Kinematics and pedagogy

[davieddy]

I used to wonder why many found elemntary Newtonian mechanics
much more difficult/mysterios than it is.
I realized (very soon after I started teaching) that they didn't
understand what acceleration meant, because kinematics (simple
but not trivial) just doesn't get taught.

Even constant acceleration: they are told to learn by rote 4 equations,
and forget how they were derived (even if they were ever shown).
If you don't believe this, glance at Physicsforums Homework and the
advice dished out by the "tutors"

These days, 12 year olds know about vectors, either as arrows or number
pairs.

My schedule:

Lesson one: What is meant by "change" in anything, and vectors in
partcular? (Final = Initial + Change).
"Average velocity" is change in position/time.
Homework one: an object is launched from (0,0) and after one second it is at (10,30). What is the average velocity for the first second?
The average velocity changes by (0,-10) each second. Plot the
trajectory on graph paper.

Lesson two: What is mass? When the cue ball hits the target ball,
the velocity of both is changed. If one changes by 2 m/s and the other
by 3 m/s, which has more mass? How much more?
Now demonstrate by experiment that those changes in velocity are
invariably in opposite directions and in the same ratio*.
In my view, that is no more or less than Newtonian mechanics
in a nutshell. (Force, momentum and energy are merely convenient
definitions). Just add the superposition principle (aka triangle of forces).

Now for you grownups:

Take a curved orbit r(t) = k₁f((t))
acceleration is a₁(t)

If we change k₁ to k₂, do you agree that
a₂(t)/k₂ = a₁(t)/k₁ ?

If so, I would be grateful for some critical appraisal of my musings
on the L4,5 Lagrangian points in This thread

David

* If the ratio for A and B is _Ak_B, one more crucial fact needs empirically verifying before we can consistently define mass, namely: _Ak_B * _Bk_C = _Ak_C
How many ever get taught that?

[lavalamp]

Originally Posted by davieddy

Even constant acceleration: they are told to learn by rote 4 equations,
and forget how they were derived (even if they were ever shown).

'Twas 5 equations when I did them.

I seem to recall Mr. Smith threatening that if we did not know them off by heart within a week, he would hang us upside down by our ankles and beat us with a stick.

It was never explained to us how they were derived, but it seemed fairly obvious regardless.

On the other hand, I had an excellent physics teacher who always showed the derivation of new formulae, and it did help cement my understanding of them. So I can certainly see how derivations can help students to learn.

[davieddy]

Originally Posted by lavalamp

'Twas 5 equations when I did them.

Yep.
s = (u+v)t/2 (no a)
v = u + at (no s)
Now use these to eliminate t, v and (if you insist), u.

Since these first two equations are simple cases of things
they should know: distance = area under a v/t graph and definition
of acceleration, it would do them good to forget the rest, and
give them much-needed practice in manipulating equations!

Wait till they can understand a = (1/2)dv²/ds and can integrate
v = u + at. Then they might understand why KE = (1/2)mv²
is generally applicable.

Now how about my 3 bodies indefinitely forming an equilateral triangle?

David

[xilman]

Originally Posted by davieddy

Now how about my 3 bodies indefinitely forming an equilateral triangle?

Ok, if you must.

It follows trivially from the initial symmetry and the fact that (Newtonian) gravitation is a purely central force.


Paul

[davieddy]

Originally Posted by xilman

Ok, if you must.

It follows trivially from the initial symmetry and the fact that (Newtonian) gravitation is a purely central force.


Paul

Symmetry?
Are you perchance assuming the masses are equal?

And some consideration as to the initial velocities would not go amiss
either.

But I did think the answer obvious enough to refrain from over-elaborating.

Then again, no Newspaper man ever went broke by underestimating
the intelligence of his readership.

David

[S34960zz]

Originally Posted by lavalamp

I seem to recall Mr. Smith threatening that if we did not know them off by heart within a week, he would hang us upside down by our ankles and beat us with a stick.

"Wrong, Do it again!"
"If you don't eat yer meat, how can you have any pudding? How can you have any pudding if you don't eat yer meat?"

[davieddy]

Originally Posted by lavalamp

It was never explained to us how they were derived, but it seemed fairly obvious regardless.

Yes, we can recognize s = ut + at²/2 as a parabola
starting from where the gradient is u, or the integral of v = u + at,
but let's face it, we actually think "This is trivial. I know it backwards
because Mr Smith told me it was true and I had to learn it - OR ELSE".

Just thought of another difficulty folk may have with these formulae:
none of them have t, a or u as the "subject". How could they
be expected to find them?


Looks like I'm going to have to give the answer to that 3/4 equidistant
body problem myself. Simple yes. Trivial no. Beautiful certainly.
After all, it seems to have escaped the attention of Newton, Euler,
Lagrange, Wiki and even Xilman
(I won't risk insulting Poincare)

David
(Couldn't resist!)

[davieddy]

Originally Posted by davieddy

Take a curved orbit r(t) = k₁f((t))
velocity is v₁(t), acceleration is a₁(t)

If we change k₁ to k₂, do you agree that
v₂(t)/k₂ = v₁(t)/k₁ and
a₂(t)/k₂ = a₁(t)/k₁ ?

If so, I would be grateful for some critical appraisal of my musings
on the L4,5 Lagrangian points in This thread

Originally Posted by xilman

It follows trivially from the initial symmetry and the fact that (Newtonian) gravitation is a purely central force.

Paul

This response simply proves my point in post #1:
the geometry and kinematics of a problem may be easy
or difficult but not trivial and not to be ignored.

In general, the gravitational field of more than one mass is
neither directed toward, nor proportional to 1/(distance from)²
the Centre of Mass.

But if three (or four!) masses are an equal distance (a) from each other,
it is simple to show that the acceleration of the ith mass is -GMr_i/a³ (see the post I linked to). r_i is the position of the ith mass relative to the C of M,
M is the sum of the masses.

This is why I find it difficult to believe that Euler was aware of
L1, L2 and L3, but not L4 and L5. And why do folk imply that one of
the three masses needs to be small or the orbits circular for L4 and L5
to be well-defined? (I'm not discussing stability here).

Note that if we posit that the equilateral triangle rotates and scales
about the C of M, the "orbits" of the masses are similar in the sense
of my quote above.

Now r_i = k_ia ("a" here = side of triangle, k_i the constant ratio)
so the acceleration of the ith mass is GMk_i³/r_i² towards the C of M.
Familiar?

David

[davieddy]

T² is proportional to a³, for acceleration = G(Mass of Sun)/r².
(This "a" is the semi major axis of the elliptical orbit ;=)

Just as well the constant of proportionality differs for our three
masses so that the speed of the masses is proportional to the
distance they have to travel, making the periods identical.

David