Video Transcript
In this video, we’re going to learn
about sources of centripetal force. In particular, we’ll see what are
the physical mechanisms that cause the center-seeking force.
In all kinds of different physical
situations, whether it be a planet orbiting a star or a ball being swung in a circle
held by a string or a car navigating a turn on the road, the force that causes these
objects to move towards the center of their circular path is mediated by a physical
mechanism. In the case of the orbiting planet,
that force is the force of gravity. In the case of the ball being swung
around on a string, it’s the tension force that creates the center-seeking
force. And for our car making a turn in
the road, the friction force pushes that car towards the center of its circular
path.
The centripetal or center-seeking
force is always due to some physical cause. This means that whenever we draw a
free body diagram of an object that’s moving in a circular path, we should be able
to identify the force pushing it towards the center of the circle as something other
than 𝐹 sub 𝑐, the centripetal force. The centripetal force in a given
situation can be caused by gravity, tension force, spring force, friction, and the
list goes on. The name “centripetal force” is the
name we give to one of these physically caused forces, which happens to make an
object move towards the center of a circular path.
For an object moving in a circle,
the centripetal or center-seeking force acting on it is equal to the mass of the
object times its linear speed squared divided by the radius of the circular arc it’s
moving in. This relationship is true
regardless of what force is physically causing the center-seeking force. This is helpful to us because it’s
a bit like having a completed Newton’s second law expression before we’ve even
included the effects of the physical force causing the circular motion. Let’s get some practice now,
connecting up this expression for centripetal force with the physical forces that
create the circular motion.
Railroad tracks follow a circular
curve of radius 500.0 meters and are banked at an angle of 5.00 degrees. For trains of what speed are these
tracks designed?
We can call the speed we want to
solve for 𝑣. And we’ll start off with a diagram
of this train track. If we draw a cross section of the
train track with a train car on it, we’re told that the distance of the train car
from the center of its curvature—we’ve called 𝑟—is 500.0 meters and that the track
is banked at an angle—we’ve called 𝜃—of 5.00 degrees. Based on the condition of the
track, we wanna solve for the speed of the train 𝑣 at which it could navigate this
turn without any external forces being applied, like the force of friction, to keep
the wheels from sliding on the track. Under this constraint, there are
only two forces that are acting on the train car. There’s the force of gravity acting
straight down on it. And then there’s the normal force
pushing up on the car perpendicular to the track surface.
If we decide that motion vertically
upward from this cross section is positive and that motion down the bank is positive
in the 𝑥-direction, then we can break the force of gravity into its 𝑥- and
𝑦-components. And since the angle at the top
corner of this right triangle is the angle 𝜃, then we can write that the
𝑥-component of the gravitational force is 𝑚 times 𝑔 times the sin of 𝜃, where
𝑔, the acceleration due to gravity, we treat as exactly 9.8 meters per second
squared. Since this component of the
gravitational force is the only force acting in the 𝑥-direction in our scenario, we
can take advantage of the fact that Newton’s second law says that this force is
equal to the mass of the car times its acceleration in the 𝑥-direction. When we write this expression,
we’ll label that acceleration 𝑎 sub 𝑥.
There’s another clue we can use to
solve for the speed of the train car 𝑣. And that’s the fact that the train
car is moving in a circular arc. That means that the force pushing
it to stay in this arc, in this case a component of the gravitational force, is a
centripetal or center-seeking force equal to the mass of the train car times its
speed squared over the radius of the circular path in which it moves. So in our expression, we can
replace 𝑚 𝑎 sub 𝑥 with 𝑚𝑣 squared over 𝑟. And we see that the mass of the car
cancels out from both sides of the equation. We then rearrange to solve for the
speed 𝑣. We find it’s equal to the square
root of the radius 𝑟 times 𝑔 times the sin of the angle 𝜃.
Since we know all three of these
values, we’re ready to plug in and solve for 𝑣. When we enter this expression on
our calculator, we find that 𝑣 is 20.7 meters per second. That’s the speed the train should
maintain through this curve to most easily navigate this circular arc.
Now let’s look at an example
involving centripetal force in a vertical arc.
A ball of mass 1.3 kilograms at the
end of a 1.5-meter string swings in a vertical circle. At its lowest point, the ball is
moving with a speed of 9.5 meters per second. What is the speed of the ball at
the top of its circular path? What is the magnitude of the
tension in the string when the ball is at the top of its circular path? What is the magnitude of the
tension in the string when the ball is at the bottom of its circular path?
In this three-part problem, we can
call the speed of the ball at the top of its path 𝑣 sub 𝑡, the tension at that
point 𝑇 sub 𝑡, and the tension in the string at the bottom of the ball’s path 𝑇
sub 𝑏. Let’s start by sketching out this
vertical circular path. This ball, with a mass of 1.3
kilograms we’ve called 𝑚, moves in a circular vertical arc on a string of length
𝑟, which is 1.5 meters. We know that when the ball is at
the bottom of the circular arc, its speed is given as 9.5 meters per second. Given all this information, we
wanna solve for the speed of the ball when it’s at the top of its arc.
Considering this system as a whole,
since no energy is added to it or taken from it, we can use the conservation of
energy to solve for the velocity of the ball at the top of its arc. We can say that the initial kinetic
plus potential energy of our system is equal to the final kinetic plus potential
energy, where our initial point is at the bottom of the circular arc and the final
point is at the top. Recalling that kinetic energy is
given as one-half an object’s mass times its speed squared and that gravitational
potential energy is 𝑚 times 𝑔 times ℎ, where the acceleration due to gravity, 𝑔,
we’ll treat as exactly 9.8 meters per second squared.
We can write that one-half the
ball’s mass times 𝑣 sub 𝑏 squared is equal to one-half its mass times 𝑣 sub 𝑡
squared plus 𝑚 times 𝑔 times two 𝑟, the radius of the circular arc. We see that the mass 𝑚 cancels
from this expression. And when we rearrange to solve for
𝑣 sub 𝑡, we find it’s equal to the square root of 𝑣 sub 𝑏 squared minus four
times 𝑔 times 𝑟. Since we’re given 𝑣 sub 𝑏 and 𝑟
and know 𝑔 is a constant, we’re ready to plug in and solve for 𝑣 sub 𝑡. Entering this expression on our
calculator, we find that 𝑣 sub 𝑡 is 5.6 meters per second. That’s the speed of the ball at the
top of its circular arc.
Next, we wanna solve for the
tension of the string when the ball is at point 𝑓, at the top of its arc. If we draw a free body diagram
showing the forces acting on the ball at this point, the two forces acting on the
ball at this point are the gravitational force, 𝐹 sub 𝑔, and the tension force, 𝑇
sub 𝑡. Recalling Newton’s second law of
motion that the net force on an object is equal to its mass times its acceleration
and letting motion in the downward direction be positive in our example. We can write that 𝑚𝑔, the
gravitational force, plus 𝑇 sub 𝑡, the tension force from the balls at the top of
its arc, is equal to the mass of the ball times its acceleration, 𝑎.
Because the ball is moving in a
circle, we know that it accelerates centripetally, that is, towards the center of
the circular path, and that the magnitude of this acceleration is equal to 𝑣
squared over 𝑟. So we can write that 𝑚𝑔 plus 𝑇
sub 𝑡 is equal to 𝑚 times 𝑣 sub 𝑡 squared over 𝑟. This expression on the right-hand
side is the centripetal force acting on this ball. We want to rearrange and solve for
𝑇 sub 𝑡. When we do, we see it’s equal to 𝑚
times the quantity 𝑣 sub 𝑡 squared over 𝑟 all minus 𝑔. When we plug in for 𝑚, 𝑣 sub 𝑡,
𝑟, and 𝑔 and enter this expression on our calculator, we find that, to two
significant figures, 𝑇 sub 𝑡 is 15 newtons. That’s the tension in the string
when the ball is at the top of its path.
Lastly, we wanna solve for the
tension in the string when the ball is at the bottom of its arc. To do this, we can once again draw
a free body diagram of the forces on the ball at this point. There are again two forces on the
ball: the tension force, 𝑇 sub 𝑏, and the weight force. But now they point in opposite
directions. We can write that the gravitational
force minus 𝑇 sub 𝑏 is equal to negative 𝑚𝑣𝑏 squared over 𝑟, where the
centripetal acceleration is negative because it points up. When we rearrange this expression
to solve for 𝑇 sub 𝑏, we find it’s equal to 𝑚 times the quantity 𝑣 sub 𝑏
squared over 𝑟 plus 𝑔. When we plug in for these values
and enter this expression on our calculator, we find that 𝑇 sub 𝑏 is 91
newtons. That’s the tension in the string
when the ball is at the bottom of its arc.
Let’s summarize what we’ve learned
about sources of centripetal force.
We’ve seen that centripetal force
always has a physical cause, such as the tension in a string or the pull of
gravity. For an object moving in a circle,
that object experiences a centripetal force equal to the mass of the object times 𝑣
squared over 𝑟, its linear speed over the distance from the object to the center of
the circle. We’ve seen that this quantity, 𝑣
squared over 𝑟, is equal to the net acceleration acting on the object. And finally, we’ve seen that free
body diagrams and Newton’s second law of motion are two tools that help us solve for
centripetal force.
