Energy

Energy is defined as the ability to do work.
Energy is stored in many forms, some examples are

Chemical
In the form of fuel like petrol or diesel.
Kinetic
A moving mass stores kinetic energy.
Potential
A mass at altitude has the potential to release the energy imparted to it in raising it up, by descending.
Rotational
A spinning mass has energy the magnitude of which is dependant on
  • the mass
  • the concentration of the mass relative to the spin axis
  • the rate of spin

Energy cannot be created or destroyed, but it can be converted from one form to another.
An internal combustion engine takes chemical potential energy stored in the form of fuel and turns it into heat, sound and rotational kinetic energy. It should be noticed that not all the energy transformed is useful to us.

Mathematically

Units

Energy is measured in units called Joules.
Energy has dimensions of kg m2 s-2
1J is equivalent to 1Nm.

Kinetic Energy

 KE = ½ m v2
where
KE
Kinetic Energy in kg m2 s-2 [J].
m
Mass [ kg].
v
Velocity squared [ m2s-2].

The ½ comes from integrating momentum with respect to velocity, the kinetic energy is the sum of the work done in changing the momentum from 0 to the value which is dependant upon V.
Worked example 

Taking the H269C as an example @ 75kts at MAUM m of 2050lbs, which converted to kgs is 932kg
m = 932kg
75kts in ft s-1
126 fts-1 in ms-1 is 38.4
v = 38.4 m s-1 v2 = 38.4 * 38.4
v2 = 1475
mv2 =

Gravitational Potential Energy

PE = m g h
where
PE
Potential Energy [J]
m
Mass [kg].
g
Acceleration due to gravity [m s-2]
h
Height [m].
Worked example 
Taking the H269C as an example @ 1000ft at MAUM m of 2050lbs, which converted to kgs is 932kg
m= 932kg
g = 9.8 ms-2
altitude is 1000ft which in m is 305m
h=305m
PE = mgh = 932 * 9.8 * 305
PE = 2785748 J.

Rotational Energy

RE = ½ Ι ω2
where
RE
Rotational Energy [J]
Ι
Moment of Inertia. [kg m2]
ω2
Angular Velocity Squared. [s-2]
Worked example 
Taking the H269C as an example RRPM in cruise is 450RPM.
450RPM = 450 * 2 * Pi / 60
15 * Pi Rad S-1
ω = 47.12 Rad S-1

From pre calculated tables we find that Ι for a solid rod spun around one end with const mass per unit volume is
Ι = 1/3 M L2
M ≈ 9Kg
L = 4.09m
Ι = 50.55 kgm2
RE = ½ * 50.55 * 47.12 2
RE = 56130.65356 J * 3 [No. of blades]
RE = 168391.9607 J

An assumption is made that the rod is sufficiently similar to to a rotor blade for simplicity.

Mechanical Energy

ME = KE + PE + RE

Web references