AP Physics 1 review of Centripetal Forces | Physics | Khan Academy

- [Voiceover] What does
period and frequency mean?

The period is the number of seconds

it takes for a process to complete

an entire cycle, circle, or revolution.

So, if there's some repeating process,

the time it takes that process
to reset is the period,

and it's measured in seconds.

The frequency is the number of cycles,

or circles, or revolutions
completed in one second.

So, if there's some
process that's repeating,

the number of times the process repeats

in one second would be the frequency.

This means it has units
of one over second,

which is just called, the hertz.

And because the period
and frequency are defined

in this inverse way as seconds per cycle

or cycles per second, each one

is just the inverse of the other.

In other words, the period is

just one over the frequency, and

the frequency is equal
to one over the period.

One example of a repeating process

is an object going in a
circle at a constant speed.

If this is the case, you can relate

the speed, the radius of the circle,

and the period of the motion since speed

is just distance per
time, and the distance

the object travels in
one cycle is two pi R

the circumference, the speed would

just be two pi R per the period,

or since one over the
period is the frequency,

you could write the speed as
two pi R times the frequency.

Since time is not a
vector these quantities

are not vectors and
they cannot be negative.

So, what's an example involving period

and frequency look like?

Let's say moon travels around a planet

in a circular orbit of radius
R at a constant speed S.

And we wanna know what the
period and frequency are

in terms of given quantities and

fundamental constants, so
we'll use the relationship

between the speed, the
period, and the frequency.

We know that for object in circular motion

the speed is two pi R over the period.

And that means the period here

would be equal to two pi R over the speed.

And since frequency is
one over the period,

if we take one over this quantity

we just flip the top and bottom

and we get that this is
the speed over two pi R.

But we can't leave our
answer in terms of V.

We had to express this in
terms of given quantities.

We were given S, so our
answer for the period

has to be two pi R over
S, and for frequency

it would be S over two pi R, which is C.

What is centripetal acceleration?

The centripetal acceleration of an object

is the acceleration that's causing

that object to go in a circle.

And it's important to note that

this centripetal
acceleration always points

toward the center of the circle.

The formula to find the
centripetal acceleration

is speed squared divided by the radius

of the circle the object is travelling in.

Even though this has a
bit of an exotic formula

for the acceleration, it's
still an acceleration,

so it still has units of
meters per second squared,

and it is a vector, which means it does

have a direction, i.e toward
the center of the circle.

But this centripetal acceleration

does not cause the object
to speed up or slow down.

This centripetal
acceleration is only changing

the direction of the velocity.

If the object going in the circle

is also speeding up or slowing down,

there's also gotta be a component

of the acceleration that's
tangential to the circle,

in other words, if the object

is going in a circle and speeding up,

there's gotta be a
component of acceleration

in direction of the velocity,

and if the object is slowing down,

there's gotta be a
component of acceleration

in the opposite direction to the velocity.

So, centripetal acceleration
changes the direction

of the velocity, and
tangential acceleration changes

the magnitude or size of the velocity.

But this formula of V squared over R

is only giving you the magnitude

of the centripetal acceleration.

This does not account for
any tangential acceleration.

So, what's an example problem

involving centripetal
acceleration look like?

Let's say particle A is
travelling in a circle

with a constant speed S and a radius R.

If particle B is travelling in a circle

with twice the speed of A
and twice the radius of A,

what's the ratio of the acceleration

of particle A compared to particle B.

So, particle A is gonna have
a centripetal acceleration

of the speed squared over the radius,

and particle B is also
gonna have an acceleration

of the speed squared, but this speed

is twice as much of the
speed of particle A,

and is travelling in a circle

with twice the radius of particle A.

When we square the two
we'll get four over two,

gives us a factor of two times the speed

of A squared over the radius of A.

So, the ratio of the
acceleration of particle A

compared to particle
B is gonna be one half

since the acceleration of particle A

is half the acceleration of particle B.

Centripetal forces are
not a new type of force,

centripetal forces are just one

of the any other forces
that we've already met

that happen to be pointing
towards the center

of the circle making an
object travel in a circle.

So, for a moon going around the Earth,

gravity is the centripetal force.

For a yo-yo going around on a string,

the tension is the centripetal force.

For a skateboarder doing a loopty loop,

the normal force is the centripetal force.

And for a car going around a roundabout,

the static frictional force
is the centripetal force.

And these forces still
follow Newton's second law,

but using centripetal forces means you're

also going to have to use the expression

for the centripetal acceleration.

Now, if a force is
directed radially inward

toward the center of the circle,

you would count that force as positive

since it points in the same direction

as the centripetal acceleration.

And if a force points radially out

from the center of the circle,

you would count that as a negative force.

And if force is directed
tangential to the circle,

you wouldn't include it in
this calculation at all.

You can include those forces in their own

Newton's second law equation,

but you wouldn't be using V squared over R

for that acceleration.

Those tangential forces change
the speed of the object,

but the centripetal force changes

the direction of the object.

So, what's an example problem involving

centripetal forces look like?

Imagine a ball of mass M rolling over

the top of the hill of
radius R at a speed S.

And we wanna know, at a top of the hill,

what's the magnitude of the normal force

exerted on the ball by the road.

So, we'll draw our force diagram.

There's gonna be an upward normal force

on the ball from the road, and there's

gonna be a downward force of gravity

on the ball from the
Earth, and these two forces

are not going to be equal and opposite.

If they were equal and opposite

they would balance, and
if the forces are balanced

the object would maintain its velocity

and keep travelling in a straight line.

But this ball doesn't
travel in a straight line,

it starts accelerating downward.

So, this normal force
is gonna have to be less

than the force of gravity.

To figure out how much less, we can use

Newton's second law with the formula

for centripetal acceleration.

The speed is S, the radius is R,

the force of gravity
is gonna be a positive

centripetal force since
it's directed toward

the center of the circle.

The normal force is gonna be
a negative centripetal force

since it's directed radially away from

the center of the circle.

Then we divide by the mass, which,

if you solve this for normal force,

gives you the force of gravity

minus M, S squared over
R, which makes sense

'cause this normal force has to be less

than the force of gravity.

Newton's universal law
of gravity states that

all masses in the
universe pull, i.e attract

every other mass in the universe

with gravitational force.

And this force is proportional

to each mass, and inversely proportional

to the square of the
center to center distance

between the two masses.

In mathematical form, it just says

that the force of gravity
is equal to big G,

a constant, which is 6.67
times 10 to the negative 11th

multiplied by each mass in kilograms,

and then divided by the
center to center distance

between the two masses, in other words,

not the surface to surface distance,

but the center to center distance.

And even if these two
objects have different

masses, the magnitude of the force

they exert on each other
is gonna be the same.

This is illustrated by the formula

since you could swap these two masses

and you get the same number.

And it's also something we
know from Newton's third law.

This force of gravity is a vector

and it has a direction, the
direction is always such

that it attracts every other mass,

and since this is a force,
the unit is in Newtons.

So, what's an example problem involving

Newton's universal law
of gravity look like?

Let's say two masses, both of mass M,

exert a gravitational
force F on each other.

If one of the masses is
exchanged for a mass 3M

and the center to center distance between

the masses is tripled, what would

the new gravitational force be?

We know the gravitational force

is always big G times one
of the masses multiplied

by the other mass divided
by the center to center

distance squared.

So, the initial force
between the two masses

would be big G M times M over R squared,

but the new force with the
exchanged values would be

big G times 3M times M divided

by three times the radius squared.

The factor of three squared on the bottom

gives nine, and three divided by nine

is one over three times
big G M M over R squared.

So, we can see that the
force with the new values

1/3 of the force with the old values.

What's gravitational field mean?

The gravitational field
is just another word

for the acceleration due
to gravity near an object.

You can visualize a gravitational field

as vectors pointing
radially in toward a mass.

All masses create a
gravitational field that points

radially in toward them and dies off like

one over R squared the farther
you get away from them.

So, the formula for the
gravitational field little G

created by a mass M is big G times

the mass creating the field
divided by the distance

from the center of the
mass to the point where

you're trying to determine
the value of the field.

And again, this value for
the gravitational field

is going to be equal to the value

for the acceleration due to gravity

of an object placed at that point.

The gravitational field is a vector

since it has a direction,
i.e. toward the center

of the object creating it.

And since gravitational
field is equivalent

to acceleration due to gravity,

the units are meters per second squared,

but you could also write
that as newtons per kilogram,

which is another way
of thinking about what

gravitational field means.

Not only is it the
acceleration due to gravity

of an object placed at that point,

but it's the amount of
the gravitational force

exerted on a mass M placed at that point.

So, you could think of
the gravitational field

as measuring the amount
of gravitational force

per kilogram at a point in space,

which when rearranged gives
you the familiar formula

that the force of gravity
is just M times G.

So, what's an example problem involving

gravitational field look like?

Let's say a hypothetical
planet X had three times

the mass of Earth and
half the radius of Earth.

What would be the
acceleration due to gravity

on planet X, i.e. the gravitational field

on planet X, in terms of the acceleration

due to gravity on Earth, which is GE.

So, we know that the
gravitational field on Earth

has to be big G times mass of the Earth

over the radius of Earth squared,

which we're calling G sub E, and

the gravitational field
on planet X would be

big G times three times
the mass of the Earth

divided by half the radius
of the Earth squared,

and when we square this factor of a half

we'll get 1/4, which
is in the denominator,

so three divided by a
1/4 is 12 times big G

mass of the Earth over
radius of the Earth squared,

and since this entire term
here is the acceleration

due to gravity on Earth,
the acceleration due

to gravity on planet
X is gonna be 12 times

the acceleration due to gravity on Earth.

Sometimes when you're solving
gravitational problems,

you'll be given the density
instead of the mass.

The density is the
amount of mass per volume

for a given material.

The symbol for density
is the Greek letter rho,

and you can find it by taking the mass

divided by the volume.

So, the units of density are
kilograms per meter queued.

And it's not a vector
since it has no direction,

but it does let you solve for mass.

If you know the density you could say

that the mass is the
density times the volume.

So, what's an example problem
involving density look like?

Let's try the hypothetical
planet problem again,

but this time instead of being told

that planet X has three
times the mass of Earth,

let's say that planet X has three times

the density of Earth, and
again, half the radius of Earth.

What would be the
acceleration due to gravity

on planet X in terms of the acceleration

due to gravity on Earth GE.

We could write down the formula

for gravitational acceleration
or gravitational field,

which is big G M over R squared,

but this time we don't know the mass,

we just know the density,
so we want to rewrite

this formula in terms of density,

which we can do by rewriting
the M as rho times V,

since density is mass per volume,

and mass is density times volume.

But we don't know the
volume of this planet,

we just know the radius, so
we need to rewrite volume

in terms of radius, which we could do

since planets are spherical
and the volume of a sphere

is four thirds pi R cubed.

We can substitute this expression in

for the volume and
finally get an expression

for the acceleration
due to gravity of big G

times rho four thirds pi R
cubed divided by R squared.

And we can cancel an R squared
on the top and the bottom,

which leaves this little
G as equaling big G

rho four thirds pi R.

So, the gravitational
acceleration on Earth

would be big G rho of Earth four thirds pi

times the radius of Earth.

And the gravitational acceleration

on planet X would be big G times

the density of planet
X, which is three times

the density of Earth times four thirds pi

times the radius of planet
X, which is one half

the radius of Earth,
which when we pull out

the three and factor of a half,

gives us three halves times the expression

for the acceleration
due to gravity on Earth.

So, the gravitational acceleration

on planet X is gonna be three halves

the gravitational
acceleration on planet Earth.

Gravitational orbits
are just a special case

of centripetal acceleration
where some object

is orbiting another object due

to the gravitational force.

And if that orbit is a circle,

we can relate the speed,
the radius of the orbit,

and the larger mass to each other

using Newton's second law
and centripetal acceleration.

You just plugin the acceleration

as the centripetal
acceleration, V squared over R,

and since the centripetal
force is the force of gravity,

you can plugin the expression for

the force of gravity as
the centripetal force,

which is big G M M over the
distance between them squared.

And since the mass of the
orbiting object cancels

we get an expression
that relates to the speed

of the orbiting object, the larger mass

that's pulling that object in,

and the center to center
distance between the objects,

which, if we solve this for V,

gives us the square root
of big G times the mass

pulling in the object divided by

the center to center
distance between the objects.

Note that this formula
does not depend on the mass

that's in orbit since
that mass canceled out

in the calculation.

So, what's an example problem involving

gravitational orbits look like?

Well, imagine a space station of mass MS

is orbiting at an altitude of 3R

above a planet of mass MP and radius R

as seen in this diagram, and then imagine

a different space station of mass 3MS

is orbiting at an altitude
of 2R above a planet

of mass 4MP and a radius of
2R as seen in this diagram,

and we wanna know, if the speed

of the space station of mass MS is V,

then in terms of V, what's the speed

of the space station of mass 3MS?

Well, we just showed that the speed

of an orbiting object is gonna be equal

to the square root of big G times

the mass of the larger object pulling in

the smaller object divided by the

center to center distance
between the objects.

And since this formula
doesn't involve the mass

of the orbiting object, it doesn't matter

that the objects have different masses,

but the mass of the planet
can make a difference.

So, to get the speed of
the space station MS,

we could say that it's
the square root of Big G

the mass of planet P over the
center to center distance,

which is not gonna be the radius

of the planet or the altitude,

it's gonna be the radius of
the planet plus the altitude

since this has to be the
center to center distance,

which in this case will
be 3R plus R, which is 4R.

And now to get the speed
of the space station

of mass 3MS, we'll use the same formula,

which is big G mass of the planet, which

in this case is 4MP divided by

the center to center distance,

which in this case would be 2R plus 2R,

and again that's 4R, and if we compare,

the only difference
between these expressions

is that there's an extra
factor of four within

this square root.

So, if we take that factor
out, the square root

of four is two, we'd get
two times the expression

for the speed of the space station MS.

So, the space station 3MS
is travelling two times

the speed of the space station MS.