Posts Tagged ‘kinetic energy’

Coulomb’s Frictional Force

Monday, April 4th, 2016

    Humans have been battling the force of friction since the first cave man built the wheel.   Chances are his primitive tools produced a pretty crude wheel that first go-around and the wheel’s surfaces were anything but smooth, making its usefulness less than optimal.   As an engineering expert, I encounter these same dynamics when designing modern devices.   What held true for the cave man holds true for modern man, friction is often a counterproductive force which design engineers must work to minimize.   Today we’ll learn about frictional force and see how it impacts our example broken coffee mug’s scattering pieces, and we’ll introduce the man behind friction’s discovery, Charles-Augustin de Coulomb.

Coulomb's Work on Friction

Charles-Augustin de Coulomb

   

    Last time we learned that the work required to shatter our mug was transformed into the kinetic energy which propelled its broken pieces across a rough concrete floor.   The broken pieces’ energetic transformation will continue as the propelling force of kinetic energy held within them is transmuted back into the work that will bring each one to an eventual stop a distance from the point of impact.   This last source of work is due to the force of friction.

    In 1785 Charles-Augustin de Coulomb, a French physicist, discovered that friction results when two surfaces make contact with one another, and that friction is of two types, static or dynamic.   Although Leonardo Da Vinci had studied friction hundreds of years before him, it is Coulomb who is attributed with doing the ground work that later enabled scientists to derive the formula to calculate the effects of friction.   Our example scenario illustrates dynamic friction, that is to say, the friction is caused by one of the surfaces being in motion, namely the mug’s ceramic pieces which skid across a stationary concrete floor.

    While in motion, each of the mug’s broken pieces has its own unique velocity and mass and therefore a unique amount of kinetic energy.   The weight of each piece acts as a vertical force pushing the piece down “into” the floor, this due to the influence of Earth’s gravitational pull, that is, the force of gravity.

    Friction is created by a combination of factors, including the ceramic pieces’ weights and the surface roughness of both the pieces themselves and the concrete floor they skid across.   At first glance the floor and ceramic mug’s surfaces may appear slippery smooth, but when viewed under magnification it’s a whole different story.

    Next time we’ll examine the situation under magnification and we’ll introduce the formula used to calculate friction along with a rather odd sounding variable, mu.

Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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Kinetic Energy to Work, Work to Kinetic Energy

Thursday, March 24th, 2016

    Last time we watched our example ceramic coffee mug crash to a concrete floor, where its freefall kinetic energy performed the work of shattering it upon impact.   This is a scenario familiar to engineering experts like myself who are sometimes asked to reconstruct accidents and their aftermaths, otherwise known as forensic engineering.   Today we’ll take a look at what happens when the shattered mug’s pieces are freed from their formerly cozy, cohesive bond, and we’ll watch their transmutation from kinetic energy to work, and back to kinetic energy.

    As we watch our mug shatter on the floor, we notice that it breaks into different sized pieces that are broadcast in many directions around the point of impact.   Each piece has its own unique mass, m, travels at its own unique velocity, v, and has a unique and individualized amount of kinetic energy.   This is in accordance with the kinetic energy formula, shown here again:

KE = ½ × m × v2

    So where did that energy come from?

 The Scattering Pieces Have Kinetic Energy

The Scattering Pieces Have Kinetic Energy

   

    According to the Work-Energy Theorem, the shattered mug’s freefalling kinetic energy is transformed into the work that shatters the mug.   Once shattered, that work is transformed back into kinetic energy, the energy that fuels each piece as it skids across the floor.

    The pieces spray out from the point of the mug’s impact until they eventually come to rest nearby.   They succeed in traveling a fair distance, but eventually their kinetic energy is dissipated due to frictional force which slows and eventually stops them.

    The frictional force acting in opposition to the ceramic pieces’ travel is created when the weight of each fragment makes contact with the concrete floor’s rough surface, which creates a bumpy ride.   The larger the fragment, the more heavily it bears down on the concrete and the greater the frictional force working against it.   With this dynamic at play we see smaller, lighter fragments of broken ceramic cover a greater distance than their heavier counterparts.

    The Work-Energy Theorem holds that the kinetic energy of each piece equals the work of the frictional force acting against it to bring it to a stop.   We’ll talk more about this frictional force and its impact on the broken pieces’ distance traveled next time.

Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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When Kinetic Energy Meets With Opposing Force

Tuesday, March 1st, 2016

    Objects in motion inevitably meet with opposing forces, a theme which I frequently encounter in my work as an engineering expert.   Today we’ll calculate the opposing force our exemplar coffee mug meets when it falls into a pan of kitty litter, thus transforming its freefalling kinetic energy into the work required to move through clay litter.

    Let’s revisit the Work-Energy Theorem formula, whose terms were explained in last week’s blog,

F × d = – ½ × m × v12          (1)

    The left side of this equation represents the mug’s work to move through the litter, while the right side represents its kinetic energy, which it gained through freefall.   To solve for F, the amount of force acting in opposition to the mug’s mass m as it plows a depth d into the litter, we’ll isolate it on one side of the equation, as shown here,

F = [- ½ × m × v12 ] ÷ d          (2)

    So how do we solve for F when we don’t know the value of v1, the mug’s freefall velocity at impact?   We’ll use the fact that The Law of Conservation of Energy tells us that all energies are equal, and we’ll eliminate the part of Equation (2) that contains this unknown variable, that is, the right side of the equation which deals with kinetic energy.   In its place we’ll substitute terms which represent the mug’s potential energy, that is, the latent energy held within it as it sat upon the shelf prior to falling.  Equation (2) then becomes,

F = [- m × g × h] ÷ d           (3)

where g is the Earth’s acceleration of gravity factor, a constant equal to 9.8 meters/sec2 , and h is the height from which the mug fell.

Kinetic Energy Meets With Opposing Force

Kinetic Energy Meets With Opposing Force

   

    So if we know the mug’s mass, the distance fallen, and the depth of the crater it made in the litter, we can determine the stopping force acting upon it at the time of impact.   It’s time to plug numbers.

    Let’s say our mug has a mass of 0.25 kg, it falls from a height of 2 meters, and it makes a crater 0.05 meters deep.   Then the stopping force acting upon it is,

F = [- (0.25 kg) × (9.8 meters/sec2) × (2 meters)] ÷ (0.05 meters)

=  – 98 Newtons

    The mug was subjected to -98 Newtons, or about -22 pounds of opposing force when it fell into the litter, that resistance being presented by the litter itself.

    Next time we’ll see what happens when our mug strikes a hard surface that fails to cushion its impact.    Energy is released, but where does it go?

Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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Combining the Law of Conservation of Energy and Work-Energy Theorem

Thursday, February 18th, 2016

     It’s not uncommon in my work as an engineering expert to encounter a situation in which I’m missing information.   At that point I’ve got to find a creative solution to working the problem.   We’ll get creative today when we combine the  Law of Conservation of Energy and the Work-Energy Theorem to get around the fact that we’re missing a key quantity to calculate forces exerted upon the falling coffee mug we’ve been following in this blog series.

    Last time we applied the Work-Energy Theorem to our mug as it came to rest in a pan of kitty litter.   Today we’ll set up the Theorem formula to calculate the force acting upon it when it meets the litter.   Here’s where we left off,

 F × d =½ × m × v12

where, F is the force acting to slow the progress of the mug with mass m inside the litter pan.   The mug eventually stops and comes to rest in a crater with a depth, d.   The left side of the equation represents the mug’s work expenditure, as it plows through the litter, which acts as a force acting in opposition to the mug’s travel.

Work-Energy Theorem Applied To A Falling Coffee Mug

Kinetic Energy Meets Up With Displacement

   

    The right side of the equation represents the mug’s kinetic energy, which it gained in freefall, at its point of impact with the litter.   The right side is in negative terms because the mug loses energy when it meets up with this opposing force.

    Let’s say we know the values for variables d and m, quantities which are easily measured.   But the kinetic energy side of the equation also features a variable of unknown value, v1, the mug’s velocity upon impact.   This quantity is difficult to measure without sophisticated electronic equipment, something along the lines of a radar speed detector used by traffic cops.   For the purpose of our discussion we’ll say that we don’t have a cop standing nearby to measure the mug’s falling speed.

    If you’ll recall from past blog discussions, the Law of Conservation of Energy states that an object’s — in this case our mug’s — kinetic energy is equal to its potential energy.   It’s this equivalency relationship which will enable us to solve the equation and work around the fact that we don’t have a value for v1.

    We’ll do the math and plug in the numbers next time.

Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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Applying the Work-Energy Theorem to Falling Objects

Monday, February 8th, 2016

    So far we’ve applied the Work-Energy Theorem to a flying object, namely, Santa’s sleigh, and a rolling object, namely, a car braking to avoid hitting a deer.   Today we’ll apply the Theorem to a falling object, that coffee mug we’ve been following through this blog series.   We’ll use the Theorem to find the force generated on the mug when it falls into a pan of kitty litter.   This falling object scenario is one I frequently encounter as an engineering expert, and it’s something I’ve got to consider when designing objects that must withstand impact forces if they are dropped.

   

Applying the Work-Energy Theorem to Falling Objects

Applying the Work-Energy Theorem to Falling Objects

   

    Here’s the Work-Energy Theorem formula again,

F × d = ½ × m × [v22v12]

where F is the force applied to a moving object of mass m to get it to change from a velocity of v1 to v2 over a distance, d.

    As we follow our falling mug from its shelf, its mass, m, eventually comes into contact with an opposing force, F, which will alter its velocity when it hits the floor, or in this case a strategically placed pan of kitty litter.   Upon hitting the litter, the force of the mug’s falling velocity, or speed, causes the mug to burrow into the litter to a depth of d.   The mug’s speed the instant before it hits the ground is v1, and its final velocity when it comes to a full stop inside the litter is v2, or zero.

    Inserting these values into the Theorem, we get,

F × d = ½ × m × [0 – v12]

F × d = – ½ × m × v12

    The right side of the equation represents the kinetic energy that the mug acquired while in freefall.   This energy will be transformed into Gaspard Gustave de Coriolis’ definition of work, which produces a depression in the litter due to the force of the plummeting mug.   Work is represented on the left side of the equal sign.

    Now a problem arises with using the equation if we’re unable to measure the mug’s initial velocity, v1.   But there’s a way around that, which we’ll discover next time when we put the Law of Conservation of Energy to work for us to do just that.

 

Copyright 2016 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

   

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The Math Behind the Work-Energy Theorem

Friday, January 1st, 2016

    As an engineering expert I’ve applied the Work-Energy Theorem to diverse situations, but none as unique as its most recent application, the progress of Santa’s sleigh.   Last week we saw how Santa and his reindeer team encountered a wind gust which generated enough force to slow them from an initial velocity of v1 to a final velocity, v2, over a distance, d.   Today we’ll begin using the Work-Energy Theorem to see if Santa was able to keep to his Christmas delivery schedule and get all the good boys and girls their gifts in time.

Santa and the Work Energy Theorem

    Before we can work with the Work-Energy Theorem, we must first revisit the formula it’s predicated upon, de Coriolis’ formula for kinetic energy,

KE = ½ × m × v2                            (1)

where, KE is kinetic energy, m is the moving object’s mass, and v its velocity.

    The equation behind the Work-Energy Theorem is,

W = KE2 KE1                                    (2)

where W is the work performed, KE1 is the moving object’s initial kinetic energy and KE2 its final kinetic energy after it has slowed or stopped.   In cases where the object has come to a complete stop KE2 is equal to zero, since the velocity of a stationary object is zero.

    In order to work with equation (2) we must first expand it into a more useful format that quantifies an object’s mass and initial and final velocities.   We’ll do that by substituting equation (1) into equation (2).   The result of that term substitution is,

W =× m × v22 ] – [½ × m × v12]      (3)

    Factoring out like terms, equation (3) is simplified to,

W = ½ × m × [v22v12]                        (4)

    Now according to de Coriolis, work is equal to force, F, times distance, d.   So substituting these terms for W in equation (4), the expanded version of the Work-Energy Theorem becomes,

F × d = ½ × m × [v22v12]                 (5)

    Next time we’ll apply equation (5) to Santa’s delivery flight to calculate the strength of that gust of wind slowing him down.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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The Work-Energy Theorem — Background

Friday, December 18th, 2015

      My work as an engineering expert sometimes involves computations of energy expended, as when I must determine how much energy is required to move something.   But sometimes the opposite needs to be calculated, that is, how much energy is required to stop something already in motion.   That’s the subject of today’s discussion, which we’ll approach by way of the Work-Energy Theorem.

      The Work-Energy Theorem states that the work required to slow or stop a moving object is equal to the change in energy the object experiences while in motion, that is, how its kinetic energy is reduced or completely exhausted.   Although we don’t know who to attribute the Theorem to specifically, we do know it’s based on the previous work of Gaspard Gustave de Coriolis and James Prescott Joule, whose work in turn built upon that of Isaac Newton’s Second Law of Motion.

      Consider the example shown here.  A ball of mass m moves unimpeded through space at a velocity of v1 until it is met by an opposing force, F.   This force acts upon the ball over a travel distance d, resulting in the ball’s slowing to a velocity of v2.

 The Work-Energy Theorem

The Work – Energy Theorem Illustrated

 

      Does the illustration make clear the Work-Energy Theorem dynamics at play?   If not, return for the second part of this blog, where we’ll clarify things by getting into the math behind the action.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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Converting Kinetic Energy to Electrical Energy

Tuesday, November 3rd, 2015

     When acting as an engineering expert I’m often called upon to investigate incidents where energy converts from one form to another, a phenomenon that James Prescott Joule observed when he built his apparatus and performed his experiments with electricity.   Today we’ll apply Joule’s findings to our own experiment with a coffee mug when we convert its kinetic energy into electrical energy and see how the units used to express that energy also change. 

     We had previously calculated the kinetic energy contained within our falling coffee mug to be 4.9 kg • meter2/second2, also known as 4.9 Joules of energy, by using de Coriolis’ Kinetic Energy Formula.   Now most of us don’t speak in terms of Joules of energy, but that’s easily addressed.   As we learned in a previous blog on The Law of Conservation of Energy, all forms of energy are equivalent and energy can be converted from one form to another, and when it does, the unit of energy used to express it also changes.

     Let’s say we want to put our mug’s 4.9 Joules of kinetic energy to good use and power an electric light bulb.   First we must first find a way of converting the mug’s kinetic energy into electrical energy.   To do so, we’ll combine Joule’s apparatus with his dynamo, and connect the mug to this hybrid device with a string.

Converting Kinetic Energy to Electrical Energy

                     Converting Kinetic Energy to Electrical Energy

     As the mug falls its weight tugs on the string, causing the winding drum to rotate.   When the drum rotates, the dynamo’s magnet spins, creating electrical energy.   That’s right, all that’s required to produce electricity is a spinning magnet and coils of wire, as explained in my previous blog, Coal Power Plant Fundamentals – The Generator.

     Now we’ll connect a 5 Watt bulb to the dynamo’s external wires.   The Watt is a unit of electrical energy named in honor of James Watt, a pioneer in the development of steam engines in the late 18th Century.

     Now it just so happens that 1 Watt of electricity is equal to 1 Joule of energy per a specified period of time, say a second.   This relationship is expressed as Watt • second.   Stated another way, 4.9 Joules converts to 4.9 Watt • seconds of electrical energy.   Let’s see how long we can keep that 5 Watt bulb lit with this amount of energy.    Mathematically this is expressed as,

Lighting Time = (4.9 Watt • seconds) ÷ (5 Watts) = 0.98 seconds

     This means that if the mug’s kinetic energy was totally converted into electrical energy, it would provide enough power to light a 5 Watt bulb for almost 1 second.

     Next time we’ll see what happens to the 4.9 Joules of kinetic energy in our coffee mug when it hits the floor and becomes yet another form of energy.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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James Prescott Joule and the Joule Apparatus

Tuesday, October 6th, 2015

      As an engineering expert I’ve often witnessed energy change forms, something our example coffee mug has been experiencing as it moves from a shelf to the floor.   The mug’s various energies were proven to be mathematically equivalent, expressed as, 4.9 kg•meter2 /second2 , which is read as, “kilogram meter squared per second squared.”   This mouthful led to the renaming of the measurement to the Joule, in honor of James Prescott Joule, a scientist who successfully demonstrated the interrelationship between different forms of energy.   We’ll focus on one of his experiments, the Joule Apparatus, today.

      Back in the 1840s Joule built his Apparatus, a device which demonstrated the interrelationship between different forms of energy.

The Joule Apparatus

The Joule Apparatus

      The Joule Apparatus consisted of a weight suspended by string over a pulley, which in turn was wound around a winding drum.   As long as the drum remained stationary, the weight remained motionlessly suspended.   While motionless, the weight’s potential energy lay latent within it, just as it had in our example coffee mug resting on a shelf.

      But when the pressure keeping the winding drum stationary was released, the weight was set free to fall, and its potential energy began converting to kinetic.   In the process, the string the weight was attached to unwound from the drum, which caused the drum to turn and along with it the paddle wheel it was attached to.

      Joule’s Apparatus followed energy through many forms.  From the quiet of potential energy to the kinetic energy demonstrated by the falling weight.   The kinetic energy in turn was converted into mechanical energy, made manifest by the interaction between the moving drum and paddle wheel.   The rotating paddles agitated the water, causing its temperature to rise.    Observing this, Joule concluded that the mechanical energy of the spinning paddle wheel had been converted into heat energy, which temperature measurement proved was transferred into the water.   Joule’s experiment thus proved the link between potential, kinetic, mechanical, and heat energies.

      Joule’s work paved the way to make possible the later development of a host of modern mechanical devices that also converted heat energy into mechanical energy, or vice versa.   These devices include a car’s engine and your kitchen’s refrigerator.

      Next time we’ll see how Joule demonstrated a link between electrical and other forms of energy, including mechanical and heat.   We’ll then use his discoveries to convert our falling coffee mug’s kinetic energy into yet another form.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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Calculating Velocity — de Coriolis’ Kinetic Energy Formula

Monday, September 28th, 2015

      Last time we introduced Gaspard-Gustave de Coriolis’ formula to compute kinetic energy.  Today we’ll use it to determine the speed of descent, or velocity, of the coffee mug we’ve been watching closely in the last few blogs.  To calculate the mug’s velocity, we must bear in mind physicist Julius Robert von Mayer’s assertion that all forms of energy are interrelated, and in fact interchangeable, because energy can neither be created nor destroyed, it can only change forms.  For a refresher, see The Law of Conservation of Energy.

      Let’s now put a practical spin on this concept and apply it to our coffee mug’s free fall to the floor.  Once again, de Coriolis’ formula,

KE = ½ × m × v2                                               (1)

where m is the mass of our falling object and v its velocity.  The ½ is an unchanging, constant term that’s present due to the mathematical Rules of Integration governing integral calculus.  Calculus and its derivations are beyond the scope of this blog, but if you’re interested in pursuing this, follow this link to, The Physics Hypertextbook – Kinetic Energy.

      According to von Mayer’s Law, at the precise instant before the mug hits the floor its kinetic energy, KE, is equal to the potential energy, PE, it possessed when it rested passively on the shelf.  Stated another way, the instant before the mug makes contact with the floor, all its potential energy will have been converted into kinetic.

      The mug’s PE was calculated previously to be equal to 4.9 kg • meter2/second2.  See Computing Potential Energy for a review.  Knowing this, the mathematical relationship between the mug’s potential and kinetic energies is expressed as,

PE = KE = 4.9 kg • meter2/second2                         (2)

      By substituting this mathematical representation for KE into equation (1) we arrive at,

4.9 kg • meter2/second2 = ½ × m × v2                   (3)

We also know the mug’s mass, m, to be equal to 2.6 kilograms, so integrating that into the right side of equation (3) it becomes,

4.9 kg • meter2/second2 = ½ × ( 0.25kg) × v2         (4)

That leaves the mug’s velocity, v2, as the only remaining unknown term.  We’ll use algebra to isolate this variable by dividing both sides of equation (4) by ½ × ( 0.25kg).

(4.9 kg • meter2/second2) ÷ [½ × ( 0.25kg)] = v2

39.20 meter2/second2 = v2

Finally, we’ll take the square root of the equation to place it in terms of v.

6.26 meters/second = v

The mug’s velocity an instant before impact equates to 6.26 meters/second, or almost 21 feet per second.

mechanical engineering expert witness falling objects

      Next time we’ll discuss a metric unit used to measure energy known as the Joule and discover the man behind it.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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