- Good morning.
Today, we are going to take physics we have already learned
and begin applying it to circular motion.
♫ Flipping Physics ♫
- Let's start with a sphere on a string,
which is moving along a circular path
with a constant radius.
Let's pause with the sphere right there
and let's discuss how to describe the location
of the sphere.
One way to describe the sphere's location
would be to set the end of the string
which is opposite the sphere as the origin,
and to locate the sphere using the x, y coordinate system.
- That is called the Cartesian coordinate system,
which was introduced by Rene Descartes in the early 1600s.
- Sure.
- Billy, you are correct.
Now, notice what happens when we put the sphere
in motion again.
Notice how both the x and y positions of the sphere
change as a function of time
as the sphere moves along the circle.
This makes it more complicated to work with,
so typically in physics we will describe the location
of an object moving along an arc using the radius
and the angular position
- That would be using polar coordinates.
Right, Mr. P?
- Correct, Bobby.
Rather than using the Cartesian coordinates, x and y,
we can use the polar coordinates r and theta.
And notice when the sphere is moving in a circle
how the angle theta changes.
However, the radius stays constant.
So only one variable changes when we describe
the sphere's location using the radius and theta.
Now, we can relate Cartesian coordinates
and polar coordinates using trig functions.
Bo, how do we do that?
- Uh,
so.
- SOH is a good place to start.
- That's funny.
- Yeah.
- Uh,
sine theta equals opposite over hypotenuse,
or y over r,
and multiply by r to get y equals the radius
times the sine of the angle.
- And we can do the same thing with cosine.
Cosine of theta equals adjacent over hypotenuse,
or x over r, and therefore x equals r times cosine theta.
This is how we can go back and forth
between Cartesian and polar coordinates.
Now let's discuss how we describe an object
moving from one location to another
when it is moving along a circle with a constant radius.
Again, we have the sphere on a string, only this time
we're going to discuss having it move
from this initial position
to this final position.
When it does, it goes from an initial angular position
to a final angular position,
which means it goes through an angular displacement.
The angular displacement, or delta theta,
equals angular position final
minus angular position initial.
But the sphere does not only go through
an angular displacement,
the sphere also travels a linear distance.
Billy, what is the linear distance traveled by an object
moving along an arc called?
- Umm.
- The linear distance when moving along an arc?
- Uh, uh.
- It's arc length.
- Oh, yeah.
Yeah, it's arc length, yeah.
The linear distance an object travels through
when moving along an arc.
Oh, and lowercase s is the symbol for arc length.
- And the equation for arc length,
arc length, or lowercase s,
equals the radius times the angular displacement.
When you use the arc length equation,
the units for the angular displacement
must be in radians.
In a future lesson, I'll show you
why you must use radians in this equation.
Okay.
Bobby, if we reduce the radius of the string
attached to the sphere,
what happens to s, the arc length, traveled by the sphere?
- The radius is made smaller,
so if the angular displacement stays the same,
then, because the arc length equals the radius
times the angular displacement, then the arc length
must be reduced, because s equals r delta theta.
- That is correct.
A smaller radius means a smaller arc length.
Class, circumference equals?
- Pi r squared. - It's two pi r.
- Circumference equals two pi r.
You can tell because the units for pi r squared
would be squared,
and area has squared units.
- Right.
- Circumference equals two pi r, or,
if you rearrange it, circumference equals
the radius times two pi,
which shows you that the circumference equation
is just a special case of the arc length equation
where the angular displacement is for one full circle.
Which means the angular displacement is two pi radians.
So, realize, most likely you have already worked with
an arc length equation,
you just didn't know it.
Thank you very much for learning with me today.
I enjoyed learning with you.