Welcome to our lesson on
tangential and normal components of acceleration.
Acceleration can be broken down into two parts.
One part is tangent to the path of motion
and the other is perpendicular to the path of motion.
The tangent part or component is called the tangential
component, and the perpendicular part
is the normal component.
The tangential component represents the
acceleration of the object in a straight line,
while the normal component of acceleration will increase
as the speed and curvature increase.
Then lastly, the tangential and normal components
of acceleration will be perpendicular
for circular motion at a constant rate.
Let's go and take a look at a graph to visualize this.
At this point here, this red vector
would be our acceleration vector.
This blue vector here would be the tangent component
of the acceleration vector.
In other words, it's pointing in the direction
the object would move in a straight line from this point.
Then, in green here, is our
normal component of acceleration.
We will be finding both the blue and the green
components of acceleration in this video.
So, taking a look at this here,
the acceleration vector valued function
would be equal to the tangential component of acceleration
times the unit tangent vector valued function,
plus the normal component of acceleration,
times the unit normal vector valued function.
You can see below, there are several formulas
to determine these components,
but we are going to be using these
last formulas here, in this video.
Notice they both require us to find
the velocity vectored valued function,
as well as the acceleration vector valued function.
For the tangential components, we will dot them,
and divide them by the magnitude of V.
For the normal component,
we will cross V and A, then find the magnitude
and divide by the magnitude of V.
Remember, to determine V, we need to find R prime,
and to determine A, we need to find
V prime or R double prime.
Let's go ahead and take a look at our examples.
We want to determine the tangential and normal components
of acceleration when T is equal to two,
for this vector valued function.
Let's go ahead and start by determining the
velocity vector valued function,
and the acceleration vector valued function.
Remember, the velocity vector valued function
will be the derivative of our given vector valued function,
so that will have an X component of one,
and a Y component of one half T.
The acceleration vector valued function
is going to be equal to the vector valued function
with the X component of zero,
and the Y component of one half.
Notice they also both require the magnitude
of the velocity vector valued function.
Let's go ahead and find that now.
It's going to be the square root of one squared,
it will be one, plus one half T squared
will be one fourth, T to the second.
Okay, let's go ahead and see if we can find
the tangential component now.
We're going to dot V and A,
and then divide by the magnitude of V.
We already found the magnitude.
We'll have one time zero, that is going to be zero,
and then one half T times one half,
that will be one fourth T.
Then, of course, our denominator is going to stay the same.
Now, we could simplify this,
but since we only need to know the value of this
when T is equal to two, we will go ahead
and sub in two for T now.
Will have one fourth times two,
that's going to be one half,
and the denominator is going to be
the square root of one plus one fourth times two squared.
Well, two squared is four, one fourth times four
would be one, so we will have a square root of two.
Looks like our tangential component is
going to be one over two square root two.
Here's our tangential component
of acceleration at T equals two.
Now, let's go ahead and determine
the normal component of acceleration.
On the next slide, we will go ahead and
determine V crossed with A,
and then determine that magnitude,
and then divide by the magnitude of V.
Let's go ahead and determine V crossed with A.
We will have, the first row will be I J K.
The second row will be the compliments of V,
so we'll have one, one half T,
and the Z component would be zero.
The third row will come from A,
so will have zero, one half, zero.
We'll go ahead and use the cofactor
expansion method to determine this.
So, eliminate row one, column one
for this first two by two determinant.
This will give us one half T, zero, one half, zero.
Here, we will have one, zero, zero, zero,
and here, we will have one, one half T, zero, one half.
This is going to work out pretty well.
Here, we will have zero minus zero.
Then, for the Z component,
we're going to have one half minus zero.
Now, when we determine the magnitude of this,
it's going to be pretty easy to do, again.
This is going to be the square root of one half squared,
or the square root of one fourth, and that will be one half.
Okay, so now we are set to determine
the normal component of acceleration.
We have the magnitude of the
crossed product, that's one half.
Our denominator is the magnitude of V,
which we found on the previous screen,
the square root of one plus one fourth T squared.
We'll go ahead and evaluate this, when T is equal to two.
Again, when T is two, this is going
to be the square root of two.
It looks like we have one over two, square root two.
It looks like both our normal component
and our tangential component of acceleration
have the same value.
If we go back to the graph, that should make sense,
because if we take a look at the magnitude
of these two vectors, the tangent vector
and the normal vector, they look like
they are the same length, and even also visualize
if you were to take this blue vector here,
move the initial point to the terminal point
of the green vector, the sum of those two vectors
would be the acceleration vector.
Let's go ahead and take a look at one more example.
Here's the same type of problem with a
different vector valued function,
and we are concerned about the components
of acceleration when T is equal to one.
The velocity vector valued function
is going to be one, three T squared,
and of the acceleration vector valued function
is going to be zero, six T,
and the magnitude of the velocity vector valued function
is going to be the square root of one squared,
that will be one, plus three T squared to the second power,
that will be nine T to the fourth.
We will go straight to determining the tangential component.
We'll dot these, and divide by this magnitude.
V dotted with a will have one time zero, that's zero,
and then three T squared times six T,
that'll be 18 T to the third.
I will divide it by the square root of
one plus nine T to the fourth.
When T is equal to one,
I think we will have 18 over the square root of one plus,
that will be nine, so the square root of 10.
Now, it's going to take a little more work
to determine the normal component.
Let's go ahead and do that on the next screen.
Okay, so I've already set this up to save some time.
Here's the set up for V crossed with A.
Again, the first row is I J K,
the second row is from V, and the third row is from A.
Were using the cofactor expansion method.
Again, it's going to work out pretty well.
This will be zero minus zero for the X component,
and zero minus zero for the Y component,
and the Z component is going to be
six T minus zero, so we have six T.
The magnitude of the crossed product,
we have the square root of six T to the second,
or 36 T squared, which will give us six T.
So, the normal component of acceleration
is going to be equal to six T,
divided by the square root of one plus nine T to the fourth.
Now, we will sub in T equals one.
We are going to have six over the square root of 10.
So, here is our normal component,
and here is our tangential component.
Notice the tangent component is
three times that of the normal component.
Let's go and take a look at this graphically.
In red, we see the acceleration vector.
Notice the tangent component of the acceleration vector
is much longer than the normal component,
as you see here in green,
and it does look like it's three times longer
than the normal component, which is a
good indication that we did our work correctly.
I hope you found this video helpful.
Thank you for watching.