neglecting frictional resistance).

N m s kg m s m s

kg m s

2

2 1

= ×

=

–

–

kg m s N m s2 1– =

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Solution

torque rate of change of angular momentum

c

=

= hhange of angular momentum

time

to

∴ =T 8 20

π

rrque N m

1.257 N m

=

=

0 4. π

Change in angular momentum =

= ( )

I I

I

ω ω

ω ω

2 1

2 1

–

–

== ( )

=

2 16 12

8

π π

π

–

–kg m s2 1

Initial angular velocity

rad s

1ω = ×

=

360 2 60

12

π

π ––1

2Final angular velocity

rad

ω = ×

=

480 2 60

16

π

π ss 1–

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Alternative solution

T I

I t

=

= ( )

= ( )

=

α

ω ω2 1

2 16 12

20

1 257

–

–

.

π π

N m

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________________________________________________________________________________________

FUNCTION OF A FLYWHEEL ________________________________________________________________________________________

A flywheel used in a machine serves as a reservoir which stores energy during

the period when the supply of energy is more than the requirement and releases

it during the period when the requirement of energy is more than the supply.

Therefore, the function of a flywheel is to smooth out the rotation of machines

subjected to an erratic torque. We use a piston engine as an example to show

how this works.

CYCLIC VARIATION IN ENGINE TORQUE

Consider a single cylinder 4-stroke engine, as shown in FIGURE 5, in which a

complete cycle takes two revolutions (4π radians) divided into four equal strokes: power stroke, exhaust stroke, suction stroke and compression stroke.

In the power stroke (or working stroke) shown in FIGURE 6, the pressure

force P = pA acting on the piston due to the high pressure in the cylinder is

transmitted to crankpin C, creating a moment Tθ about O which turns the

crankshaft. Tθ is called the engine torque and varies with θ due to variation in gas pressure, position of crank, and inertia of piston and connecting rod. The

work output from the engine accelerates the flywheel attached to the same

shaft as the crank.

Note that the engine torque during the compression stroke is in the opposite

direction to the rotation of the shaft. During this part of the cycle, the flywheel

decelerates slightly and the negative engine torque is overcome by the reaction

torque exerted by the flywheel due to its inertia.

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FIG. 5 A single cylinder 4-stoke heat engine operation procedure

Exhaust valve open

To exhaust pipe

The piston moves up again to

exhaust the burned gases

Intake valve open

Exhaust valve

Intake gas

Both valves closed

Cylinder

The piston then moves up,

compressing the gas for

ignition

Both valves closed

Both valves closed

The expanding gas

moves the piston down, a stage called

the power stroke

Gas vapour and air mixture

A mixture of gasoline

vapour and air enters the combustion chamber as the piston

moves down

Exhaust stroke (5)

Spark plug

Piston

Connecting rod Crankshaft

Compression stroke (2)

When the gas ignites it expands

Ignition (3)

Power stroke (4)

Intake stroke (1)

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FIG. 6 A single cylinder 4-stroke engine in working stroke

For a given nominal speed N or angular velocity ω, which is controlled by the intake and exhaust valves, the engine torque Tθ actually varies considerably

throughout the cycle. Due to the cyclic variation in Tθ, there will be

fluctuations in engine speed and energy or work transmitted. FIGURE 7

shows the engine torque change with the crank angle.

Crank

(a) Contruction of the engine

Shaft

Piston Connecting rod

Crankpin C

p

ω

(b) Force and torque on the crank

P

T θ

ω

θ

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FIG. 7 The T – θ diagram for the single cylinder 4-stroke engine

For a 4-stroke engine,

where Tm is the mean torque (N m). This mean value is the torque that is

measured.

Note that the angle θ in one cycle is not always 4π but whatever angle corresponds to the cycle, e.g. for a single cylinder two-stroke engine, 4π is replaced by 2π in the above equation.

If the inertia of piston and connecting rod, and friction are neglected then work

done by gas in cylinder is equal to work done by engine torque.

cyclic work d= = ⌠

⌡ ⎮ ⎮

0

4

4 π

πT Tmθ θ

Compressor stroke

Mean torque T m

T θ

Fluctuation of energy

T

0 2π

3π 4π π θ

A B

Power stroke

Exhaust stroke

Intake stroke

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ENERGY CONSERVATION IN FLYWHEEL

Now, consider a rotating flywheel, as shown in FIGURE 8, in which Tθ is the

torque transmitted from the engine and Tr is the resisting torque.

FIG. 8

The resisting torque Tr can be the reaction torque exerted by a load that is

driven by the engine. Or, if a brake force is applied to the engine, Tr will be

the brake torque. Anyway, it shall be assumed that Tr does not fluctuate over

one cycle, i.e. is constant.

Then the resultant torque on the flywheel is

where α = angular acceleration of the flywheel (rad s–2) I = moment of inertia of the flywheel (kg m2)

If the mean speed of the flywheel is invariant, then Tr must be equal to the

mean value of Tθ , which is Tr = Tm.

T T T I= =θ α– r

Flywheel

T r

T θ

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Thus, T = Tθ – Tm = Iα and the work done by the resultant torque T on the flywheel over any interval is Tθ, which is equal to the corresponding change in its kinetic energy or energy fluctuation ∆KE, which is expressed as

Since the greatest fluctuation of energy, ,

occurs over AB shown in FIGURE 7, then the corresponding minimum and

maximum values of the rotational speed, ωmin, ωmax, over the cycle are at

points A and B.

Hence, we can say that the speed fluctuations are minimised through the use of

a flywheel. The more massive the flywheel or the larger the moment of inertia

I, the smaller the fluctuations in speed.

Now, we use a coefficient φ to describe the fluctuation of speed. It is defined as

where

Since ∆KE I

I

max max 2

min 2

max min max

= ( )

= ( ) +

1 2

1 2

ω ω

ω ω ω

–

– ωω

ω ω ω

min

max min

( )

= ( )I –

φ

ω ω

=

=

coefficient of fluctuation of speed

max ++ = ω min –1the mean speed (rad s

2 )

φ ω ω

ω = max min

–

∆KE Imax max min= ( ) 1 2

2 2ω ω–

T KE Iθ ω ω= = ( )∆ 1 2 2

2 1 2–

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After introducing φ, we have

Another coefficient β is used to characterise the energy fluctuation. It is called the coefficient of fluctuation of energy and is defined as the ratio of the

maximum fluctuation in energy in a cycle, ∆KEmax, to the work done per cycle, W. Thus:

Example 4

During the working stroke of a single cylinder 4-stroke engine, 10 kJ of work

is done by the expanding gas on the piston and during the compression stroke

the work done on the gas is 5 kJ. The energy involved in the exhaust and

intake strokes is negligible. The coefficient of fluctuation of energy, β, is 1.2. If the mean speed of the flywheel attached to the engine shaft is 300 rev min–1

and the fluctuation of speed is not to exceed 2% of the mean, what is the

minimum power that the engine is required to deliver and the moment of

inertia required for the flywheel? If the radius of gyration of the flywheel is

0.6 m, what is the mass required?

β φ ω

= = ∆KE

W

I

W max

2

∆KE Imax = φ ω 2

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Solution

For a single cylinder 4-stroke engine, the full working cycle involves the crank

rotating through 4π radians (see FIGURE 7). Thus, the net work done during the full working cycle is given by

as W = work in intake stroke + work in compression stroke + work in power

stroke + work in exhaust stroke.

From the question we know that the work involved in the exhaust and intake

strokes is negligible. Also, note that the work done by the gas in the power

stroke is in the shaded area above the x axis in FIGURE 7, which means the

work is positive; whereas the work done on the gas in the compression stroke

is negative since the corresponding area is below the x axis. Therefore, we

have

therefore, the mean torque is

Hence, the engine power is given by

P T= = × ×⎛

⎝⎜ ⎞ ⎠⎟

= =

m

W kW

ω 397 89 300 2

60

12 500 12 5

.

.

π

Tm

N m

= ×

=

5 10 4

397 89

3

π

.

W T

T

= × × =

∴ × =

10 10 5 10 4

5 10 4

3 3

3

– m

m

π

π

W T= θ

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In this question, the maximum energy fluctuation is

Applying ∆KEmax = φIω2 and φ = 2% = 0.02, the moment of inertia required for the flywheel is calculated by

Since I = mk2 and k = 0.6 m, then the mass required is

m I

k =

=

=

2

2

304 0 6

844 4

.

. kg

I KE

=

= ×

× ×⎛

⎝⎜ ⎞ ⎠⎟

=

∆ max

k

φω 2

3

2

6 10

0 02 300 2

60

304

. π

gg m2

∆KE Wmax

J

= = × ×

= ×

β 1 2 5 10

6 10

3

3

.

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. A flywheel has a mass of 20 kg and a radius of gyration of 0.4 m. Its

speed is increased uniformly from 240 rev min–1 to 360 rev min–1 by

means of an applied torque of 2 N m. Determine:

(a) the change in angular momentum

(b) the time taken for this speed to increase.

2. A steel cylindrical flywheel is 500 mm in diameter, has a width of

100 mm and is free to rotate about its polar axis. It is uniformly

accelerated from rest and takes 15 seconds to reach an angular velocity of

800 rpm. Acting on the flywheel is a constant friction torque of 1.5 N m.

Taking the density of the steel to be 7800 kg m–3, determine the torque

which must be applied to the flywheel to produce motion.

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3. The torque-crank angle, T–θ, diagram for a single cylinder 4-stroke engine is shown in FIGURE 9.

FIG. 9

The work involved in each stroke is expressed by the shaded areas A1, A2,

A3 and A4. If the maximum energy fluctuation is A2 = 800 J, determine

the moment of inertia for a flywheel attached to the engine shaft, which

keeps the speed within the range 412 to 420 rpm. Also, find the mass of a

suitable flywheel with a radius of gyration of 0.5 m.

T m

T θ

T

θ

A 1

A 4A

2

A 3

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. (a) Moment of inertia of flywheel I mk=

= ×

=

2

220 0 4.

33 2

240 2 601

. kg m

Initial angular velocity

2

ω = × π

==

= ×

8

360 2 602

π

π

rad s

Final angular velocity

1–

ω

==

=

12

2

π rad s

Change in angular momentum

1–

–I Iω ωω

ω ω

1

2 1

3 2 12 8

40 21

= ( )

= ( )

=

I –

. –

. –

π π

kg m s2 1

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(b)

Alternative solution:

T I=

∴ =

=

=

α

α

α

Angular acceleration 2

2 3 2

3 2

0

.

.

..

–

. –

.

–625

0 625 12 8

4 0 625

2

2 1

rad s 2

α ω ω

π π

π

=

=

=

=

t

t

t

00 1

20 1

.

.

s

Time taken s=

Torque change in angular momentum

time

=

∴ 2 ==

∴ =

=

=

40 21

40 21 2

20 1

20 1

.

.

.

.

t

t

s

Time taken s

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2.

Therefore the angular acceleration is

The cylinderical flywheel in this question can be treated as a solid disc,

then from FIGURE 4 we know that

In order to find I we need to find the mass of the flywheel first.

Since mass = density × volume, and the volume of the flywheel is

volume width= × π 4

2d

I mk mr

r

= =

= =

2 21 2

500 2

0 25 mm m.

α ω ω

ω

=

=

=

=

2 1

2

83 78 15

5 59

–

.

.

t

t

rad s–2

Initial angular velocity

Final angular

ω1 0=

vvelocity rad s–1ω 2 800 2

60 83 78=

× =

π .

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then

The moment of inertia is

The torque applied to the flywheel, T, is to accelerate the wheel, Ta, and to

overcome the friction torque, Tf:

T T T

T T I

T

= +

= =

∴ = ×

a f

f a N m,

1 5

4 79

.

.

α

55 59 1 5

28 28

. .

.

+

= N m

I mr=

= × ×

=

1 2

1 2

153 15 0 25

4 79

2

2. .

. kg m2

m = × ×( ) × × ×( )

=

7800 100 10 4

500 10

153

3 3 2– –π

..15 kg

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3.

The moment of inertia is

I = mk2 and k = 0.5 m

∴ = = = kgm I

k 2 2 21 85 0 5

87 4 .

. .

I KE

= = ×

= ∆ max

2 2 kg m

φω 800

0 0193 43 56 21 85

2. . .

ω

ω

max –1

min

rpm rad s= = ×

=

=

420 420 2

60 43 98

412

π .

rpm rad s

–1

max m

= ×

=

∴ = +

412 2 60

43 14 π

.

ω ω ω iin –1

m

rad s

then

2 43 98 43 14

2 43 56=

+ =

=

. . .

φ ω aax min– – =

ω ω

= 43 98 43 14

43 56 0 0193

. . .

.

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

In this lesson, we have analysed the angular momentum and kinetic energy of

flywheels and explained the need for flywheels in machines subjected to an

erratic torque. The coefficient of fluctuation of speed and the coefficient of

fluctuation of energy are used to describe the fluctuations in speed and energy

under the action of this torque. As an example, a single cylinder four-stroke

engine has been used to show the application of a flywheel.

In our next lesson we will consider the effects of coupling, which also involves

the conservation of angular momentum.

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