Archive for the ‘Professional Malpractice’ Category

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 vAangular 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?

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

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Copyright 2017 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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Two Types of Velocity Associated With a Spinning Flywheel

Thursday, October 19th, 2017

    Anyone who has spun a potter’s wheel is appreciative of the smooth motion of the flywheel upon which they form their clay, for without it the bowl they’re forming would display irregularities such as unattractive bumps.   The flywheel’s smooth action comes as a result of kinetic energy, the energy of motion, stored within it.   We’ll take another step towards examining this phenomenon today when we take our first look at calculating this kinetic energy.   To do so we’ll make reference to the two types of velocity associated with a spinning flywheel, angular velocity and linear velocity, both of which engineers must negotiate anytime they deal with rotating objects.

    Let’s begin by referring back to the formula for calculating kinetic energy, KE.  This formula applies to all objects moving in a linear fashion, that is to say, traveling a straight path.   Here it is again,

KE = ½ × m × v2

where m is the moving object’s mass and v its linear velocity.

    Flywheels rotate about a fixed point rather than move in a straight line, but determining the amount of kinetic energy stored within a spinning flywheel involves an examination of both its angular velocity and linear velocity.   In fact, the amount of kinetic energy stored within it depends on how fast it rotates.

    For our example we’ll consider a spinning flywheel, which is basically a solid disc.   For our illustrative purposes we’ll divide this disc into hypothetical parts, each having a mass m located a distance r from the flywheel’s center of rotation.   We’ll select a single part to examine and call that A.

Two Types of Velocity Associated With a Spinning Flywheel

Two Types of Velocity Associated With a Spinning Flywheel

   

    Part A has a mass, mA, and is located a distance rA from the flywheel’s center of rotation.   As the flywheel spins, part A rides along with it at an angular velocity, ω, following a circular path, shown in green.   It also moves at a linear velocity, vA, shown in red.   vA represents the linear velocity of part A measured at any point tangent to its circular path.  This linear velocity would become evident if part A were to become disengaged from the flywheel and cast into the air, whereupon its trajectory would follow a straight line tangent to its circular path.

    The linear and angular velocities of part A are related by the formula,

vA = rA × ω

    Next time we’ll use this equation to modify the basic kinetic energy formula so that it’s placed into terms that relate to a flywheel’s angular velocity.   This will allow us to define a phenomenon at play in the flywheel’s rotation, known as the moment of inertia.

 

 

Copyright 2017 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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Another Specialized Application of the Euler-Eyelewein Formula

Tuesday, June 13th, 2017

    Last week we saw how friction coefficients as used in the Euler-Eyelewein Formula, can be highly specific to a specialized application, U.S. Navy ship capstans.   In fact, many diverse industries benefit from aspects of the Euler-Eytelwein Formula.   Today we’ll introduce another engineering application of the Formula, exploring its use within the irrigation system of a hydroponics plant.

Another Specialized Application of the Euler-Eyelewein Formula

Another Specialized Application of the Euler-Eyelewein Formula

   

    Pumps conveying water are an indispensable part of a hydroponics plant.   In the schematic shown here they are portrayed by the symbol ⊗.

    In our simplified scenario to be presented next week, these pumps are powered by a mechanical power transmission system, each consisting of two pulleys and a belt.   One pulley is connected to a water pump, the other pulley to a gasoline engine.   A belts runs between the pulleys to deliver mechanical power from the engine to the pump.

    The width of the belts is a key component in an efficiently running hydroponics plant.   We’ll see how and why that’s so next time.

 

Copyright 2017 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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Friction Results in Heat and Lost Work Within a Compound Pulley

Saturday, January 7th, 2017

    Last time we saw how the presence of friction reduces mechanical advantage in an engineering scenario utilizing a compound pulley.   We also learned that the actual amount of effort, or force, required to lift an object is a combination of the portion of the force which is hampered by friction and an idealized scenario which is friction-free.   Today we’ll begin our exploration into how friction results in reduced work input, manifested as heat energy lost to the environment.   The net result is that work input does not equal work output and some of Mr. Toga’s labor is unproductive.

Friction Results in Heat and Lost Work Within a Compound Pulley

Friction Results in Heat and Lost Work Within a Compound Pulley

    In a past blog, we showed how the actual force required to lift our urn is a combination of F, an ideal friction-free work effort by Mr. Toga, and FF , the extra force he must exert to overcome friction present in the wheels,

FActual = F + FF                        (1)

    Mr. Toga is clearly working to lift his turn, and generally speaking his work effort, WI, is defined as the force he employs multiplied by the length, d, of rope that he pulls out of the compound pulley during lifting.   Mathematically that is,

WI = FActual × d                      (2)

    To see what happens when friction enters the picture, we’ll first substitute equation (1) into equation (2) to get WI in terms of F and FF,

WI = (F + FF ) × d                   (3)

Multiplying through by d, equation (3) becomes,

WI  =  (F × )+ (FF  × d)        (4)

    In equation (4) WI is divided into two terms.   Next time we’ll see how one of these terms is beneficial to our lifting scenario, while the other is not.

 Copyright 2017 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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Pulleys Make Santa’s Job Easier

Monday, December 19th, 2016

  Happy Holidays from EngineeringExpert.net, LLC and the Engineering Expert Witness Blog.Pulleys Make Santa's Job Easier

Pulleys Make Santa’s Job Easier

    

 Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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Rope Length Tradeoff in a Compound Pulley

Friday, November 18th, 2016

    We’re all familiar with the phrase, “too much of a good thing.”  As a professional engineer, I’ve often found this to be true.   No matter the subject involved, there inevitably comes a point when undesirable tradeoffs occur.   We’ll begin our look at this phenomenon in relation to compound pulleys today, and we’ll see how the pulley arrangement we’ve been working with encounters a rope length tradeoff.   Today’s arrangement has a lot of pulleys lifting an urn a short distance.

    We’ll be working with two distance/length factors and observe what happens when the number of pulleys is increased.   Last time we saw how the compound pulley is essentially a work input-output device, which makes use of distance factors.   In our example below, the first distance/length factor, d1, pertains to the distance the urn is lifted above the ground.   The second factor, d2, pertains to the length of rope Mr. Toga extracts from the pulley while actively lifting.   It’s obvious that some tradeoff has occurred just by looking at the two lengths of rope in the image below as compared to last week.   What we’ll see down the road is that this also affects mechanical advantage.

    The compound pulley here consists of 16 pulleys, therefore it provides a mechanical advantage, MA, of 16.   For a refresher on how MA is determined, see our preceding blog.

Rope Length Tradeoff in a Compound Pulley

Rope Length Tradeoff in a Compound Pulley

   

    With an MA of 16 and the urn’s weight, W, at 40 pounds, we compute the force, F, Mr. Toga must exert to actively lift the urn higher must be greater than,

F > W ÷ MA

F > 40 Lbs. ÷ 16

 F > 2.5 Lbs.

    Although the force required to lift the urn is a small fraction of the urn’s weight, Mr. Toga must work with a long and unwieldy length of rope.   How long?   We’ll find out next time when we’ll take a closer look at the relationship between d1 and d2.

 Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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More Pulleys Increase Mechanical Advantage

Thursday, October 27th, 2016

    Last time we saw how compound pulleys within a dynamic lifting scenario result in increased mechanical advantage to the lifter, mechanical advantage being an engineering  phenomenon that makes lifting weights easier.   Today we’ll see how the mechanical advantage increases when more fixed and movable pulleys are added to the compound pulley arrangement we’ve been working with.

More Pulleys Increase Mechanical Advantage

More Pulleys Increase Mechanical Advantage

   

    The image shows a more complex compound pulley than the one we previously worked with.   To determine the mechanical advantage of this pulley, we need to determine the force, F5, Mr. Toga exerts to hold up the urn.

    The urn is directly supported by four equally spaced rope sections with tension forces F1, F2, F3, and F4.   The weight of the urn, W, is distributed equally along the rope, and each section bears one quarter of the load.   Mathematically this is represented by,

F1 = F2 = F3 = F= W ÷ 4

    If the urn’s weight wasn’t distributed equally, the bar directly above it would tilt.   This tilting would continue until equilibrium was eventually established, at which point all rope sections would equally support the urn’s weight.

    Because the urn’s weight is equally distributed along a single rope that’s threaded through the entire pulley arrangement, the rope rule, as I call it, applies.   The rule posits that if we know the tension in one section of rope, we know the tension in all rope sections, including the one Mr. Toga is holding onto.   Therefore,

F1 = F2 = F3 = F4 = F5 = W ÷ 4

    Stated another way, the force, F5 , Mr. Toga must exert to keep the urn suspended is equal to the weight force supported by each section of rope, or one quarter the total weight of the urn, represented by,

F5 = W ÷ 4

    If the urn weighs 40 pounds, Mr. Toga need only exert 10 pounds of bicep force to keep it suspended, and today’s compound pulley provides him with a mechanical advantage, MA, of,

MA = W ÷  F5

MA = W ÷  (W ÷ 4)

MA = 4

    It’s clear that adding the two extra pulleys results in a greater benefit to the man doing the lifting, decreasing his former weight bearing load by 50%.   If we added even more pulleys, we’d continue to increase his mechanical advantage, and he’d be able to lift far heavier loads with a minimal of effort.   Is there any end to this mechanical advantage?  No, but there are undesirable tradeoffs.   We’ll see that next time.

 Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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Calculating the Force of Friction

Wednesday, April 27th, 2016

    Last time we introduced the frictional force formula which is used to calculate the force of friction present when two surfaces move against one another, a situation which I as an engineering expert must sometimes negotiate.   Today we’ll plug numbers into that formula to calculate the frictional force present in our example scenario involving broken ceramic bits sliding across a concrete floor.

   Here again is the formula to calculate the force of friction,

FF = μ × m × g

where the frictional force is denoted as FF, the mass of a piece of ceramic sliding across the floor is m, and g is the gravitational acceleration constant, which is present due to Earth’s gravity.   The Greek letter μ, pronounced “mew,” represents the coefficient of friction, a numerical value predetermined by laboratory testing which represents the amount of friction at play between two surfaces making contact, in our case ceramic and concrete.

    To calculate the friction present between these two materials, let’s suppose the mass m of a given ceramic piece is 0.09 kilograms, μ is 0.4, and the gravitational acceleration constant, g, is as always equal to 9.8 meters per second squared.

   

Calculating the Force of Friction

Calculating the Force of Friction

   

    Using these numerical values we calculate the force of friction to be,

FF = μ × m × g

FF = (0.4) × (0.09 kilograms) × (9.8 meters/sec2)

FF = 0.35 kilogram meters/sec2

FF = 0.35 Newtons

    The Newton is shortcut notation for kilogram meters per second squared, a metric unit of force.   A frictional force of 0.35 Newtons amounts to 0.08 pounds of force, which is approximately equivalent to the combined stationary weight force of eight US quarters resting on a scale.

    Next time we’ll combine the frictional force formula with the Work-Energy Theorem formula to calculate how much kinetic energy is contained within a single piece of ceramic skidding across a concrete floor before it’s brought to a stop by friction.

Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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Willem Gravesande’s Experimentation on Kinetic Energy

Friday, September 11th, 2015

      Last time we introduced The Law of Conservation of Energy, which holds that energy can neither be created nor destroyed.    We then applied the concept to a mug resting on a shelf, brimming with latent gravitational potential energy.    Today we’ll continue our discussion with a focus on kinetic energy and how Willem Gravesande’s experimentation contributed to our understanding of the subject.

      The concept of kinetic energy was first posited by mathematicians Gottfried Leibniz and Johann Bernoulli in the early 18th Century when they theorized that the energy of a moving object is a factor of its mass and speed.    Their theory was later proven by Willem Gravesande, a Dutch lawyer, philosopher, and scientist.

      Gravesande conducted experiments in which he dropped identical brass balls into a soft clay block.  See Figure 1.

Gravesande's Experiment

Figure 1

      Figure 1 shows the results obtained when balls of the same mass m are dropped from various heights, resulting in different velocities as they fall and different clay penetrations.    The ball on the left falls at velocity v and penetrates to a depth d.    The center ball falls at twice the left ball’s velocity, or 2v, and penetrates four times as deep, or 4d.    The right ball falls at three times the left ball’s velocity, 3v, and it penetrates nine times deeper, 9d.    The results indicate an exponential increase in clay penetration, dependent on the balls’ speed of travel.

      In fact, all the kinetic energy that the balls exhibited during freefall was converted into mechanical energy from the instant they impacted the clay until their movement within it stopped.    This change in forms of energy from kinetic to mechanical demonstrates what Julius Robert von Mayer had in mind when he derived his Law of Conservation of Energy.   For a refresher on the subject, see last week’s blog, The Law of Conservation of Energy.

      As a result of his experimentation, Gravesande was able to conclude that the kinetic energy of all falling objects is a factor of their mass multiplied by their velocity squared, or m × v2.

      We’ll see next time how Gravesande’s work paved the way for later scientists to devise the actual formula used to calculate kinetic energy and then we’ll apply it all to our coffee mug falling from the shelf.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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Gravity and the Mass of the Sun

Friday, December 12th, 2014

      As a young school boy I found it hard to believe that scientists were able to compute the mass of our sun.  After all, a galactic-sized measuring device does not exist.  But where there’s a will, there’s a way, and by the 18th Century scientists had it all figured out, thanks to the work of others before them.  Newton’s two formulas concerning gravity were key to later scientific discoveries, and we’ll be working with them again today to derive a third formula, bringing us a step closer to determining our sun’s mass.

      Newton’s Second Law of Motion allows us to compute the force of gravity, Fg, acting upon the Earth, which has a mass of m. It is,

Fg = m × g                     (1)

      Newton’s Universal Law of Gravitation allows us to solve for g, the sun’s acceleration of gravity value,

g = (G × M) ÷  r2              (2)

where, M is the mass of the sun, r is the distance between the sun and Earth, and G is the universal gravitational constant.

Force of gravity, falling objects, engineering expert witness

      You will note that g is a common factor between the two equations, and we’ll use that fact to combine them.  We’ll do so by substituting the right side of equation (2) for the g in equation (1) to get,

Fg = m × [(G × M) ÷  r2]

then, using algebra to rearrange terms, we’ll set up the combined equation to solve for M, the sun’s mass:

M = (Fg × r2) ÷ (m × G)           (3)

      At this point in the process we know some values for factors in equation (3), but not others.  Thanks to Henry Cavendish’s work we know the value of m, the Earth’s mass, and G, the universal gravitational constant.  What we don’t yet know is Earth’s distance to the sun, r, and the gravitational attractive force, Fg, that exists between them.

      Next time we’ll introduce some key scientists whose work contributed to a method for computing the distance of our planet Earth to its sun.

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