Flywheel Torque and Distribution of Mass

    Last time we began our discussion on moment of inertia and how it affects a flywheel’s storage of kinetic energy.   That inertia is a function of the flywheel’s mass, in particular how the mass is distributed.   Today we’ll continue our discussion and see how an engineering principal known as torque affects things.

Flywheel Torque and Distribution of Mass

Flywheel Torque and Distribution of Mass


    We learned in a previous blog that torque is most simply defined as a measure of how much force acts upon an object to cause it to rotate around a pivot point or center of rotation, shown as a small black dot in the illustration.   For our discussion we’ll focus on two parts of the flywheel, the hub, part A, and the rim, part E.

    Part A has a mass mA located a distance rA from the flywheel’s center, while part E has a mass mE located a distance rE 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.

    Because of this relationship, we can calculate the kinetic energy contained within a flywheel using the kinetic energy formula,

KE = ½ × ∑[m × r2] × ω2               (1)

    As the flywheel’s angular velocity increases or decreases in response to the engine’s energy output, parts A and E reflect acceleration or deceleration of aA and aE.   Since parts 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,

FA = mA ×  aA                                  (2)

FE = mE ×  aE                                  (3)

    Part A is small compared to part E, therefore mE is greater than mA and accordingly FE is greater than FA.   Forces FA and FE act as torques, because they cause parts A and E to rotate around the flywheel’s center of rotation, so they are designated as Torque A, TA, and Torque E, TE.    These torques are computed by,

TA = FA ×  rA                                   (4)

TE = FE ×  rE                                   (5)

    Part E’s greater mass will contribute more torque than part A, and it will also contribute more to the flywheel’s kinetic energy content.

    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 develop an equation which allows us to quantify the horsepower required to accelerate a flywheel.

opyright 2017 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



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