Roller coaster and centripetal acceleration

In summary: Fc = m*v^2/r-m*g(y2 - y1) = 0.5*m*v2^2Apply what you have learned about pure potential and pure kinetic energy, as well as about a mix of both.Fc = m*v^2/r-m*g(y2 - y1) = 0.5*m*v2^2The first of those two is the general expression for centripetal force. It does not mention v2 as such. Isn't there another equation using v2 instead of the generic v?
  • #1
annamal
381
33
Homework Statement
The frictionless track for a toy car includes a loop-the-loop of radius R. How high, measured from the bottom of the loop, must the car be placed to start from rest on the approaching section of track and go all the way around the loop?
Relevant Equations
Fc = m*v^2/r
-m*g(y2 - y1) = 0.5*m*v2^2
See attached image.
The solution to this problem calculates v2 at the top of the roller coaster ride. Why is that? Shouldn't you calculate v2 at the bottom of the roller coaster ride as you require the maximum velocity there to get around the loop?
 

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  • #2
annamal said:
The solution to this problem calculates v2 at the top of the roller coaster ride. Why is that? Shouldn't you calculate v2 at the bottom of the roller coaster ride as you require the maximum velocity there to get around the loop?
And what does it do with the velocity it calculated?
 
  • #3
haruspex said:
And what does it do with the velocity it calculated?
That's just the velocity going around the roller coaster.
 
  • #4
annamal said:
That's just the velocity going around the roller coaster.
That does not answer my question. What equations does the solution have which involve v2?
 
  • #5
annamal said:
That's just the velocity going around the roller coaster.
Velocity on the roller coaster is not the relevant question (and is also not constant in the loop for obvious reasons). The relevant question is: When and why would the car risk falling off the roller coaster?
 
  • #6
annamal said:
...
The solution to this problem calculates v2 at the top of the roller coaster ride. Why is that? Shouldn't you calculate v2 at the bottom of the roller coaster ride as you require the maximum velocity there to get around the loop?
If the toy is properly designed, placing the car at the top end of the ramp will be sufficient.
Now, imagine that you are the designer of the toy and that the only information given to you are the diameter of the loop and the requirement to save material and package volume by making the ramp as short as possible.
Bonus question: Could you manipulate the weight of the car to reduce the ramp height?
 
  • #7
haruspex said:
That does not answer my question. What equations does the solution have which involve v2?
Fc = m*v^2/r
-m*g(y2 - y1) = 0.5*m*v2^2
 
  • #8
And the incline to the loop at the beginning is negligible why?
 
  • #9
annamal said:
And the incline to the loop at the beginning is negligible why?
Apply what you have learned about pure potential and pure kinetic energy, as well as about a mix of both.
 
  • #10
annamal said:
Fc = m*v^2/r
-m*g(y2 - y1) = 0.5*m*v2^2
Still something missing.
The first of those two is the general expression for centripetal force. It does not mention v2 as such. Isn't there another equation using v2 instead of the generic v?
 

1. What is centripetal acceleration in relation to roller coasters?

Centripetal acceleration is the acceleration that is directed towards the center of a circular path. In the context of roller coasters, it is the acceleration that keeps the riders moving in a circular path as the roller coaster moves along the track.

2. How is centripetal acceleration calculated for roller coasters?

Centripetal acceleration can be calculated using the formula a = v^2/r, where a is the centripetal acceleration, v is the velocity of the roller coaster, and r is the radius of the circular path the roller coaster is traveling on.

3. What factors affect the centripetal acceleration of a roller coaster?

The centripetal acceleration of a roller coaster is affected by the velocity of the roller coaster, the radius of the circular path, and the mass of the roller coaster and its riders. The steeper the curve and the faster the roller coaster is moving, the greater the centripetal acceleration will be.

4. How does centripetal acceleration impact the experience of riding a roller coaster?

The centripetal acceleration experienced by riders on a roller coaster can create a feeling of weightlessness or a sensation of being pushed into the seat or held against the harness. The intensity of these sensations depends on the speed and curvature of the track.

5. Can centripetal acceleration be dangerous on roller coasters?

In most cases, the centripetal acceleration experienced on roller coasters is not dangerous. However, if the acceleration is too high or if there are sudden changes in direction, it can cause discomfort or even injury to riders. This is why roller coasters are carefully designed and tested to ensure safe levels of centripetal acceleration.

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