Calculus proof of centripetal acceleration formula Video Lecture - Class 11

What I want to do in this video is a calculus proof of the famous
centripetal acceleration formula that tells us the magnitude
of centripetal acceleration, the actual direction will change
it's always going to be pointing inwards, but the magnitude of centripetal acceleration is equal to the
magnitude of the velocity-squared divided by the radius
I want to be very clear, this is a scalar formula right over here; we're talking about the magnitude
of the acceleration and the magnitude of the velocity. If these were vectors,
we would have arrows drawn over it. So this really, I don't want people to get confused
because this is a v, this is really referring to the speed-squared.
and this is the magnitude, and these are all
scalar quantities. So to do that, let's imagine some
object, maybe it's some object in orbit around a planet or something,
so let's say that's the planet, and you have some object
that is in orbit around the planet, and it is going in a counter-clockwise
direction, and so let's specify its position vector as
a function as time.
So that is its position vector and it is going to
change as a function of time as this thing spins around
We're going to assume, for the purposes of this
proof.
So that is our
y-axis, and this is our x-axis.
We're going to define theta as the angle between the positive
x-axis and our vector.
And we're going to assume this thing is in orbit with the radius of r. So the
magnitude of our position vector, even though the direction is going to change
the magnitude of our position vector is not going to change. It's always going to have
length r. So this is going in a circle with radius r.
The magnitude of our position vector,
which is changing as a function of time, is going to be
r. So how can we write
the position vector in terms of its components at any given time?
We can write the position vector, and I'll do it in engineering notation,
and so you might want to review those videos if some of this looks foreign and I'll
do a bit of basic trigonometry in breaking down the vector into its components
and I encourage you to review some of those videos if some of that looks a bit
daunting. If you take the position vector at any time,
the magnitude is r,
this angle is theta,
its x-component, in blue,
this vector right over here, the magnitude of the vector, I should say,
is going to be r cosine of theta
We learned that this came from basic trigonometry when we started
two-dimensional projectile motion, we saw how to break these
vectors down into its components, and the y-component of this vector
is going to be r sine of theta
So this is going to be r sine of
theta. So the position vector at any time
can be written as a sum of its x- and y-components.
So it's the magnitude of its x-component, it's going to be
r cosine of theta, and I could write theta as a function of
time if I'd like, but I'm just going to write r cosine of theta
actually, let me write it that way, so it shows theta is a function of time
This thing is moving, and there's going to be that times the
i-unit vector, we're in engineering notation over here.
So that's the i-unit vector, it tells us that the x-component
is going in the positive x direction. Plus
the magnitude of the y-component, which is
r sine of theta, which is going to be a function of time.
So to be clear, the function of time applies to the theta.
And that is going
in the j-direction.
So that is our j-unit vector.
So now we have position as a function of theta
which is actually a function of time. So let's take the derivative of
this thing right here. So what is
the derivative of our position vector with respect to time
Well that's just going to be our velocity vector,
as a function of time, and it's going to be equal to
we just have to take the derivative of each of these parts with respect to
time. And you just do the chain rule. So you're going to have
the r sit outside cause that's just a constant. So you're going to have r
the derivative of cosine of theta t with respect to theta t
So I'm just doing the chain rule right over here. That's going to be
negative sine of theta t
and then as the chain rule, we also have to multiply that
times the derivative of the theta of t
with respect to t. So times d-theta, dt
so this is just the chain rule right over here.
So that's going to be how it's changing in the x-direction, and in the y-direction
we do something very similar. In the y-direction
we take the same derivative. We have the r scalar out front
r, and then the derivative of sine of theta
with respect to theta is going to be cosine of theta
and I'll write it as a function of time, and then do the chain rule
you'll also have to multiply that by the rate at which theta is changing with respect to t,
times d-theta, dt,
and this is all going to be times our j-unit vector. Now,
there's something you might already realize, and you should rewatch the video
on angular velocity if this is foreign to you, but
d-theta, dt, this is our angular velocity.
That's why I said to rewatch that video. This right over here, the rate at which
the angle changes with respect to time, that is angular velocity.
So this right over here is
angular velocity. And for the sake of this
video, this is an assumption we'll have to make for this formula right over here
we're going to assume, that omega, which is the rate of change
of our angle with respect to time, we're going to assume
that this is constant.
So this is an assumption we're making for this proof. This is we are going
to assume that omega is constant. And if omega is
constant, then we can treat it as a constant and we can factor
it out of this expression. So let's factor out
a negative omega-r from this expression over here. So we can rewrite
our velocity as a function of time is equal to
I'm going to factor out a negative omega
times r, and if you factor out a negative omega-r,
what you're left with is
this first term,
sine of theta-t
And I didn't have to make it explicit that theta is a function of t, but this makes it explicit
that theta is a function of t, and then times
our i-unit vector,
plus, so if we're factoring out
a negative omega-r, this becomes negative
cosine of theta,
which is a function of t, and that
is times our j-unit vector.
So we factored out
a negative omega-r. Now let's take the derivative of
this with respect to time. So if we take
the derivative of velocity
with respect to time, this is clearly just what the acceleration is
as a function of time, and we're going to assume that the magnitude of this thing
is constant, but the actual direction is changing, so this is the acceleration
as a function of time, is going to be equal to
this negative
omega-r, so what is the derivative of
this thing right over here? So the derivative of sine with respect to theta,
we're just doing the chain rule here,
the derivative of sine with respect to theta
is going to be cosine of theta
as a function of t.
And then chain rule, we also have to take that and multiply it with the derivative of theta
with respect to t.
I could write d-theta, dt here
But that once again, is just omega.
So that is just omega, and that of course
is in the i-direction. And from that,
and next to that, we take the derivative of
cosine of theta of t with respect to theta, so that's going to be
that would be negative sine of theta, so we would have a negative out front
so it becomes positive sine of
theta as a function of t.
And then we have to do the chain rule, the derivative of theta with respect to t.
We have to multiply by this, and for that we could write
d-theta, dt right there, but that again is the same thing
as omega. And all that is being multiplied by the
j-unit vector.
So now let's factor out this other omega, and we get something interesting,
we get the acceleration vector as a function of
time is equal to, and if we factor out another omega, we get
negative omega-squared r,
I'm just factoring out another negative omega,
times, and I'll write it in parentheses here,
cosine of theta
as a function of t
times our i-unit vector
plus
sine of theta, which is a function of t
times our j-unit vector.
Now what is all of this business right over here?
Just look at this part right over here, well r times this,
especially if you distributed the r, that is exactly this thing right over here
If you distribute the r, you get exactly r cosine theta as a function
of t times our i-unit vector plus our sine theta as a
function of t times the j-unit vector. So everything that I squared-off in orange right over here,
this is our position vector as a function of time.
So all that work we did, we just got a
very interesting result. We got that our acceleration vector
as a function of time is equal to the negative
of our constant angular velocity-squared
times our position vector
And just to be clear, angular velocity is kind of the pseudo vector,
it tends to be treated like a scalar, especially when you're dealing with two-dimensionals
like this, it's really a pseudo scalar, but let's just go with this.
We're assuming this right over here is a constant scalar quantity.
Now, we're very very very very close here.
Now what we want to do is to relate this
this is essentially the scalar version of it, so if we wanted
to take the magnitudes of both sides,
so we're saying the acceleration vector is equal to this constant times
the position vector, so let's take the magnitude of both sides of this thing
So then we get the magnitude of the acceleration vector,
which I'm just going to call a sub c,
is going to be equal to
you could say the magnitude of this negative omega-squared
but when you take the magnitude, it's like taking the absolute value
in fact, absolute value is just the one-dimensional version of magnitude,
that's just going to be positive omega-squared
we don't care about the direction, sign gives us the direction, we just care about
the actual size. So this is going to be
the magnitude of negative omega-squared
times the magnitude of our position vector
the magnitude of omega-squared is just going to be omega-squared
you can get rid of the sign, and the magnitude of our position vector
we saw at the beginning of this video, is just r,
our radius
so this right over here is just going to be equal to
the radius of the circle that we're going around.
Now, we also know the angular velocity, or
the magnitude of the angular velocity, is equal to the
magnitude of our velocity, or the speed of our
object, divided by the radius of the circle
that it is going around. So we could substitute that right over here.
So if we square it, this is going to be (v over r)-squared, now we saw that in the video on angular velocity,
times r
and this is all going to be the magnitude of our acceleration,
which is really our centripetal acceleration, our inward directed acceleration.
So this is going to be equal to, and I think you see where this is going,
This is equal to v-squared over r-squared
times r, but this r cancels out
with the r-squared, so you're just left with v-squared
over r, and you're done!
The magnitude of the centripetal acceleration is equal to your speed,
the magnitude of your velocity, squared, divided by
your radius. And, we are done.