Last time we arrived at a general formula to compute the made up of numerous parts. Today we’ll discuss the terms in that formula, which encompasses a phenomenon of flywheelknown as flywheels .moment of inertia
The kinetic energy formula we’ve been working with is, again,
ω (1)^{2} The bracketed part of this equation makes reference to spinning Our illustration shows we have five parts to consider: a hub, three spokes and a rim, and label them angular velocity is represented by ω.flywheel’s For our A through E our expanded equation becomes,Σ[ m × _{A}r] + [_{A}^{2}m × _{B}r] + [_{B}^{2}m × _{C}r] + [_{C}^{2}m × _{D}r] + [_{D}^{2}m × _{E}r] (2)_{E}^{2} Equation (2) represents the sum total of It’s a numerical representation of the flywheel’s degree of resistance to changes in motion.flywheel. The more mass a When at rest this greater moment of inertia means it will take more effort to return it to motion. But once in motion the flywheel’s greater moment of inertia will make it harder to stop. That’s because there’s a lot of moment of inertia.stored within its spinning mass, and the heavier a flywheel is, the more kinetic energy it contains. In fact, for any given angular velocity kinetic energy ω, a large and heavy flywheel stores more kinetic energy than a smaller, lighter flywheel. But there’s more to a flywheel’s Copyright 2017 – Philip J. O’Keefe, PE Engineering Expert Witness Blog ____________________________________ |