I have some circles today and they're
made of brass; and the wonderful thing
about circles, as we all know, is they
roll, which is why as an invention
they're up there just behind fire and
ahead of sliced bread. So if I set this
going, in theory- look at that! Right? Rolls
right across the table.
And the reason a circle rolls, and sorry
if I'm breaking this down a bit too much,
is because as it moves its centre of
mass, so if I draw that over here-
if only I had a template I
could have drawn around! As a circle
rolls its centre of mass stays at that
same height. Which is why, if I tried to
roll a square, I've got a square here,
there's its centre of mass in the middle.
As it rolls the centre of mass has to go
up and then it has- then it goes down
right? And so up and then down. So it will
roll but you've got to- you've got to
drive it, you've got to be constantly
forcing it because when it's here its
centre of mass is there, but then if it
lifts up the centre of mass has
gone up. So this- that will stay the same
whereas as you turn a square the centre of
mass has to be raised. The thing about
circles is you have- you still have to drive it
because we've got friction and, you know,
as always we live in a messy physical
reality; but the actual centre of mass
doesn't go up or down so that means it
rolls quite neatly. It's not the only way
you can roll a circle so its centre of
mass stays at the same height. I've got
two circles here, and they've both got
notches cut into them, and that means I
can slot them together. Now they are
brass, and they really grip quite- let me
just make sure I got that right in. So I've
now got two firmly locked together
circles and they're at right angles, they're
orthogonal to each other. And if I
put them down, the centre of mass is kind-
of well probably there-ish? And it's a
certain height above the table. The
amazing thing is, as it rolls that centre
of mass stays at exactly the same height.
The centre of mass of each individual
circle, because circles are quite
predictable it's in the middle. Actually
I've got some cardboard ones; so the
centre of mass, we can draw it on, in fact
it's got a picture of the disks on the
disk which I find hugely entertaining. So
the centre of the mass is there in
the middle; and that kind of makes sense
because it's, you know, the same distance
away. That's the, you know, symmetric
centre of that circle. When you've got two
of them together it gets a bit more
complicated. And so now you think, well
where's the exact centre of this? And you
can calculate it, which I did just to be
thorough, or you can just say, look,
symmetrically where's the- that is the
point there that that point right in
there is the kind of symmetric point
right in the centre of all the mass,
which is why we give it the strange name
the centre of mass. And so as long as
that spot stays the same height above
the surface, this nice flat table, then
you're not adding or receiving or taking
out any potential energy as it rolls. So
that's why, if this is just the right
intersection and that dot stays the same
height above the table, that's why you
get the rolling effect, there we go.
If you want to make these yourself then
you'll see I made them by cutting a notch
into a piece of- card's easier, you can make
them out of CDs or anything else. These
are made of brass,
so they roll a bit nicer but they're a
bit harder to machine up. But if you want
to make some of these yourself, you want to
make that cut. And you want to work out
how far in you have to cut for the whole
thing to work, and you can calculate
mathematically what you have to do. In
fact the easiest way to do the maths,
which is of course why we're here, is to
consider when you put them together the
centre of this disk- so that's the exact
middle of that disk there, that's the
middle of that disk there; normally
they'd be- if we didn't have grooves in
them they would sit like that. And so
that's the radius of this disk, that's
the radius of this disk. And so the
centres would be 2 times the radius
apart, at all times. By having those slots
though they get closer together; and the
question is, how far apart are these two
centres now that we've put the slots in
there so that height doesn't change? And
we can work that out with some triangles.
So, here we go, now there are two
important states we have to worry about
with the wobbler disk; and one is when
you're looking at it perfectly end on
and they're both at 45 degrees to the
surface, so that's that position there
which I'm going to draw end on here.
There, so that that's that's you know
looking at it that way, straight down. So
the centre of mass is there, it is now-
we're going to call this the height
above the table. That there is the radius
of the disks, and that there is our 45
degrees, or pi on 2 depending on what you
want to say. I mean you could say that is
tau on 4, but then you'd be an idiot. So
if- because that's 45, it means this here
is also the same as the height because
we have a wonderful 45-45 right angle
triangle there. And so we know, using our
friend Pythagoras, that r-squared equals
h squared plus h squared, which equals
2h squared. And so if we take the square
root of that
we know the radius equals square root of
two times h. So that's our ratio in
that position. The important thing is we
want that h to stay exactly the same
height even when it's rolling. So once
it's rolled for a bit and it ends up in
that position; so now you've got one disk
coming straight up and the other one is
pretty much- well it's not flat flat but
you know what I mean there. So let's
sketch that, so we've got one disk like that
and then there's the centre. Now that
centre is going to- line down to the
surface and there's going to be the
other disk there like that. Okay so
that's in that position. So once again
there is the centre of that disk there,
there's the centre of that disk there. What
we want to calculate is this distance
that the centres are apart. So that's d,
the distance we want to know, because
that tells us how big to make the
notches so that they sit at the right
distance. And that d we can calculate
because we know that's the centre; down
to where it touches that must be the
radius of that disk; and this here, that
height there, must be the radius because
it's the circle standing straight up; the
centre of mass is there which is halfway
between the two centres. And that is the
h which is all-important; and that
distance there, from that centre to there,
well that's half of this. So that little
distance there that's d on 2, because
it's half of that distance. And so now we
can use our good friends similar
triangles, which means that for this
small triangle here, so to get from the
bottom to there that's r plus d on 2
divided by h, equals the big triangle
which is r plus d over r. All right, so
now we're off and racing. All we have to
do is solve this to work out what d is.
We want to get rid of that annoying h
there, but we know from over here that h
equals r divided by root 2, so we can
substitute that in over here, so that now
becomes r plus d on 2 over r over
root 2, and that still equals r plus d
over r. To tidy this up a little bit I'm
going to multiply both sides by r,
because it's going to cancel out with
these ones underneath. And so that ends
up being root 2 outside of r plus d on
2 equals r plus d. And to get rid of this
mess at the front here I'm going to
multiply both sides by root 2. So if I
multiply this by root 2 I get a 2 at the
front, so that becomes 2r plus d because
that 2 cancels with the 1/2. And if I multiply
this by root 2 I get root 2 r plus root
2 d. And now I can put all the same terms
on the same side; so this one over here, if I
subtract that root 2 d from both
sides I get 1 d minus the root 2 outside
of d equals, and I subtract the 2 r
from both sides that's going to be root
2 minus 2r - have I made any mistakes? I've
probably made a mistake in there
somewhere. So all we want to do now is
work out the ratio that d is compared to
the radius. And so what we have to do is
rearrange this to get d in terms of r.
And because I know where we're going, and
because I'm picking a very careful path
through the algebraic, you know, field we
could be working our way across - I mean
if you try this, if you set up different
triangles, you did all these steps
differently you'd still get the same
answer at the end because that's why
maths is wonderful. Multiply both sides
by 1 plus root 2. 1 times 1 is 1,
and so we're going to multiply that by
that as well. Negative 2 times 1 is going
to be minus 2, so that cancels out,
that's lovely.
And so all of this has shown us that the
wobbling disks work only when those two
centres are root 2 times the radius of
the disks apart. There are far more
complicated ways through this terrain,
but no matter what you do, no matter what
algebra you're dancing around with, long
as you're careful, everything drops
out at the very end it's so lovely. No
matter what you do everything settles
and cancels out and you always end up
with root 2 times the radius. And I love
the fact that such a quite a complicated
system, and a decent amount of maths to
get there, is such a nice simple neat
answer. And that for me is mathematics.
First time I did this I didn't work out
that distance between the centres; I
worked out, because I'm a pragmatic man at heart,
- actually I won't pull those apart because
they're really wedged in - as a pragmatic
man I worked out what is that distance
there. So I did that calculation to try
and calculate that. And you can work it
out now, either you can work backwards
from this or I can just ruin the
surprise and tell you that that distance
there is 1 minus 1 over root 2 times the
radius. So in fact if you want to make
one of these, and for once I will
actually sketch one properly; so there's
your disk, and that's got some radius r.
That distance there is 1 minus 1 over
root 2 times whatever that radius r is.
If you're a lazy person that is about
29%, that is 0.29 times the radius,
give or take. 29% that's- it's
about- it's about 1/3. But if you want it
to work really well then you need to go
for 0.29 times the radius, or you
can work it out.
If you want to hear and
see even more mathematics from Matt
Parker, more than we could ever fit in
100 Numberphile videos, he's got a
new book out: Things to Make and Do in
the Fourth Dimension. Here are the front
covers of the book in the UK and the
United States; you can already order it
from wherever you normally get books, or
if you want a copy hand signed by Matt
himself go to his website. And I'm going
to put the links in the video
description. I think if you get the one
signed by Matt you're going to get the
UK one no matter where you live, check it
out. People like Matt give loads of time
to Numberphile, we filmed this video on a
Sunday afternoon when you could have
been out with family and friends. There's
not much we can do in return, but
checking out his book is one of those
things. And don't do it just to be nice!
Do it because it's an excellent book and
worth reading.