Monthly Archives: March 2019

Ut tensio sic vis

And if two men ride of a horse, one must ride behind.

Shakespeare, Much Ado About Nothing

Sir Isaac Newton stands in popular estimation as the foremost intellect of his age; or perhaps, of any age. If a person is never truly dead while their name is spoken, then Sir Isaac stands with us still: partially overshadowed by Einstein at the dawn of the twentieth century, maybe, but never totally eclipsed.

But in the roiling intellectual cauldron of the Age of Enlightenment, even such a venerable polymath as Newton had some serious competition. As Newton himself modestly observed in a letter to a contemporary in 1676: “If I have seen a little further it is by standing on the shoulders of Giants.”

Except that one interpretation has it that the letter was not intended to be modest, but was rather a combative dig at the man to whom it was addressed: Robert Hooke, a man of but “middling” stature and, as a result of a childhood illness, also a hunchback. Not one of the “Giants” with broad philosophic shoulders to whom Newton felt indebted to, then.

Robert Hooke, as painted by Rita Greer in 2007. The painting is based on contemporary descriptions of Robert Hooke. No undisputed contemporary paintings or likenesses of Hooke have survived, possibly because of malicious intent on the part of Newton.

In popular estimation, therefore, it would appear that Hooke is fated always to sit behind Newton. At GCSE and A-level, students learn of Newton’s Laws of Motion, the eponymous unit of force, and his Law of Universal Gravitation.

And what do they learn of Hooke? They learn of his work on springs. They learn of “Hooke’s Law”: that is, the force exerted by a spring is directly proportional to its extension.

Ut tensio, sic vis.

[As extension, so is the force.]

— Robert Hooke, Lectures de Potentia Restituvia [1678]

Newton has all the laws of motion on Earth and in Heaven in the palm of his hand, and Hooke has springs. Perhaps, then, Hooke deserves to be forever second on the horse of eternal fame?

But look closer. To what objects or classes of object can we apply Hooke’s Law? The answer is: all of them.

Hooke’s Law applies to everything solid: muscle, bone, sinew, concrete, wood, ice, crystal and stone. Stretch them or squash them, and they will deform in exact proportion to the size of the force applied to them.

That is, if one power stretch or bend it one space, two will bend it two, and three will bend it three, and so forward.

The major point being that Hooke’s Law is as universal as gravity: it is baked into the very fabric of the universe: it is a direct consequence of the interactions between atoms.

Graph of interatomic force against distance between two atoms. Hooke’s Law applies in the red circle.

Now before I wax too lyrical, it must be pointed out that Hooke’s Law is a first-order linear approximation: it fails when the deforming force increases beyond a certain limit, and that limit is unique to each material. But within the limits of its domain indicated by the red circle above, it reigns supreme.

How do you calculate how much a steel beam will bow when a kitten walks across it? Hooke’s Law. How could we model the stresses on the bones of a galloping dinosaur? Hooke’s Law. How can we calculate how much Mount Everest bends when it is buffeted by wind? Hooke’s Law.

Time to re-evaluate the seating order on Shakespeare’s horse, mayhap?

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Why does kinetic energy = 1/2mv^2?

Why does kinetic energy Ek=½mv2?

Students and non-specialist teachers alike wonder: whence the half?

This post is intended to be a diagrammatic answer to this question using a Singapore Bar Model approach: so pedants, please avert your eyes.

I am indebted to Ben Rogers’ recent excellent post on showing momentum using the Bar Model approach for starting me thinking along these lines.

Part the First: How to get the *wrong* answer

Imagine pushing an object with a mass m with a constant force F so that it accelerates with a constant acceleration a so that covers a distance s in a time t. The object was initially at rest and ends up moving at velocity v.Screenshot 2019-03-09 at 14.24.59.png

(On the diagram, I’ve used the SUVAT dual coding conventions that I suggested in a previous post.)

So let’s consider the work done on the object by the force:

Step 1:    work done = force x distance moved in the direction of the force

Step 2:    W= F x s

But remember s = v x t so:

Step 3:    W= F x vt

And also remember that F = m x a so:

Step 4:    W= ma x vt

Also remember that a = change in velocity / time, so a = (v – 0) / t = v / t.

Step 5:        Wd = (v / t) x vt

The ts cancel so:

Step 6:    W= mv2

Since this is the work done on the object by the force, it is equal to the energy transferred to the kinetic energy store of the object. In other words, it is the energy the object has gained because it is moving — its kinetic energy, no less: E= mv2.

On a Singapore Bar Model diagram this can be represented as follows:

Screenshot 2019-03-09 at 15.14.17

The kinetic energy is represented by the volume of the bar.

But wait: Ek=mv2!?!?

That’s just wrong: where did the half go?

Houston, we have a problem.

Part the Second: how to get the *right* answer

The problem lies with Step 3 above. We wrongly assumed that the object has a constant velocity over the whole of the distance s.

Screenshot 2019-03-09 at 17.35.43.pngIt doesn’t because it is accelerating: it starts off moving slowly and ends up moving at the maximum, final velocity v when it has travelled the total distance s.

So Step 3 should read:

But remember that s = (average velocity) x t.

Because the object is accelerating at a constant rate, the average velocity is (v + u) / 2 and since u = 0 then average velocity is v / 2.

Step 3:    Wd= F x (v / 2) t

And also remember that F = m x a so:

Step 4:    Wd= ma x (v / 2) t

Also remember that a = change in velocity / time, so a = (v – 0) / t = v / t.

Step 5:        Wd = (v / t) x (v / 2) t

The ts cancel so:

Step 6:    Wd= ½mv2

Based on this, of course, E= ½mv2
(Phew! Houston, we no longer have a problem.)

Screenshot 2019-03-09 at 17.58.45.png

Using the Bar Model representation, the volume of the bar which is above the blue plane represents the kinetic energy of an object of mass m moving at a velocity v.

The reason it is half the volume of the bar and not the full volume (as in the incorrect Part the First analysis) is because we are considering the work done by a constant force accelerating an object which is initially at rest; the velocity of the object increases gradually from zero as the force acts upon it. It therefore takes a longer time to cover the distance s than if it was moving at a constant velocity v from the very beginning.

So there we have it, E= ½mvby a rather circuitous method.

But why go “all around the houses” in this manner? For exactly the same reason as we might choose to go by the path less travelled on some of our other journeys: quite simply, we might find that we enjoy the view.

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Filed under Bar Model, Dual Coding, Education, Physics