Calculating Roller Coaster Design for Physics Class

[Shekels]

I have an assignment in my high school physics class to create a roller coaster that has at least one loop and a number of turns and drops. I have been searching the web to find a way to calculate if my car i made of K-Nex weighing 3oz, 85.04g, will have enough speed to clear a loop. The drop is about .8382m high, 2'9'', and i did find on another sight that the top of the loop can't be higher that half of the drop height. I guess i know it will clear the loop but i would like to know how fast it will be going, what angle the drop should be, and how fast the car will be traveling after it does clear the loop, if you have any information that would help me with my task i would apriciate that very much.

thank you very much for your time.

shekels

 

[Danger]

Welcome to PF, Shekels.
Try doing a forum search on this subject to start with. There are some pretty good responses to previous threads about it. It seems to be a pretty popular school project these days.

 

[sir-pinski]

Hi Shekels,

It's great to hear that you are working on a fun and challenging project for your physics class! Creating a roller coaster with specific design parameters is a great way to apply the concepts you have learned in class to a real-life scenario.

To calculate the speed of your roller coaster car, you will need to use the principles of energy conservation. This means that the total amount of energy at the top of the first drop (known as potential energy) is equal to the total amount of energy at the bottom of the loop (known as kinetic energy).

The formula for potential energy is PE = mgh, where m is the mass of the car, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the drop. In your case, the potential energy at the top of the first drop would be 85.04g x 9.8 m/s^2 x 0.8382m = 699.8 joules.

The formula for kinetic energy is KE = 1/2mv^2, where m is the mass of the car and v is the velocity. To find the velocity, you can rearrange the equation to v = √(2KE/m). In this case, the kinetic energy at the bottom of the loop would also be 699.8 joules, since energy is conserved. Plugging in the mass of the car (0.08504 kg) and solving for v, we get a velocity of approximately 11.33 m/s.

To calculate the angle of the drop, you can use trigonometry. The sine of the angle is equal to the height of the drop divided by the length of the drop. In this case, the length of the drop is 2'9'' or 0.8382m. Therefore, sinθ = 0.8382m/0.8382m = 1. Taking the inverse sine, we get an angle of 90 degrees, which makes sense since the drop is straight down.

After the car clears the loop, its velocity will depend on the design of the loop (such as the radius and height) and any other turns or drops that come after it. To calculate the final velocity, you can use the same formula as before, but with the new kinetic energy and mass of the car.

I hope this information helps you with your project. Best of luck with