### Video Transcript

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.