Flywheels  Kinetic Energy
The kinetic energy stored in flywheels  the moment of inertia.
A flywheel can be used to smooth energy fluctuations and make the energy flow intermittent operating machine more uniform. Flywheels are used in most combustion piston engines.
Energy is stored mechanically in a flywheel as kinetic energy.
Kinetic Energy
Kinetic energy in a flywheel can be expressed as
E _{ f } = 1/2 I ω^{2}(1)
where
E _{ f } = flywheel kinetic energy (Nm, Joule, ft lb)
I = moment of inertia (kg m^{2}, lb ft^{2})
ω = angular velocity ( rad /s)
Angular Velocity  Convert Units
 1 rad = 360 ^{ o } / 2 π =~ 57.29578 ^{ o }
 1 rad/s = 9.55 rev/min (rpm) = 0.159 rev/s (rps)
Moment of Inertia
Moment of inertia quantifies the rotational inertia of a rigid body and can be expressed as
I = k m r^{2}(2)
where
k = inertial constant  depends on the shape of the flywheel
m = mass of flywheel (kg, lb _{ m } )
r = radius (m, ft)
Inertial constants of some common types of flywheels
 wheel loaded at rim like a bicycle tire  k =1
 flat solid disk of uniform thickness  k = 0.606
 flat disk with center hole  k = ~0.3
 solid sphere  k = 2/5
 thin rim  k = 0.5
 radial rod  k = 1/3
 circular brush  k = 1/3
 thinwalled hollow sphere  k = 2/3
 thin rectangular rod  k = 1/2
Moment of Inertia  Convert Units
 1 kg m^{2}= 10000 kg cm^{2}= 54675 ounce in^{2}= 3417.2 lb in^{2}= 23.73 lb ft^{2}
Flywheel Rotor Materials
Material  Density (kg/m ^{ 3 } )  Design Stress ( MPa)  Specific Energy ( kWh/kg ) 

Aluminum alloy  2700  
Birch plywood  700  30  
Composite carbon fiber  40% epoxy  1550  750  0.052 
Eglass fiber  40% epoxy  1900  250  0.014 
Kevlar fiber  40% epoxy  1400  1000  0.076 
Maraging steel  8000  900  0.024 
Titanium Alloy  4500  650  0.031 
"Super paper"  1100  
Sglass fiber/epoxy  1900  350  0.020 
 1 MPa = 10 ^{ 6 } Pa = 10 ^{ 6 } N/m^{2}= 145 psi
 Maraging steels are carbon free ironnickel alloys with additions of cobalt, molybdenum, titanium and aluminum. The term maraging is derived from the strengthening mechanism, which is transforming the alloy to martensite with subsequent age hardening.
Example  Energy in a Rotating Bicycle Wheel
A typical 26inch bicycle wheel rim has a diameter of 559 mm (22.0") and an outside tire diameter of about 26.2" (665 mm) . For our calculation we approximate the radius  r  of the wheel to
r = ((665 mm) + (559 mm) / 2) / 2
= 306 mm
= 0.306 m
The weight of the wheel with the tire is 2.3 kg and the inertial constant is k = 1 .
The Moment of Inertia for the wheel can be calculated
I = (1) (2.3 kg) (0.306 m)^{2}
= 0.22 kg m^{2}
The speed of the bicycle is 25 km/h ( 6.94 m/s) . The wheel circular velocity (rps, revolutions/s)  n _{ rps }  can be calculated as
n _{ rps } = (6.94 m/s) / (2 π (0.665 m) / 2)
= 3.32 revolutions /s
The angular velocity of the wheel can be calculated as
ω = (3.32 revolutions /s) (2 π rad/ revolution )
= 20.9 rad/s
The kinetic energy of the rotating bicycle wheel can then be calculated to
E _{ f } = 0.5 (0.22 kg m^{2}) ( 20.9 rad/s )^{2}
= 47.9 J
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