Video: Determining the Reaction Force Required to Provide a Centripetal Force

Railroad tracks follow a circular curve of radius 500.0 m and are banked at an angle of 5.00°. For trains of what speed are these tracks designed?

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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.

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