I think what Marc /meant/ to say was KE = 1/2 (Iw^2), where KE is the kinetic
energy of
the rotating mass, I is the moment of inertia and w is the angular velocity (^2
means
squared).
What this equation tells us is that increasing either the moment of inertia of
the
flywheel or it's angular velocity (rpm) will increase the amount of stored
energy it
represents. One can also see that it takes more energy input to raise the
angular
velocity to a given value if the value of I is higher. I is representative of
not only
the mass, but its distribution, and as Marc stated in the full text of his
post, mass that
is located further from the rotational center of the flywheel resists rotation
more than
the same mass closer to the center (or words to that effect), in addition to
the thermal
issues to be considered.
Isn't physic fun?
Gary McCormick
San Jose, CA
Marc Sayer wrote:
> <snip>The energy from the rotation of the flywheel (e=MC2) is added to the
>torque
> produced by the engine to help overcome the inertia of the vehicle and get it
> moving. You can compensate for a reduction in that rotational inertia caused
>by
> a reduction in mass, by simply raising the engine speed (a reduction in M can
>be
> offset by a corresponding increase in C).<snip>
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