Roller coaster drop physics problem

In summary, the problem is to determine the minimum height for a roller coaster drop in order to provide 2.3 g's for the riders on a circular path with a radius of 46 meters. The formula for acceleration in a circular path is R\[\omega^2\], with \[\omega\] as the angular velocity in radians per second. To achieve 2.3 g's, the speed of the car must be 32.2 m/s at the bottom, which requires an angular velocity of 0.7 radians/sec. The height of the drop can be calculated using this velocity and the formula for distance traveled.
  • #1
Dragoon
17
0
i don't even know how to derive an equation for this problem i am confused if someone would just show me how to derive the equation that would be appreciated not looking for an answer just how to do it.

A roller coaster is designed so that after a large drop, the cars enter a circular path, radius = 46 meters, which is to provide 2.3 g's for the riders. What is the minimum height the drop can be to achieve this effect?
 
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  • #2
Do you know a formula for acceleration going around a circle at a constant speed? If I remember correctly, the centripetal acceleration is R\[\omega^2\] where \[\omega\] is the angular velocity in radians per second. 1 radian per second corresponds to a speed of R m/s.

I'm unsure as to whether the "2.3 g's" includes the 1 g they would feel if the car were not moving but I'm going to assume it does not. That means that the motion of the car itself must provide 2.3 g's. That is, we want R\[\omega^2\]= 46\[\omega^2\]= 2.3 g= 2.3*9.8= 22.54 so \[\omega^2\]= 22.54/46= 0.49 and so \[\omega\]= 0.7 radians/sec (the square root). Since 1 radian per second corresponds to 46 meters/sec, the speed of the car, to provide that acceleration, must be 46(0.7)= 32.2 m/s. Now, what height must the car drop from in order to have that speed at the bottom?
 
  • #3
thanks for the help i was working on that problem forever and wasnt sure how to handle it i appreciate it. thanks
 

1. How does the height of a roller coaster affect the speed of the drop?

The height of a roller coaster affects the speed of the drop due to potential energy. The higher the coaster is at the top of the drop, the more potential energy it has. As the coaster descends, this potential energy is converted into kinetic energy, resulting in a faster drop.

2. What role does gravity play in a roller coaster drop?

Gravity is the force that pulls the roller coaster towards the ground. As the roller coaster drops, gravity accelerates it towards the ground, increasing its speed. This acceleration due to gravity is what makes the drop thrilling for riders.

3. How does the shape of a roller coaster track impact the drop?

The shape of the track impacts the drop by affecting the forces acting on the riders. A steep drop with a sharp curve at the bottom can result in a sudden increase in g-forces, making the drop feel more intense. A smoother, gradual drop can result in a smoother and less intense experience.

4. What factors determine the maximum speed of a roller coaster drop?

The maximum speed of a roller coaster drop is determined by several factors, including the height of the drop, the mass of the roller coaster and its riders, and the amount of friction between the track and the wheels. The steeper and taller the drop, the faster the coaster will go, assuming there is minimal friction.

5. How does air resistance impact the speed of a roller coaster drop?

Air resistance, or drag, can impact the speed of a roller coaster drop by slowing down the coaster. As the coaster moves through the air, it experiences friction from the air particles, which can decrease its speed. This is why roller coasters often have a streamlined design to minimize air resistance and maximize the speed of the drop.

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