Welcome back.
Well, I'm now going to prove to
you that the magnitude of
centripetal acceleration is
equal to the magnitude of the
velocity when you're going
around the circle divided--
velocity squared divided
by the radius.
So let's start with the drawing,
just so that we know
what we're doing, just as much
for me as it is for you.
So that's the circle, and
you can guess that's
going to be our path.
And let's call p our
position vector.
And this is the center of
the circle right here.
Let me do the position
vector in magenta.
So let's say this is
p, my position.
Let me draw that bolder.
So that's the vector p.
And let me define a few other
things over here.
So let's say that the angle
that it's forming with the
positive x-axis-- let's
say this is the
positive x-axis-- is theta.
That's theta right there.
Let's say the radius
of the circle is r.
So we have an object.
It's right here.
This is its position defined by
this position vector, and
it's spinning around.
So its position vector is
going-- at some point, the
arrow's going to be pointed
there and then there.
It's just going to be going
around and around the circle.
But this is its position vector
at some moment in time.
So what is that position vector
in our bracket notation
of vectors?
We have to figure out its
x and y-components.
Its x-component is right here,
or you could almost say its
i-component, if we were doing
engineering notation.
That's its x-component and
that's its y-component, right?
So in our bracket notation or
whatever, I always forget what
I call things, p is this.
p, which is our position
vector, what's its
x-component?
It's the radius times
the cosine of theta.
This should be second nature
to you at this point.
Radius times cosine of theta.
What's its y-component?
It's this.
It's just this, right?
That's its y-component.
Radius times the
sine of theta.
Fair enough.
Hopefully, that makes
sense to you so far.
I just defined its position
vector and I drew it out here
visually, and then I also wrote
it analytically in its x
and y-components, as a sum
of its x y-components.
Well, that's good
and everything.
So let's see if we can
figure out what its
velocity vector is.
Well, what is velocity?
Velocity is actually just
a change in position.
Actually, now we are actually
dealing with
velocity, not speed.
We will actually get a vector.
So what is the velocity
vector?
So the velocity vector
is going to be
at any given point.
I'll do it in a different
color.
The velocity vector is going to
be tangent to the circle.
It's going to look something
like that.
That's going to be the
velocity vector.
So the velocity vector is equal
to the change in the
position over time.
So let's take the derivative
of the position vector with
respect to time.
And how do we take
a derivative?
Well, we could just take the
derivative of the x and
y-components separately, and
I'll show you how we do it in
this notation.
And if you think about
it, it should make a
little bit of sense.
So the radius is constant.
As we go around the circle,
the radius isn't changing.
Another thing to keep in mind,
we're going around.
We're spinning around the circle
at a constant rate.
So my speed isn't changing.
My velocity is obviously
changing, because the
direction is changing, but the
actual rate at which I'm
spinning, or the angular
velocity, is
going to be a constant.
That's something to keep
in mind for what we get
to in the next step.
But anyway, let's take
the derivative.
So let's take the derivative of
the x term first. Well, r
is just a constant so
it doesn't change.
We can just take the r out.
And then what's the derivative
of cosine theta
with respect to time?
Not with respect to theta.
Remember, we're taking
the derivative
with respect to time.
So we do the chain rule.
It's the derivative of theta
with respect to time times the
derivative of this term
with respect to theta.
And what's the derivative
of this?
What's minus sine of theta?
Let me put the minus out here.
Minus sine of theta.
I just didn't want to put the
minus right in front of the
sine, because you would think
it's minus sine theta.
So it's minus r times the rate
of change of the angle with
respect to time times sine of
theta, the rate at which this
term is changing with
respect to theta.
Let's do the same thing
on the y side.
r is a constant.
Chain rule: the rate at which
theta is changing
with respect to time.
you.
And then what's the derivative
of sine of theta
with respect to theta?
Well, that's cosine of theta.
This was just the chain
rule that I did.
Let's see if we can simplify
that a little bit.
So d theta, dt in both of these,
that's the same thing
as angular velocity.
And watch the video on angular
velocity if that doesn't make
sense to you.
But we can simplify this as just
w, angular velocity, and
that's going to be a constant.
And we have an r there.
Let's take a wr out
of both sides.
So we have the velocity vector
is equal to-- the velocity
vector is equal to wr--
actually, let's take minus wr
out, so minus wr, And then this
term is sine of theta.
And we're taking a minus wr, so
the wr goes away, and then
we introduce a minus
sign here, so it's
minus cosine of theta.
Good enough.
And the reason why we were able
to take-- and this w,
remember, is going
to be a constant.
It's not changing with
respect to time.
The angle is changing with
respect to time, but not the
rate of change of the angle.
It's spinning at a
constant rate.
So what's the acceleration
vector going to be?
The acceleration vector, and
I'll switch colors again to
keep it interesting.
The acceleration vector is just
the derivative of the
velocity vector with
respect to time.
And that equals-- this is this
constant term, so let's just
leave it on the outside.
It's minus wr.
And chain rule again.
If we're taking the derivative
with respect to time, first we
have to take the derivative of
theta with respect to time,
and we don't know
what that is.
That's just going to be w.
d theta dt and then times this
expression, the derivative of
this expression with
respect to theta.
The sine of theta derivative
is just cosine of theta.
And then on the y-side, what's
the derivative of theta with
respect to time?
It's just going to be omega.
And what's the derivative of
minus cosine theta with
respect to theta?
Well, that's sine theta.
And once again, this
is w and this is w.
We could take the w out
of the equation.
We get the acceleration vector
is equal to-- take the w out
of the x and y-components-- is
equal to minus w squared r
times cosine theta sine theta.
Or we could take this r and
multiply it times both of the
x and y-components, and we have
the acceleration vector
is equal to minus w squared r
cosine theta r sine theta.
Now does this thing here
look familiar?
Well, sure.
That was our original
position vector.
So we could say that the
acceleration vector is equal
to minus our angular velocity
squared times
the position vector.
And that makes actually a lot of
sense, because w squared is
going to be a positive term.
And what it's saying is that the
direction of this vector
is going to be the negative
of the direction of
the position vector.
So if our position vector
is going outward, our
acceleration vector, which I'll
draw in green, is going
to be going inward.
The acceleration vector
is going to be inward.
So it is what we
wanted to see.
Let's see if we can express
this as a function of the
magnitudes.
So we'll also say that the
magnitude of the acceleration
vector, and that's just a
without an arrow on top; I
could put brackets around it,
is equal to the negative
angular velocity squared
times the magnitude of
the position vector.
Well, what's the magnitude
of this position vector?
How long is it?
Well, its magnitude is r
by definition on the
beginning of the thing.
So this is just r.
So acceleration is equal to the
negative angular velocity
squared times r.
And what's angular velocity?
Well, we learned in that video
that angular velocity-- I'll
do it right here--
is equal to v/r.
If we just talk about the
magnitudes, not the vectors.
Remember, if we're not drawing
an arrow on top these
variables, they're just
scalar quantities.
So the acceleration is equal
to-- actually, we can get rid
of the negative sign because
we're not worried about
direction right now.
Well, we can keep it there.
It doesn't matter.
So we get v squared, because
the magnitude of this,
absolute value is essentially
magnitude.
So you get v squared over
r squared times r.
So r times r squared, you get
the acceleration is v
squared over r.
And that's what we
set out to prove.
And I'm out of time now.
So I'll see you in
the next video.
Hopefully, I didn't confuse
