(^630) „„„„„ A Textbook of Engineering Mechanics
N 2 = Minimum speed of flywheel during a cycle,
ω 1 = Maximum angular velocity of flywheel during a cycle, and
ω 2 = Minimum angular velocity of flywheel during a cycle.
We know that average speed of the flywheel,
N= 12
1
()
2
NN+ ω = 12 12
1
()0.5()
2
ω+ω = ω+ω
and average kinetic energy of the flywheel,
E=
2
2
Iω
∴ Fluctuation of energy,
E= Maximum K.E. – Minimum K.E. =
22
12 22
–(–) 12
222
IIωωI
=ωω
= ()(–) 1212
2
I
ω+ω ω ω= Iω (ω 1 – ω 2 ) ...[Q ω = 0.5 (ω 1 + ω 2 )]

(^22212) –
60 60 60
IN×π⎛⎞ππNN
⎜⎟
⎝⎠

2
12
4
(– )
3600
IN N N
π
×

2
(– ) 12
900
IN N N
π
×.
Example 31.2. A flywheel of an engine has a mass of 6.5 tonnes and radius of gyra-
tion 1.8 metres. If the maximum and minimum speeds of the flywheel are 120 r.p.m. and 118 r.p.m.
respectively, find the fluctuation of energy.
Solution. Given: Mass of flywheel (M) = 6.5 t = 6500 kg; Radius of gyration (k) = 1.8 m:
Maximum speed of flywheel (N 1 ) = 120 r.p.m. and minimum speed of flywheel (N 2 ) = 118 r.p.m.
We know that average speed of the flywheel,
N = 12
11
( ) (120 118) 119 r.p.m.
22
NN+= + =
and mass moment of inertia,
I = Mk^2 = 6500 (1.8)^2 = 21 060 kg-m^2
∴ Fluctuation of energy,
E =
2
900 IN N(– )^12 N
π
× =
2
21 060 119 (120 – 118)
900
π
×× N-m
= 54 970 N-m = 54.97 kN-m = 54·97 kJ Ans.
Example 31.3. A flywheel with a radius of gyration 0.9 m is fitted to a multicylinder
engine, which runs at a mean speed of 360 r.p.m. If the speed varies from 2% above the mean to 2%
below it and the fluctuation energy is 30 kN-m, find (i) moment of inertia of the wheel and (ii) mass
of the flywheel.
Solution. Given: Radius of gyration of flywheel (k) = 0.9 m; Mean speed of the flywheel
(N) = 360 r.p.m. and fluctuation energy (e) = 30 kN-m = 30 000 N-m.
(i) Moment of inertia of the wheel
Let I = Moment of inertia of the wheel.