### Video Transcript

In this video, we’re talking about
centripetal force. This is a force which causes an
object to move in a circular arc. We can get started discussing this
force by talking about the word centripetal. It means center seeking. So if we have an object, like this
one, that’s moving in a circular path, then this centripetal force it experiences to
keep it on that path is always pointed toward the center of the circle. In that sense, this force, usually
abbreviated 𝐹 sub c, is center seeking. Something very important to see
about centripetal force is that it’s not a new force we could say that we’re
introducing. This is simply the name we give to
any force that acts on an object towards the center of a circular arc the object is
traveling in.

For example, let’s say that this
object here that’s moving in a circular path is a car driving around a horizontal
circular road. When we think about what force is
it that keeps the car on the road as it turns to this path, we know that that force
is friction between the tires and the road surface. So in this example, it’s the
frictional force which is a centripetal or center-seeking force. Or picture this scenario
instead. Say that we have a satellite that’s
in circular orbit around the Earth. The force that draws the satellite
in to the center of its circular path is the force of gravity. So in this instance, gravity is the
centripetal force acting on the satellite. We’ll see more examples of these
various center-seeking forces as we go along.

The point is, in each case, the
centripetal force is not a new or separate force that we’re adding in to the
picture. Rather, it’s already there, present
in the particular physical circumstances. Now, whenever we talk about forces
acting on objects, we can be reminded of Newton’s second law of motion. Recall that this law tells us that
the net force acting on an object is equal to that object’s mass multiplied by its
acceleration.

So when an object moving in a
circle experiences a force, the centripetal force, then if that’s the only force
acting on the object, this force must be equal to the object’s mass multiplied by
its acceleration. But here’s something interesting
about this Newton’s second law equation. This is actually a vector
equation. That’s because force, as well as
acceleration, is a vector. So if we write this out to
recognize that, we would say that the vector of force is equal to an object’s mass
times that object’s acceleration vector.

Since force and acceleration are
the only vectors in this equation, that must mean they point in the same
direction. And therefore, this acceleration
here, caused by the centripetal force acting on our mass, must also point towards
the center of the circle that this mass is moving around. Therefore, we can call this
acceleration a centripetal acceleration. And we’ll call it 𝑎 sub c. Okay, so these objects that we’ve
identified that are moving in a circle do so because they’re experiencing a
center-seeking force. And therefore, they accelerate
towards the center of these circles too.

But here’s something
interesting. The velocity vectors of these
objects do not point in the same direction as their acceleration. That is, they don’t point towards
the center of these circular paths. Rather, they point tangent to the
part of the circular path where the object currently is. So the velocity of our car, 𝑣 sub
c, points like this. And the velocity of our satellite,
𝑣 sub s, points like this.

Notice that this means that the
velocity of each one of these objects is perpendicular to each one’s
acceleration. Now, since both of these objects
are moving in circles, there’s one other thing we can say about each scenario. In each scenario, the circular
our object is moving in has a particular radius. We’ve called the radius of our car
circle 𝑟 sub c and the radius of our satellite circle 𝑟 sub s. Now, we bring this up because
there’s another way of expressing centripetal acceleration, the acceleration of an
object moving in a circle. It’s equal to the tangential
velocity of the object, at any point along the circular path, squared and then
divided by the radius of the circle the object moves in.

If we look at this equation and
recall that velocity is a vector, we might think that our object’s centripetal
acceleration will depend on which velocity value we choose to use in this
equation. For example, considering our car
driving in a circle, what if, instead of using the car’s velocity when it’s here, we
use its velocity when it’s here in the circle or here or here or anywhere else? It’s certainly true that the
directions of these velocity vectors are not the same. But their magnitudes are. So regardless of which one we use,
since we’re squaring that term in our equation for centripetal acceleration, we’ll
get the same result no matter what. And this points to the fact that
this equation is showing us the magnitude of centripetal acceleration.

Now, if we take the right-hand side
of this equation and substitute it in for 𝑎 sub c here, then we can start to see
some interesting connections between the speed of our object and the radius of the
circle it travels in and the centripetal force it experiences. We can see, for example, that if we
were to increase our object’s speed without changing the radius of the circle it
moves in, then the center-seeking force our object experiences would go up by a
greater factor than the factor by which our speed increased. That’s because we take our speed
and square it to help us calculate the centripetal force.

On the other hand, say we were to
keep our speed the same as before. But now we decrease the radius of
the circle our object moves in. Once again, this would lead to an
increase in centripetal force. For a smaller radius and with the
same speed, it takes more force to keep an object moving in a circle. Now, so far, we’ve been talking
only about linear variables, linear speed and linear acceleration. But we know that there are angular
versions of these variables too. For example, if we have an object
moving in a circle of radius 𝑟 with some tangential, that is, linear speed 𝑣, then
we can say that this object has an angular speed symbolized using the Greek letter
𝜔. And the relationship between this
object’s linear speed 𝑣 and angular speed 𝜔 is that 𝑣 is equal to 𝑟 times
𝜔.

Now, if we consider the units
involved for these factors on the right-hand side, SI base units for linear distance
are meters. And then angular speeds are often
given in radians per second. But then, and this is an important
note, the unit of radians is dimensionless. So when we multiply the radius 𝑟
by the angular speed 𝜔, the units that result are simply meters per second, with no
unit of radians involved. These are the units of the speed
𝑣. So understanding that 𝑣 is equal
to 𝑟 times 𝜔, we can take the right-hand side of this expression and substitute it
in for 𝑣 in our equation for centripetal force. When we do this, we can see that
squaring 𝑟 times 𝜔 will give us two factors of 𝑟, one of which cancels with the
𝑟 in the denominator. And our equation simplifies to 𝐹
sub c is equal to 𝑚 times 𝑟 times 𝜔 squared.

And now let’s recall something
about this equation from earlier. Remember that in line with Newton’s
second law, this equation used to read 𝐹 sub c is equal to 𝑚 times 𝑎 sub c, the
centripetal acceleration. The form of the equation now tells
us that 𝑟 times 𝜔 squared is equal to that centripetal acceleration. And we can indicate that fact down
here. Now, before we get some practice
working with centripetal forces, let’s consider this specific scenario. Say that we’re holding one end of a
rope, right here, and tied to the other end of the rope is a small stone. And say that we then start to swing
this rope around so that it moves in a vertical circle.

Now, one characteristic of an
object moving in a circle is that its speed is constant. So the speed of our stone here is
the same as it is here and here and here and everywhere else in this arc. So recalling this form for the
centripetal force acting on our stone. If the stone’s speed is constant
all throughout, and it is, and the radius of the circle the stone moves in is
constant as well, and it is. Then the center-seeking force
experienced by the stone must also be constant in magnitude. As we mentioned earlier though,
every centripetal force has some physical cause for it, whether it’s the friction
between tires and a road or gravity between a satellite and Earth. There’s always some physical
mechanism creating a center-seeking force.

In this case, we can see that one
of those mechanisms is the tension in our rope. This tension force is always
pulling the stone towards our hand to the center of the circle. Therefore, it’s a centripetal
force. But, and here’s where things get
interesting, because this is a vertical circle that we’re swinging the stone in, the
force of gravity is also involved here. The important difference here
between our tension force and gravity is that the tension force always points
towards the center of this stone’s circular path, whereas gravity always points
straight downward. This means that when the stone is
at the bottom of its circular arc, gravity is tending to move it out of a circular
path. That is, it’s pushing it away from
the center of its circle, whereas, by contrast, when the stone is at the top of its
arc, gravity and the tension force in the rope are acting in the same direction,
downward.

In cases like these where objects
are moving in vertical circles, it’s very important to realize that so long as they
still continue to move in a circle, the center-seeking force acting on them is
constant in magnitude. If the object’s speed and the
radius of the object’s circular path don’t change, then neither does the centripetal
force. But what does change is the
magnitude of the forces involved in creating the center-seeking force. To show what we mean by that, let’s
consider the stone at these two points, at the bottom of its arc and at the top. We know that in each of these
cases, the magnitude of the centripetal force acting on it, and we can draw this
force vector in in orange, is the same. That doesn’t change.

But then, looking at the force of
gravity and the force of tension, when our stone is at the bottom point of its arc,
we can see that these forces act in opposite directions on the stone. The tension force is acting up,
while gravity is acting down. Let’s label this tension forest 𝑇
sub b, the tension in the rope when the stone is at the bottom of its path. And we’ll call the gravitational
force 𝐹 sub g. Looking at our three vectors 𝐹 sub
c, 𝑇 sub b, and 𝐹 sub g, we can say that 𝐹 sub c is the resultant of 𝑇 sub b and
𝐹 sub g.

In other words, the centripetal
force is equal to the difference between the tension force and the force of gravity
acting on the stone. If we decide that forces acting
upward are in the positive direction, then we could write that the tension force,
which acts up and therefore is positive, minus the gravitational force, the one that
points down on the stone, is equal to the centripetal force on the stone when it’s
at the bottom of its arc. So that’s our equation showing the
forces on the stone at this point in its path. Now let’s consider when the stone
is at the top of its arc. Here, all three of the forces
acting on it point downward and by our convention, therefore, are negative.

Now we call the tension force in
our rope 𝑇 sub b when our stone was at the bottom of its path. So let’s call our rope tension 𝑇
sub t when the stone is at the top of its arc. Now, as we mentioned, based on our
direction choice, all three of these forces, 𝐹 sub c, the force of gravity
represented here by this blue vector, and 𝑇 sub t, are negative. So then, the equation showing the
forces acting on our stone when it’s at the top of its arc would look like this. Looking at this equation, what if
we multiplied both sides by negative one? That would switch all the negative
signs to positive signs.

And now, look at this. We have two different expressions
for the centripetal force, which we’ve said must be the same in magnitude on our
stone at any point. And so we can equate these two
expressions. If we do that, we get a result that
looks like this. And as a last step, let’s say that
we add 𝐹 sub g to each side of the equation. Now we’re assuming that the
magnitude of the gravitational acceleration acting on the stone is constant all
throughout its path. And since the stone’s mass is also
constant all throughout, that would mean that on the left-hand side of our equation,
we’re adding and then subtracting the same amount of gravitational force. That means these terms cancel
out. While on the right-hand side, 𝐹
sub g plus 𝐹 sub g can be written as two times 𝐹 sub g.

And here’s where we see the result
of all our labor. The force of tension in the rope is
different when the stone is at the bottom of its arc versus when it’s at the
top. Indeed, we can say that the tension
force is less when the stone is at the top of its arc because we need to add two
times the force of gravity on the stone to that tension in order to equal the
tension in the rope when the stone is at the bottom. So when the object on the end of a
string or a rope or a cord is moving in a vertical circle, while the force of
gravity is essentially the same on the object all throughout. And the centripetal force certainly
is all throughout its circular movement. It’s the tension force in that rope
or string or cable that changes as the object moves around. Okay, having said all this, let’s
get to an example exercise.

What is the magnitude of the
centripetal force that must act on an object of mass 1.0 kilograms to make it move
along a circular path of diameter 1.0 meters, completing a circle every 1.0
seconds?

Well, okay, so in this scenario, we
have a circular path. And we’re told that the diameter of
this path is 1.0 meters. Along with this, we have an object
whose mass, that we can call 𝑚, is given as 1.0 kilograms. We want this object to move around
the circular path so that it completes a circle every 1.0 seconds. One time around the circle is a
complete revolution. So that means we can refer to this
time of 1.0 seconds as our period of revolution. The question is, in all this,
what’s the magnitude of the centripetal force that must act on this mass? That is, for the mass to be moving
in a circle, there must be some force making it tend towards the center of that
circle. That’s the centripetal force, and
we want to figure out how strong it is.

We can start doing this by
recalling that the centripetal force on an object is equal to the object’s mass
times its centripetal acceleration. Now, an object’s centripetal
acceleration is given by its speed squared divided by the radius of the circular
path it moves in. But then we can also recall that
for an object moving in a circle, its linear speed, its tangential speed around that
circle, is equal to the circle’s radius times the object’s angular speed 𝜔. If we substitute 𝑟 times 𝜔 in for
𝑣 in our equation for centripetal acceleration, we find it’s equal to the quantity
𝑟 times 𝜔 squared divided by 𝑟 or, more simply, 𝑟 times 𝜔 squared. We can then substitute 𝑟 times 𝜔
squared in for 𝑎 sub c in our equation for centripetal force. And now we have an expression for
this force that we can work with in terms of our given parameters.

Starting with our object’s mass,
we’re given that it’s 1.0 kilograms. And then the radius of the circle
our object is moving in is equal to the diameter divided by two. And then lastly, there’s this
factor here, the angular speed 𝜔 squared. We can recall that the units of 𝜔
are radians per second. And since we’re told that our
object completes a circle every 1.0 seconds, recalling that a full circle is equal
to two 𝜋 radians, we can say that the angular speed of our object is equal to two
𝜋 radians divided by 1.0 seconds. This is its angular speed. So then, substituting in to our
equation for centripetal force, here’s what we get. Our mass as we saw is 1.0
kilograms. Our circle’s radius is its
diameter, 1.0 meters, divided by two. That’s 0.5 meters. And then our object’s angular speed
is two 𝜋 radians in 1.0 seconds. And this factor of 𝜔 is
squared.

When we calculate all this out, we
find a result of 19.73 and so on and so forth newtons. But notice how all of the values
we’re given in our problem statement have two significant figures. That’s how many we’ll keep in our
answer. And to two significant figures, our
result is equal to 20 newtons. That’s the magnitude of the
centripetal force acting on our object.

Let’s recall some key points now
about centripetal force. We learned in this lesson that
centripetal force, true to its name, always points toward the center of a circular
arc. We also saw that for an object of
mass 𝑚 moving with a speed 𝑣 in a circular arc with radius 𝑟, the center-seeking
or centripetal force acting on that mass is equal to 𝑚 times 𝑣 squared over
𝑟.

Furthermore, we identified this
factor, 𝑣 squared over 𝑟, as a centripetal acceleration. And we also recalled the connection
between linear and angular speeds, that 𝑣 is equal to 𝑟 times 𝜔, and saw that
this implies that centripetal acceleration can be written as 𝑟 times 𝜔
squared. And lastly, we saw that when an
object moves in a vertical circular path, the effect of gravity must be
considered. This is a summary of centripetal
force.