## Posts Tagged ‘center of rotation’

### Moment of Inertia in a Flywheel

Monday, November 6th, 2017
 Last time we arrived at a general formula to compute the kinetic energy, KE, contained within the totality of a spinning flywheel made up of numerous parts.   Today we’ll discuss the terms in that formula, which encompasses a phenomenon of flywheels known as moment of inertia. Moment of Inertia in a Flywheel        The kinetic energy formula we’ve been working with is, again, KE = ½ × Σ[m × r2] × ω2                           (1)     The bracketed part of this equation makes reference to spinning flywheels comprised of one or more parts, and that’s what we’ll be focusing on today.   The symbol Σ is the Greek letter sigma, standard engineering shorthand notation used to represent the sum of all terms and mathematical operations contained within the brackets.     Our illustration shows we have five parts to consider:  a hub, three spokes and a rim, and label them A, B, C, D, and E respectively.   Each part has its own mass, m, and is a unique distance, r, from the flywheel’s center of rotation.   The flywheel’s angular velocity is represented by ω.    For our flywheel of parts A through E our expanded equation becomes, Σ[m × r2] = [mA × rA2] + [mB × rB2] + [mC × rC2] + [mD × rD2] + [mE × rE2]      (2)     Equation (2) represents the sum total of moments of inertia contained within our flywheel.  It’s a numerical representation of the flywheel’s degree of resistance to changes in motion.     The more mass a flywheel has, the greater its moment of inertia.  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 kinetic energy stored within its spinning mass, and the heavier a flywheel is, the more kinetic energy it contains.   In fact, for any given angular velocity ω, a large and heavy flywheel stores more kinetic energy than a smaller, lighter flywheel.     But there’s more to a flywheel’s moment of inertia than just mass.   What’s really important is how that mass is distributed. We’ll get into that next time when we discuss torque. Copyright 2017 – Philip J. O’Keefe, PE Engineering Expert Witness Blog ____________________________________

### How Much Kinetic Energy is Contained Within a Spinning Flywheel?

Thursday, October 26th, 2017
 Last time we introduced the fact that spinning flywheels are subject to both linear and angular velocities, along with a formula that allows us to calculate these quantities for a single part of the flywheel, designated A.   We also re-visited the kinetic energy formula.   Today we’ll build upon those formulas as we attempt to answer the question, How much kinetic energy is contained within a spinning flywheel?     Here again is the basic kinetic energy formula, KE = ½ × m × v2                                                                (1) where, m equals a moving object’s mass and v is its linear velocity.     Here again is the formula used to calculate linear and angular velocities for a single part A within the flywheel, where part A’s linear velocity is designated vA,  angular velocity by ω, and where rA is the distance of part A from the flywheel’s center of rotation. vA = rA × ω                                                                         (2)     Working with these two formulas, we’ll insert equation (2) into equation (1) to obtain a kinetic energy formula which allows us to calculate the amount of energy contained in part A of the flywheel, KEA = ½ × mA × (rA × ω)2                                                 (3) which simplifies to, KEA = ½ × mA ×  rA2  ×  ω2                                              (4)     Equation (4) is a great place to begin to calculate the amount of kinetic energy contained within a spinning flywheel, however it is just a beginning, because a flywheel contains many parts.  Each of those parts has its own mass, m, and is a unique distance, r, from the flywheel’s center of rotation, and all these parts must be accounted for in order to arrive at a calculation for the total amount of kinetic energy contained within a spinning flywheel. How Much Kinetic Energy is Contained Within a Spinning Flywheel?         Put another way, we must add together all the m × r2 terms for each and every part of the entire flywheel.   How many parts are we speaking of?  Well, that depends on the type of flywheel.  We’ll discuss that in detail later, after we define a phenomenon that influences the kinetic energy of a flywheel known as the moment of inertia.     For now, let’s just consider the flywheel’s parts in general terms.  A general formula to compute the kinetic energy contained within the totality of a spinning flywheel is, KE = ½ × ∑[m × r2] × ω2                                                  (5)     We’ll discuss the significance of each of these variables next time when we arrive at a method to calculate the kinetic energy contained within all the many parts of a spinning flywheel . Copyright 2017 – Philip J. O’Keefe, PE Engineering Expert Witness Blog ____________________________________