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>> In this video, we're going
to take a Closer Look
at Circular Motion.
Any time an object revolves
around a singular point,
or axis,
that's circular motion.
The first key property
that any object moving in
a circular motion possesses
is called tangential velocity,
represented with V
with a subscript of t.
Tangential velocity can be
thought of as the velocity
of the object as it moves
with circular motion.
The speed remains constant,
while the direction of
the object is always changing.
Centripetal acceleration is also
a key property of all objects
that undergo uniform
circular motion.
Acceleration is a measure
of an object's change
in velocity over time.
Since velocity is
a vector quantity,
which requires both
the magnitude and direction,
even an object undergoing
uniform circular motion
would experience centripetal
acceleration due to
the changing directional
components of the velocity
vector as the object turns.
This change in velocity,
the centripetal acceleration,
is always directed towards
the center of the circle.
Centripetal acceleration
of an object that is undergoing
circular motion can be
calculated if the tangential
velocity and distance the object
is from the center of
the circle, or axis, is known.
This distance is known
as the radius.
Here's the equation
for calculating
centripetal acceleration.
A sub c equals v sub t squared,
divided by the radius where
centripetal acceleration is
in meters per second squared,
tangential velocity
is in meters per second,
and the radius is in meters.
Here's a sample problem
involving the calculation of
the centripetal acceleration.
Let's write down our variables.
The radius of the circle
is 0.45 meters long,
and the tangential velocity
is 13 meters per second.
Our unknown is
our centripetal acceleration.
Since we know the distance
of the string, we can plug
that into the equation
for the radius.
The tangential velocity
is also given to us,
so we can plug that into
the equation here.
Then we square
the tangential velocity,
and we get 169 meters squared
per second squared.
Then, we divide
169 meters squared,
per second squared,
by the radius 0.45 meters,
and we get 376 meters
per second squared.
That's the centripetal
acceleration of the phone
as it moves in a circular path.
There's another important
property to consider
when investigating objects
in circular motion.
If the string holding the phone
were to break while we were
watching from above, it would
look to us like the phone is
flying off in a straight line.
That's due to its inertia.
If the string stays intact,
it provides an opposing force
directed towards the center
of the circle that keeps
the phone from flying away.
We can determine the amount
of force that the string
provides if we consider
Newton's second law,
which states the net force
acting on an object in Newtons
is equal to the objects mass
in kilograms, times its
acceleration in meters
per second squared.
In this scenario,
if we consider
just the horizontal component
of the phone in circular motion,
this net force can be called
the centripetal force,
or f sub c, which in this
example is due to the tension
in the string.
This force causes the phone
to have centripetal
acceleration, a sub c.
Centripetal force equals mass
times centripetal acceleration,
so if the phone has a mass
of 0.40 kilograms,
we can use Newton's second law
as it relates to circular motion
to calculate this force, where
we take the mass 0.40 kilograms
and multiply that by the
centripetal acceleration,
376 meters per second squared,
and find that
the centripetal force provided
by the string that kept it
in uniform circular motion
was 150 kilograms times
meters per second squared,
or 150 Newtons.
There's an equation
for centripetal force
that you can use,
which simply substitutes
centripetal acceleration
for acceleration
in Newton's second law.
It's f sub c,
or centripetal force in Newtons,
is equal to the mass,
in kilograms, of the object
in circular motion,
times the tangential velocity
of the object in meters
per second squared,
divided by the radius in meters
of the circle.
Centripetal force is just
the name we give to the class
of forces acting on
any object that's in uniform
circular motion.
The force keeping
the International Space Station
in orbit is gravitational force.
When a race car is making
a hard turn on a circular track,
the friction between the tires
and pavement acts as
the centripetal force
for the race car and driver,
so it enables the car
to maintain uniform
circular motion.
When you're on a roller coaster
and are at the bottom
of the loop, the normal force
of the seat keeps you moving
in a circle.
There are many others.
Any force that causes an object
to move in a circle
is a centripetal force.
Here's a sample problem
involving centripetal force.
In this problem, since we are
solving for the speed,
we'll need to isolate
that variable on one side
of the equation.
To do that, we would multiply
both sides of the equation
by the radius over the mass.
This will give us the radius
times centripetal force,
divided by the mass, is equal
to tangential velocity squared.
To get the tangential velocity
alone, take the square root
of both sides.
This makes the tangential
velocity equal to
the square root of the radius,
times the centripetal force,
divided by the mass.
Our radius is 0.525 kilometers.
The centripetal force is
1,850 Newtons, and the mass
is 885 kilograms,
with tangential velocity
as our unknown variable.
When we multiply 525 meters
times 1,850 Newtons where we
converted the radius to meters,
we get 971,000 meters
times Newtons.
Then we divide that by the mass.
885 kilograms, and we get
the square root of 1.10
times 10 to the third.
The square root of 1.10
times 10 to the third is 33.2,
and that would be in meters
per second, so the car
is traveling at a speed
of 33.2 meters per second
around the track, which equals
our tangential velocity.
Now here's one for you to try.
According to Newton's
second law, an object of
a given mass will accelerate
in the direction of
the net force.
For the International
Space Station, traveling in
a circle around the earth,
the net force acting is
the gravitational force.
The gravitational force
acting on an object can be
found by the following equation.
G is called the constant
of universal gravitation,
and its value is 6.673
times to 10 to the
negative eleventh Newtons,
times meters squared
per kilogram squared.
If the mass of the International
Space Station is 4.20 times
ten to the fifth kilograms,
and the mass of the Earth
is 5.97 times ten to
the 24th kilograms,
and the total distance between
the center of the Earth
and the orbiting space station
is 6.62 times
ten to the sixth meters,
what is the total
gravitational force acting on
the International Space Station
as it orbits.
Go ahead and press pause.
Give it a try, and when you
think you have an answer,
come back and check your work.
Did you get 3.82 times
ten to the sixth Newtons?
If so, great job.
If not, let's take a look
at the solution.
If you multiply
the gravitational constant
times the mass of the Earth,
times the mass of
the International Space Station,
you get 1.67 times ten
to twentieth Newtons times
meters squared.
We then need to divide that
by the radius, 6.62 times
ten to the sixth squared,
which is 4.38 times
ten to the thirteenth,
and we find that the total
gravitational force is 3.82
times ten to the sixth Newtons.
Now that we know
the centripetal force that's
acting on the space station,
let's see if we can calculate
what tangential velocity
the space station must maintain
in order to stay in orbit.
Use the equation we worked
with earlier that had
centripetal force equal to
the tangential velocity squared
times the mass,
divided by the radius.
Remember, in this example,
centripetal force is the force
due to gravity.
Go ahead and give it a try
and when you're finished,
come back and check your work.
Did you get 7,760 meters
per second?
If you did, nice work.
If not, let's take a look
at the solution.
Earlier in this segment,
we rearranged the centripetal
force equation and isolated
tangential velocity.
We found that
tangential velocity is equal
to the square root of the
product of the centripetal force
times the radius
divided by the mass.
When we plug in our values
for centripetal force,
3.82 times ten
to the sixth Newtons,
the radius, which was 6.62
times ten to sixth meters,
and the mass of
the space station,
which was 4.20 times ten
to the fifth kilograms,
and then solve, we find that
the tangential velocity
must be 7,762 meters
per second,
or 7,760 rounded to three
significant figures,
and that's the Closer Look
at Circular Motion.
For more "Physics in Motion,"
go to our home page
where the entire series
is available to you.
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