Video Transcript
In this video, we’re going to learn
about centripetal force. We’ll learn what this force is,
where it comes from, and how it got its name.
To start on this topic, imagine
that you are at an amusement park and you’ve just stepped into a new ride, called
the spinning chamber. This ride consists of a circular
room, where riders will lean up against the outside wall. When everyone’s ready, the circular
room begins to rotate about its center in faster and faster circular motion.
As the room spins faster, you begin
to feel the wall behind you press harder and harder into your back. As the ride continues to spin
faster, eventually, to your amazement, you see the floor beneath your feet drop
down. But you stay in place. You no longer need the floor to
support your weight. To better understand this spinning
chamber ride, it will be helpful to know something about the centripetal force.
Imagine that you have a large
rotating disc. And on the outer edge of that disc,
you put a mark. And what if you wanted to know the
acceleration of that point on the desk, what direction it pointed, and what its
magnitude is. We know that the velocity of that
point points tangent to the edge of the desk. But it turns out its acceleration
points in the direction we might not expect. It points inward right towards the
center of the circle.
And if we call the radius of this
rotating disc 𝑟, we can say that the acceleration experienced by our mark on the
edge of the desk is equal to its linear speed 𝑣 squared divided by the radius of
the disc 𝑟. This acceleration is called
centripetal acceleration. And it’s often symbolized 𝑎 sub
𝑐. Why, we may wonder, is this
acceleration given this name centripetal?
Well, the word centripetal simply
means center-seeking. So, the fact that this acceleration
points towards the center of the circle it moves around gives it its name. If we give our rotating disc an
angular velocity, we can call it 𝜔, then there’s another way we can write
centripetal acceleration. Recalling that linear velocity 𝑣
is equal to 𝑟 times angular velocity 𝜔, centripetal acceleration is also equal to
𝑟𝜔 squared.
Here’s something that’s interesting
about centripetal acceleration. Because acceleration is a vector
defined as the change in velocity over change in time, that means that anytime an
object is on a circular path, its direction must be changing, and, therefore, it has
a centripetal acceleration. Even if the linear speed is
constant since direction is constantly changing, the object is accelerating.
Now, let’s take this expression for
centripetal acceleration and connect it with another expression we know. Newton’s second law of motion tells
us that the net force acting on an object is equal to the object’s mass times its
acceleration. If the acceleration of our object
is a centripetal acceleration 𝑎 sub 𝑐, then we can call the net force acting on
our object 𝐹 sub 𝑐 a centripetal force. This force depends on an object’s
mass, its linear speed, and its rotational radius 𝑟. We can also write this in its
rotational analogue 𝑚𝑟 𝜔 squared.
Since the net force acting on an
object acts in the same direction as its net acceleration, that means that the
centripetal force on the mark on our rotating disc acts in the same direction as the
centripetal acceleration, towards the center of the circle. So, we see why the force is called
center-seeking itself. Let’s get some practice with these
ideas of centripetal acceleration and centripetal force through a couple of
examples.
What is the magnitude of the
acceleration of Venus toward the Sun, assuming a circular orbit with a radius of
1.082 times 10 to eleventh meters, and an orbital period of 0.6152 years. Use a value of exactly 365 for the
number of days in a year.
We can call this acceleration
magnitude of Venus toward the sun 𝑎. And if we make a sketch of the
planet Venus orbiting circularly around the sun, we know that the distance between
the center of the sun and the center of Venus is given as 𝑟 1.082 times 10 to
eleventh meters. And that the planet Venus makes it
once all the way around this orbit in a period we’ve called capital 𝑇 of 0.6152
years.
Since the planet Venus is moving in
a circular orbit, that means it will accelerate centripetally towards the center of
the circle. We recall that an object’s
centripetal acceleration is equal to its linear speed squared divided by the radius
of the circle it moves in. Recalling further that an object’s
speed is equal to the distance it travels divided by the time it takes to travel
that distance, we can say that the linear speed of Venus is equal to the distance it
travels the circumference of the circle two times 𝜋 times its radius divided by the
period 𝑇.
This means that the centripetal
acceleration of Venus is equal to two 𝜋𝑟 over 𝑇 quantity squared all divided by
the radius 𝑟. This simplifies to four 𝜋 squared
𝑟 over 𝑇 squared. Since we’re given both the radius
𝑟 and the period 𝑇, we have all the information we need. But before we plug in and solve for
𝑎, we like to convert the period 𝑇 from units of years to units of seconds.
To make that conversion, we’ll take
𝑇, which is given in years, multiply it by the number of days in a year multiply
that by the number of hours in a day and multiply that by the number of seconds in
an hour. This then will give us a time value
in units of seconds.
With the radial distance 𝑟 plugged
into our expression, we’re already to calculate 𝑎. When we do, we find a value of
1.135 times 10 to the negative two meters per second squared. That’s the magnitude of the
center-seeking acceleration of the planet Venus as it moves in its circular
orbit.
Now, let’s look at an example
involving centripetal force.
A car that has a mass of 900.0
kilograms drives along a circular unbanked curve at a constant speed of 25.000
meters per second. The radius of the curve is 500.0
meters. What magnitude force acts on the
car to maintain its speed and direction as it follows the curve? What is the minimum possible value
for the coefficient of static friction between the car’s tires and the road
surface?
We can label the force we want to
solve for capital 𝐹 and the minimum possible coefficient of static friction we’ll
name 𝜇 sub 𝑠. We can start on our solution by
drawing a diagram of the situation. Our car in this example, with a
mass of 900.0 kilograms, moves on a circular section of road with a radius of
curvature 500.0 meters. While it drives on this section, it
maintains a steady speed we’ve called 𝑣 25.000 meters per second.
Knowing all this, we wanna solve
first for the magnitude of the force acting on the car that keeps it moving in a
circular path on this unbanked road. A force that keeps an object moving
in a circular path is a centripetal force. And it’s equal to the mass of the
object times its speed squared over the radius of the circle it moves in. In our case, we’ve been given the
radius 𝑟, the speed 𝑣, and the mass of the car 𝑚. So, we’re ready to plug in and
solve for 𝐹.
When we enter this expression on
our calculator, we find that 𝐹 is 1125 newtons. That’s the center-seeking force
needed to keep this car on a circular path. If we drew a picture of our car
from behind while it was turning on this circular path, the direction of the force
𝐹 would be to the left, pointing it towards what from this perspective is the
center of the circle. The physical mechanism that creates
this force 𝐹 is the friction force on the tires of the car from the road.
Wanting to solve for the
coefficient of static friction between these two materials, we can recall that the
force of friction on an object is equal to the coefficient of friction it
experiences multiplied by the normal force acting on it. In our case, we can write that the
force of friction on the car is equal to the coefficient of static friction because
the wheels don’t slide on the road’s surface multiplied by the mass of the car times
the acceleration due to gravity 𝑔. Where 𝑔 we’ll treat as exactly 9.8
meters per second squared.
This frictional force of the road’s
surface on the tires is the physical mechanism by which the centripetal force 𝐹 sub
𝑐 is exerted on the car. So, we can say that the coefficient
of static friction times 𝑚𝑔 is equal to 𝑚𝑣 squared over 𝑟. And we see that the mass of the car
cancels out from this expression. If we then divide both sides of
this equation by 𝑔, we see that 𝜇 sub 𝑠 is equal to 𝑣 squared over 𝑟 times
𝑔.
When we plug in for the given
values of 𝑣, 𝑟, and 𝑔, we find that, to four significant figures, 𝜇 sub 𝑠 is
0.1276. That’s the minimum value that the
coefficient of static friction can be between the road’s surface and the tires in
order for this magnitude centripetal force to be exerted on the car.
Let’s summarize what we’ve learned
so far about centripetal force. We’ve seen that objects in circular
motion experience acceleration toward the center of the circle they’re moving
in. We’ve seen that this acceleration
called centripetal, or center-seeking, acceleration is equal to the linear speed of
the objects squared divided by the radius of the circle they’re moving through. We also saw that this is equal to
𝑟 times the angular velocity 𝜔 squared.
We’ve also seen that Newton’s
second law of motion says that this net centripetal acceleration corresponds to a
force. We call it the centripetal
force. By the second law, this net force
is equal to 𝑚 times 𝑎 sub 𝑐, or 𝑚 times 𝑣 squared over 𝑟. And finally, we’ve learned that
centripetal force always has a physical mechanism behind it, for example, the
tension in a string or the friction force caused by road surface on car tires.
