2.Inertia Forces-1ppt.pdf

1. Inertia Forces
DYNAMICS OF MACHINERY
Chapter 2
1

2. Introduction
2
• Dynamic forces are associated with accelerating masses. As all machines
have accelerating parts, dynamic forces are always present when the
machines operate.
• In situations where dynamic forces are dominant or comparable to the
magnitudes of external forces, and operating speeds are high, dynamic
analysis has to be carried out.
• For example, in case of rotors which rotate at speeds more than 80,000 rpm,
even the slightest eccentricity of the center of mass from the axis of rotation
produces very high dynamic forces.
• This may lead to vibrations, wear, noise or even machine failure.

3. D’ Alembert’s Principle
3

4. D’ Alembert’s Principle
4

5. Dynamic Analysis of Slider-Crank Mechanism
- Velocity and Acceleration of Piston
5
r

7. 7
Velocity of Piston

8. 8
Acceleration of Piston
• Note that this expression of acceleration has
been obtained by differentiating the
approximate expression for velocity.
• It is usually very cumbersome to differentiate
the exact expression for velocity. However,
this gives satisfactory results.

9. The driving force acting on the piston is termed as piston effort. In a vertical cylinder IC
engine, following three types of forces act:
Gas Force
The force due to variation of working fluid pressure is known as gas force, or Gas force
Fg =
πD2
4
× p …….. (1)
where
D = diameter of the piston and
p = gas pressure
Engine Force Analysis
Piston Effort

10. Inertia Force
• In an IC engine, during the first half of the stroke, the reciprocating mass accelerates
and the inertia force tends to resist the motion. Thus, the net force on the piston is
decreased.
• However, during the second half of the stroke, the reciprocating mass decelerate,
and inertia force opposes this deceleration. Thus, it increases the effective force on
the piston.
• The inertia force of the piston is given as
𝐹𝑖 = 𝑚𝑓 = 𝑚𝑟𝜔2 cos 𝜃 +
cos 2𝜃
𝑛
… (2)

11. Weight of the Reciprocating Mass
• The weight of reciprocating mass assists the piston during its movement towards
bottom dead centre (BDC). Therefore, piston effort is increased by an amount equal
to the weight of the piston.
• However, when the piston moves towards top dead centre (TDC), the piston effort is
decreased by the same amount.
Net piston effort:
𝑃 = 𝐹
𝑔 + 𝐹𝑖 ± 𝑊 … (3)

12. 12
𝐹𝐶 cos 𝜙 = 𝑃 𝑜𝑟 𝐹𝐶 =
𝑃
cos 𝜙
𝐹𝑡 = 𝐹𝐶 sin 𝜙 = 𝑃 tan 𝜙
Where, 𝑙 𝑠𝑖𝑛 𝜙 = 𝑟 𝑠𝑖𝑛 𝜃 ⇒ 𝜙 = 𝑠𝑖𝑛−1 𝑠𝑖𝑛 𝜃
𝑛

13. 13
(iv) Crank Pin Effort
Force is exerted on the crankpin as a result of the force on the piston. The component of force acting along
connecting rod perpendicular to the crank is known as crank-pin effort.
𝐹𝑐𝑡 ∗ 𝑟 = 𝐹𝑐 𝑟 𝑠𝑖𝑛 𝜃 + 𝜙
𝐹𝑐𝑡 = 𝐹𝑐 𝑠𝑖𝑛 𝜃 + 𝜙
𝐹𝑐𝑡 =
𝑃
𝑐𝑜𝑠 𝜙
𝑠𝑖𝑛 𝜃 + 𝜙
(v) Thrust on the Bearings
The component of 𝐹𝑐𝑟 along the crank in the radial direction produces a thrust on the crankshaft bearings.
𝐹𝑐𝑟 = 𝐹𝑐 𝑐𝑜𝑠 𝜃 + 𝜙
𝐹𝑐𝑟 =
𝑃
𝑐𝑜𝑠 𝜙
𝑐𝑜𝑠 𝜃 + 𝜙
𝐹𝑐𝑟
D
E
90 − 𝜃 + φ
𝑟 𝑙

14. Crank effort is the net effort applied at
the crank pin perpendicular to the crank
(𝐹𝑐𝑡) which gives the required turning
moment on the crank shaft.

15. 15

16. Graphical Method to Determine Crank Effort or Torque
T = P x distance OY
Where OY is the distance measured between centre of crank and a point of
intersection of Y axis and extension of connecting rod P2B

17. Note:
• The crank effort is a function of piston effort P and crank
rotation angle. Further, the piston effort is also a function of
crank angle .
• The diagram showing the crank effort or torque as a function of
crank rotation angle  for any reciprocating engine is called
crank-effort diagram or turning moment diagram.

18. P-V diagram of petrol engine
The turning moment diagram of any engine can be plotted if the gas
pressure p is known for all positions of the crank. The value of gas pressure
can be found from a given pressure-volume (P-V) diagram

19. Variation of Gas force and Inertia force
Using these pressure values, gas forces can be computed and plotted as
shown here:

20. Variation of Piston Effort
Further, the variation of inertia force due to mass of reciprocating parts can
be plotted as shown here:

21. T= piston effort x OY
Where OY is the crank effort arm length. The variation in crank effort arm length
for different crank position is shown below.
Variation of Crank Effort Arm Length

22. Turning Moment Diagram for a Single Cylinder Double Acting Steam Engine
Double Acting Steam Engine:
• When steam is coming in both side
of the piston and it produces double
working stroke in each revolution,
is called double acting steam
engine.
• It produces double power than
single acting steam engine.

23. Turning Moment Diagram for a Single Cylinder Double Acting Steam Engine
Double Acting Steam Engine:
• Since the work done is the product
of the turning moment and the angle
turned, therefore the area of the
turning moment diagram represents
the work done per revolution.

24. Turning Moment Diagram for a Single Cylinder
Four Stroke Engine

25. Turning Moment Diagram for a Single Cylinder
Four Stroke Engine
• Turning moment diagram
shows that torque T is entirely
positive in expansion stroke of
engine whereas in suction,
compression and exhaust
strokes, it is negative.
• This indicates that in these
strokes, power is consumed.
Thus there is large variation of
torque which may cause
fluctuation of speed.

26. Turning Moment Diagram for a Multi Cylinder
Four Stroke Engine
In multi-cylinder engine, the turning
moment diagram of each cylinder is
obtained separately and they are
superimposed over each other with starting
point shifted to phase difference of angle
between respective crank positions.

27. Flywheel
• A flywheel is an inertial energy storage device. It absorbs mechanical
energy and serves as a reservoir, storing 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.
• Internal combustion engines with one or two cylinders are a typical
example. Piston compressors, punch presses, rock crushers etc. are the
other systems that have flywheel.

28. Flywheel
Depending upon the source of power and the type of driven machinery,
there are three distinct situations where a flywheel is necessitated.
1. When the availability of energy is at a fluctuating rate but the
requirement of it for driven machinery is at uniform rate. For
example, IC Engine driven water pumps, generators, compressors,
fans, etc. In such a situation flywheel is needed to store surplus
energy.

29. Flywheel
2. In applications, namely, electric motor driven
punching, shearing and riveting machines,
rolling mills, etc., though the energy is available
at a uniform rate, the demand for it is variable.
Figure on right shows a typical energy
requirement and availability curve for an
electric motor driven rolling mill which shows
that for a small fraction of the cycle period,
there is huge requirement of energy. Thus again
a flywheel is needed.
3. In this type of situation, both the requirement
and availability of energy represent a variable
rate, e.g. IC engine driven reciprocating air
compressor or pump.
Turning moment diagram
of electric motor driven
rolling mill

30. Types of Flywheel
Three types of flywheel – disc type, web type and arm type are most
commonly used.

31. Types of Flywheel

32. Fluctuation of energy (𝑬𝒇):
• A flywheel is used to control the variations in
speed during each cycle of an engine.
• A flywheel of suitable dimensions attached to
the crankshaft, makes the moment of inertia of
rotating parts quite large and thereby it acts as a
reservoir of energy.
• During the periods when the supply of energy
is more than required, it stores energy and
during the period when the supply is less than
required, it releases the energy.
• The difference between maximum and
minimum kinetic energies of flywheel is
known as maximum fluctuation of energy, 𝑬𝒇.

33. Coefficient of Fluctuation of energy (𝑲𝒆):
Ratio of maximum fluctuation of energy and work done per cycle is known as
coefficient of fluctuation of energy.
𝐾𝑒 = Max. fluctuation of energy / work done per cycle.
Therefore, 𝐾𝑒 =
𝐸𝑓
𝐸
Work done per cycle = 𝑇𝑚𝑒𝑎𝑛 × 
where 𝑇𝑚𝑒𝑎𝑛 = Mean torque and  = angle turned by the crank in radians.
Also, work done per cycle is = 𝑃 × 60,000/𝑁
where P = Power in KW and N is in rpm.

34. Coefficient of Fluctuation of speed (𝑲𝒔):
• The difference between the greatest and the least angular speeds of the
flywheel is called the maximum fluctuation of speed.
• The ratio of greater fluctuation of speed per cycle to the mean speed is
called coefficient of fluctuation of speed.

37. Example 1
A multi-cylinder engine runs at a speed of 1500 rpm. The turning moment diagram
repeats itself for every revolution of the crankshaft. The scale of the turning moment
is 1 cm = 6,000 N-m and the crank angle is plotted to a scale of 1cm = 60°. The areas
below and above the mean turning moment line, taken in order are as follows: - 0.3,
+ 4.1, - 2.8, + 3.2, - 3.3, + 2.5, - 2.6, + 2.8 and -3.6 cm2. Find out the fluctuation of
energy. Also find out the coefficient of fluctuation of speed if the weight of the
rotating parts is 5000 N, and the radius of gyration is 0.2 m.

38. Example 1
Energy at A = 𝐸
Energy at B = 𝐸 − 0.3
Energy at C = 𝐸 − 0.3 + 4.1 = 𝐸 + 3.8
Energy at D = 𝐸 + 3.8 − 2.8 = 𝐸 + 1
Energy at E = 𝐸 + 1 + 3.2 = 𝐸 + 4.2
Energy at F = 𝐸 + 4.2 − 3.3 = 𝐸 + 0.9
Energy at G = 𝐸 + 0.9 + 2.5 = 𝐸 + 3.4
Energy at H = 𝐸 + 3.4 − 2.6 = 𝐸 + 0.8
Energy at I = 𝐸 + 0.8 + 2.8 = 𝐸 + 3.6
Energy at J = 𝐸 + 3.6 − 3.6 = 𝐸
Thus,
Greatest energy (at E) = 𝐸 + 4.2
Least energy (at B) = 𝐸 − 0.3
Therefore,
𝐸𝑓 = 𝐸 + 4.2 − 𝐸 − 0.3
⇒ 𝐸𝑓 = 4.5 𝑐𝑚2
⇒ 𝐸𝑓 = 4.5 6000 × 60 ×
𝜋
180
∴

39. Example 1
Further,
𝐸𝑓 = 𝐼𝜔𝑚𝑒𝑎𝑛
2
𝐾𝑠 = 𝑚𝑘2
𝜔𝑚𝑒𝑎𝑛
2
𝐾𝑠
28274 =
5000
9.81
× 0.22
2𝜋 × 1500
60
2
× 𝐾𝑠
Therefore, the coefficient of fluctuation of speed
Or

40. Example 2
The turning moment diagrams for a 4 stroke cycle gas engine may be assumed to be
represented by four rectangles, the areas of which are measured from the line of zero
pressure are as follows. Expansion stroke = 8.5 cm2, Exhaust = 0.8 cm2, Suction =
0.7 cm2, Compression = 2.2 cm2. Assuming the resisting torque to be uniform, find
the weight of the rim of the wheel required to keep the speed between 116 & 124
rpm. Assume that mean radius of the rim = 1m and each cm2 of area of the diagram
represents 150 N-m of energy.

41. Example 2
Net work done during the cycle in an IC engine is the difference of energy generated
during power stroke and the energy consumed in the remaining strokes.
That is, net work done during the cycle = 8.5 − 0.8 − 0.7 − 2.2 𝑐𝑚2
Therefore, net work done during the cycle = 4.8 𝑐𝑚2
= 4.8 × 150 𝑁𝑚
= 720 𝑁𝑚

42. Example 2
Therefore, maximum fluctuation of energy,
𝐸𝑓 = 8.5 × 150 −
1
4
× 720

43. Example 2
𝐸𝑓 =
1
2
𝐼𝜔𝑚𝑎𝑥
2 −
1
2
𝐼𝜔𝑚𝑖𝑛
2
1095 =
1
2
𝐼
2𝜋 × 124
60
2
−
2𝜋 × 116
60
2

44. Example 2
Also,
𝐼 =
𝑊
𝑔
𝐾2
104.01 =
𝑊
9.81
12

45. Example 3
A single cylinder, single acting 4 stroke cycle gas engine has a piston diameter of 33
cm and a stroke of 60 cm. Mean speed = 200 rpm. Mean pressures in the cylinder
above atmosphere are as follows:
Suction stroke = 0.7N/cm2 below atmosphere,
Compressions stroke = 20 N/cm2,
Expansion stroke = 70 N/cm2,
Exhaust stroke = 1.4 N/cm2.
Assuming constant resistance and minimum and maximum speed to occur at the
beginning and end of expansion stroke, determine the moment of inertia of flywheel
if total fluctuation of speed is not to exceed 1% of mean speed. Find also the drop in
speed which will then occur during a cycle in which there is no admission.

46. Example 3
Part-I: To find the moment of inertia of flywheel for 1% fluctuation in mean speed.
Work done during power stroke
= 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑝𝑜𝑤𝑒𝑟 𝑠𝑡𝑟𝑜𝑘𝑒 × 𝐴𝑟𝑒𝑎 × 𝑆𝑡𝑟𝑜𝑘𝑒
= 70 ×
𝜋 33 2
4
× 0.6
Work expended during suction, compression & exhaust strokes
= 𝑃𝑠𝐴𝐿 + 𝑃𝑐𝐴𝐿 + 𝑃𝑒𝐴𝐿
= 0.7 + 20 + 1.4
𝜋 33 2
4
× 0.6
Net work done per cycle = 35,922 − 11,341

47. Example 3
Fluctuation of energy = work done during power stroke - 1/4th of net work done per
cycle.
𝐸𝑓 = 35,922 −
1
4
× 24,581
𝐸𝑓 = 𝐼𝜔𝑚𝑒𝑎𝑛
2
𝐾𝑠
Or
29,776 = 𝐼
2𝜋 × 200
60
2
0.01
Therefore,

48. Example 3
Part-II: To find the drop in speed when there is no admission during a cycle
When there is no admission during a cycle, the engine has to perform four more idle
strokes and the energy stored in flywheel is further expended. Therefore net work done
in one cycle will have to be distributed over 8 strokes.
Therefore,
𝐸𝑓 = 35,922 −
1
8
× 24,581
Also ,
𝐸𝑓 = 𝐼𝜔𝑚𝑒𝑎𝑛
2 𝐾𝑠
⇒ 32,849.375 = 6,798 ×
2𝜋 × 200
60
2
𝐾𝑠

49. Example 3
And,
𝐾𝑠 =
𝜔𝑚𝑎𝑥 − 𝜔𝑚𝑖𝑛
𝜔𝑚𝑒𝑎𝑛
=
𝑁𝑚𝑎𝑥 − 𝑁𝑚𝑖𝑛
𝑁𝑚𝑒𝑎𝑛
Since 𝑁𝑚𝑒𝑎𝑛 = 200 𝑟𝑝𝑚
0.011 =
𝑁𝑚𝑎𝑥 − 𝑁𝑚𝑖𝑛
200

50. Flywheel of Punching Presses

51. Flywheel of Punching Presses
• From the previous discussions, it can be observed
that when the load on the crankshaft is constant or
varies and the input torque varies continuously during a
cycle, a flywheel is used to reduce the fluctuations of
speed.
• A flywheel can perform the same purpose in a punching
press or a riveting machine in which the torque available
is constant but the load varies during the cycle.
• Since punching operations occurs only during the
downward stroke and only when the punch travels
through the thickness of the plate, in the absence of a
flywheel the decrease in the speed of the crankshaft will
be very large during actual punching compared to no-
load periods.

52. Flywheel of Punching Presses
Moment

53. Formula’s Used:
Flywheel of Punching Presses
𝑃 = 𝜋𝑑𝑡𝜏𝑢

54. Example 4
A machine is required to punch 4 holes of 4cm diameter in a plate of 3 cm thickness,
per minute. The work required is 700 N-m/cm2 of sheared area. The punch has a
stroke of l0 cm. Maximum speed of flywheel at its radius of gyration is 30 m/s. Find
the weight of flywheel if speed should not fall below 28 m/s at the radius of gyration.
Also find the power of the motor.
Solution:
Work required/punching 𝑊
𝑝 = 𝜋𝑑𝑡𝑋, where 𝑋 is the work required/cm2 of sheared area.
𝑊
𝑝 = 𝜋 × 4 × 3 × 700
Therefore,
𝑃𝑜𝑤𝑒𝑟 =
𝑊
𝑝 × 𝑛𝑜. 𝑜𝑓 ℎ𝑜𝑙𝑒𝑠/𝑚𝑖𝑛
60 × 1000
=
26,390 × 4
60 × 1000

55. Example 4
𝐸𝑓 = 𝑊
𝑝 1 −
𝑡
2𝑠
= 26,390 1 −
3
2 × 10
Also,
𝐸𝑓 =
1
2
𝐼𝜔𝑚𝑎𝑥
2 −
1
2
𝐼𝜔𝑚𝑖𝑛
2
=
1
2
𝐼 𝜔𝑚𝑎𝑥
2 − 𝜔𝑚𝑖𝑛
2
𝐸𝑓 =
𝑊
2𝑔
𝑘2
𝜔𝑚𝑎𝑥
2
− 𝜔𝑚𝑖𝑛
2
𝐸𝑓 =
𝑊
2𝑔
𝜔𝑚𝑎𝑥
2 𝑘2 − 𝜔𝑚𝑖𝑛
2
𝑘2
⇒ 22,431 =
𝑊
2 × 9.81
302
− 282
Therefore, weight of the flywheel,

56. Example 5
Find weight of flywheel needed by a machine to punch 22 mm holes in 18 mm steel
plates. The machine is to make 30 rev per minute and is to be capable of punching a
hole every revolution. The hole is to be formed during 30° of rotation of crankshaft
of the punch. The crank shaft is to be connected to the fly wheel shaft by a gear train
of 12:1 ratio so that the max rpm of flywheel will be 12 times that of the machine.
Assume mechanical 𝜂 to be 85%. Minimum speed of flywheel is to be 90% of the
maximum. Mean diameter of flywheel rim is 1 m. Ultimate shear strength of the
plate is 35,000 N/cm2.
Solution:
Work required/punching 𝑊
𝑝 = Τ
1
2 𝑃𝑡 = Τ
1
2 𝜋𝑑𝑡𝜎𝑢𝑡
𝑊
𝑝 = ൗ
1
2 𝜋2.2 × 1.8 × 35,000 × 1.8
Since 𝜂𝑚𝑒𝑐ℎ is 0.85, work required/punching 𝑊
𝑝 =
3918.82
0.85

57. Example 5
𝑃𝑜𝑤𝑒𝑟 =
𝑊
𝑝 × 𝑛𝑜. 𝑜𝑓 ℎ𝑜𝑙𝑒𝑠/𝑚𝑖𝑛
60 × 1000
=
4610.3 × 30
60 × 1000
⇒
𝐸𝑓 = 𝑊
𝑝 1 −
𝜃2 − 𝜃1
2𝜋
= 4610.3 1 −
30
360
Maximum speed of flywheel = 30×12 = 360 rpm.
Therefore,
= 𝜔𝑟 =
2𝜋 × 360
60
× 0.5 =
= 0.9 × 18.84 =

58. Example 5
Now, using the equation of fluctuation of energy,
𝐸𝑓 =
1
2
𝐼𝜔𝑚𝑎𝑥
2 −
1
2
𝐼𝜔𝑚𝑖𝑛
2
=
𝑊
2𝑔
𝜔𝑚𝑎𝑥
2 𝑘2 − 𝜔𝑚𝑖𝑛
2
𝑘2
⇒ 4226 =
𝑊
2 × 9.81
18.242 − 16.92
The weight of the flywheel is found as,

59. Example 6
A machine Press is worked by an electric motor delivering 2.25 kW continuously. At
the commencement of an operation, a flywheel of moment of inertia 600 Nm2 on the
machine is rotating at 250 rpm. The pressing operation requires 5,300 N-m of energy
and occupies 0.75 seconds. Find the maximum number of pressings that can be made
in 1 hour and the reduction in speed of the flywheel after each pressing. Neglect
friction losses.
Solution:
Work required/punching 𝑊
𝑝 = Τ
1
2 𝑃𝑡 = Τ
1
2 𝜋𝑑𝑡𝜎𝑢𝑡
𝑊
𝑝 = ൗ
1
2 𝜋2.2 × 1.8 × 35,000 × 1.8
Since 𝜂𝑚𝑒𝑐ℎ is 0.85, work required/punching 𝑊
𝑝 =
3918.82
0.85

60. Example 7
In a machine, the intermittent operations demand the torque to be applied as follows:
• During the first half revolution, the torque increases uniformly from 800 Nm
to 3000 Nm.
• During next one revolution, the torque remains constant.
• During next one revolution, the torque decreases uniformly from 3000 Nm to
800 Nm.
• During last 1½ revolutions, the torque remains constant.
Thus a cycle is completed in 4 revolutions. The motor to which the machine is
coupled exerts a constant torque at a mean speed of 300 rpm. A flywheel of mass
1500 kg and radius of gyration of 450 mm is fitted to the shaft. Determine:
i. The power of the motor
ii. The total fluctuation of speed of the machine shaft.

61. Example 8
A single cylinder four stroke petrol engine develops 20 kW at 600 rpm. The
mechanical efficiency of the engine is 80 % and the work don by the gases during
expansion stroke is three times the work consumed in compression stroke. If the total
fluctuation of speed is not to exceed ±1.5 percent of the mean speed and turning
moment diagram during expansion stroke is assumed to be triangular. Determine the
mass of the flywheel if the diameter of it is 1 m. Also determine the cross section of
flywheel rim. The width to thickness ratio is 1.5 and mass density of flywheel
material is 7100 kg/m3.

62. Example 8
Solution:
Permissible speed variation of ±1.5% implies that the coefficient of fluctuation of speed,
𝐾𝑠 = 0.03
Angular velocity,
𝜔 =
2𝜋 × 600
60
𝜔 = 62.83 Τ
𝑟𝑎𝑑 𝑠
Indicated power,
𝐼𝑃 =
𝐵𝑃
𝜂𝑚𝑒𝑐ℎ
=
20
0.8
𝐼𝑃 = 25 𝑘𝑊

63. Example 8
IP = Indicated work done/cycle × Number of explosions/sec
25 × 103 = 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒/𝑐𝑦𝑐𝑙𝑒 ×
600
2 × 60
Therefore, work done per cycle
𝑊𝐷 = 5000 𝑁𝑚
Since the work during suction and exhaust strokes is neglected, the net work done during
the cycle is given by (Given: 𝑊
𝑒𝑥𝑝 = 3𝑊
𝑐𝑜𝑚𝑝)
𝑊𝐷 = 𝑊
𝑒𝑥𝑝 − 𝑊
𝑐𝑜𝑚𝑝
𝑊𝐷 = 𝑊
𝑒𝑥𝑝 − 𝑊
𝑒𝑥𝑝/3
𝑊𝐷 = 2/3 𝑊
𝑒𝑥𝑝
Therefore,
𝑊
𝑒𝑥𝑝 = 3/2 𝑊𝐷 = 3/2 × 1500 = 7,500 𝑁𝑚

64. Example 8
𝑊
𝑒𝑥𝑝 = 7,500 𝑁𝑚 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐴𝐷𝐹 =
1
2
𝐷𝐹 × 𝐴𝐸
⇒ 𝐴𝐸 = 𝑇𝑚𝑎𝑥 =
2𝑊
𝑒𝑥𝑝
𝐷𝐸
=
2 × 7500
𝜋
⇒ 𝐴𝐸 = 4,774.6 𝑁𝑚
Mean torque
𝑇𝑚𝑒𝑎𝑛 =
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒/𝑐𝑦𝑐𝑙𝑒
4𝜋
=
5000
4𝜋
⇒ 𝑇𝑚𝑒𝑎𝑛 = 397.88 𝑁𝑚

65. Example 8
Excess torque = 𝐴𝐺 = 𝐴𝐸 − 𝐺𝐸 = 4774.6 − 397.88 = 4376.72 𝑁𝑚
In similar triangles ABC and ADF,
𝐵𝐶
𝐷𝐹
=
𝐴𝐺
𝐴𝐸
⇒ 𝐵𝐶 = 𝐷𝐹
𝐴𝐺
𝐴𝐸
= 𝜋 ×
4376.72
4774.6
⇒ 𝐵𝐶 = 0.9167𝜋
Fluctuation of energy 𝐸𝑓 = area of ABC
⇒ 𝐸𝑓 =
1
2
× 0.9167𝜋 × 4376.72 = 6302.2 𝑁𝑚

66. Example 8
Also, fluctuation of energy
𝐸𝑓 = 6302.2 = 𝑚𝑘2𝜔2𝐾𝑠 = 𝑚 × 0.52 × 62.832 × 0.03
Therefore mass of the flywheel
Assuming that 90 per cent of mass is contributed by rim and remaining 10 per cent by
hub, arms, etc.,
0.9𝑚 = 𝜋𝑑 × 𝑏 × 𝑡 × 𝜌 ⇒ 𝑏 × 𝑡 =
0.9𝑚
𝜋𝑑 × 𝜌
=
0.9 × 212.86
𝜋 × 0.5 × 7100
⇒ 𝑏 × 𝑡 = 0.0086 𝑚2
Since 𝑏 = 1.5𝑡, 1.5 × 𝑡2 = 0.0086. Therefore,

67. Thank you