Last time we began our discussion on moment of inertia and how it affects a flywheel’s in particular how the mass,is mass . Today we’ll continue our discussion and see how an engineering principal known as distributed affects things.torque
We learned in a previous blog that the hub, part flywheel,A, and the rim, part E. Part r from the flywheel’s center, while part _{A}E has a mass m located a distance _{E}r from it. When an engine applies mechanical power to the flywheel by way of its rotating shaft, the revolutions per minute, RPM, increase and along with it the angular velocity, ω, also increases. For a refresher on this, follow the link._{E}Because of this relationship, we can calculate the kinetic energy contained within a flywheel using the kinetic energy formula,
(1)^{2} As the flywheel’s angular velocity increases or decreases in response to the engine’s energy output, parts a. Since parts _{E}A and E exhibit both mass and acceleration, they are subject to Newton’s Second Law of Motion, which states that force equals mass times acceleration. Using that relationship we can calculate the force exerted on each part by,
a (2)_{A}
a (3)_{E} Part m and accordingly _{A}F is greater than _{E}F. Forces _{A}F and _{A}F act as torques, because they cause parts _{E}A and E to rotate around the flywheel’s center of rotation, so they are designated as Torque A, Tand Torque _{A, }E, T These torques are computed by,_{E. }
r (4)_{A}
r (5)_{E} Part Most flywheels are designed with heavy rims supported by small hubs and slender spokes, because the more mass that’s distributed away from the flywheel’s center of rotation, the greater the flywheel’s moment of inertia and torque, and the more kinetic energy it can store. Next time we’ll discuss how flywheel moment of inertia, torque, and kinetic energy come into play when designing machines with varying power output. opyright 2017 – Philip J. O’Keefe, PE Engineering Expert Witness Blog ____________________________________ |