Posts Tagged ‘integral calculus’

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

____________________________________