Roller coaster loop problem

In summary, a roller coaster at Six Flags Great America uses a teardrop-shaped vertical loop design to keep cars on track at high speeds. The biggest loop is 40.0 m high with a maximum speed of 31.0 m/s. Given a speed of 12.0 m/s at the top and a centripetal acceleration of 2g, the radius of the loop can be calculated using the equations n + mg = mv^2/r and Ac = mv^2/r.
  • #1
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Homework Statement


A roller coaster at the Six Flags Great America amusement park in Gurnee, Illinois, incorporates some clever design technology and some basic physics. Each vertical loop, instead of being circular, is shaped like a teardrop. The cars ride on the inside of the loop at the top, and the speeds are fast enough to ensure that the cars remain on the track. The biggest loop is 40.0 m high, with a maximum speed of 31.0 m/s (nearly 70 mph) at the bottom. Suppose the speed at the top is 12.0 m/s and the corresponding centripetal acceleration is 2g.

What is the radius of the arc of the teardrop at the top?


Homework Equations


Ac=mv^2/r
Fr= n+mg


The Attempt at a Solution


I think n + mg = mv^2/r, but I don't know what to do with the given value Ac=2g or how to go any further from here. Any help is greatly appreciated.
 
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  • #2
nvm, I got it.
 
  • #3


I would approach this problem by first understanding the concepts of centripetal acceleration and forces involved in circular motion. From the given information, we know that the roller coaster is designed in a teardrop shape to maintain the cars on the track, and that the speed at the top of the loop is 12.0 m/s and the corresponding centripetal acceleration is 2g.

To find the radius of the arc at the top, we can use the equation Ac=mv^2/r, where Ac is the centripetal acceleration, m is the mass of the car, v is the velocity, and r is the radius of the arc. We also know that at the top of the loop, the net force acting on the car is equal to the sum of the normal force (n) and the force of gravity (mg).

Therefore, we can set up the equation n + mg = mv^2/r and solve for r. By plugging in the given values of v=12.0 m/s, Ac=2g, and g=9.8 m/s^2, we can solve for r and find that the radius of the arc at the top of the loop is approximately 22.5 meters.

Overall, this problem demonstrates the application of fundamental physics principles in the design and operation of roller coasters, highlighting the importance of understanding forces and motion in real-world scenarios.
 

1. What is the "Roller coaster loop problem"?

The "Roller coaster loop problem" refers to the challenge of designing a roller coaster loop that is safe and comfortable for riders.

2. What factors are considered when designing a roller coaster loop?

Some of the key factors that are taken into account when designing a roller coaster loop include the speed of the train, the radius and angle of the loop, the forces exerted on riders, and the structural integrity of the loop.

3. What makes a roller coaster loop successful?

A successful roller coaster loop is one that is able to provide a thrilling experience for riders while still being safe and comfortable. This means that the loop should not cause excessive forces on riders, and it should also be able to support the weight of the train and riders without any structural issues.

4. What happens if a roller coaster loop is designed incorrectly?

If a roller coaster loop is designed incorrectly, it can result in a range of issues such as uncomfortable or even dangerous forces on riders, structural failures, or an inability for the train to complete the loop. This can lead to accidents and injuries for riders.

5. How do scientists and engineers ensure the safety of roller coaster loops?

To ensure the safety of roller coaster loops, scientists and engineers use computer simulations, physical models, and real-life testing to analyze and optimize the design. They also follow strict safety regulations and standards set by governing bodies to ensure the loop is safe for riders.

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