A different type of roller coaster loop problem

In summary, the loop at the amusement park Physics Hill has a radius of 31.89 meters and a height of 63.78 meters.
  • #1
JustinT
1
0

Homework Statement


A roller coaster at the nearby amusement park Physics Hill, has a large circular loop where cars travel at the top of the loop at 25.0 m/s causing the riders to enjoy an acceleration of 3g. Assuming the height of this loop is the same as the diameter, how high is the loop?

Homework Equations


g = 9.8 m/s^2
v^2/r = acentripetal

The Attempt at a Solution


I figured at the top of the loop the acceleration acting on the car is all downward. There is already 1g of acceleration on the car so the centripetal acceleration plus the existing 1g needs to equal the 3g's of acceleration the riders are experiencing. However my answer for the radius seems unrealistic.

v^2/r + g = 3g

v^2/r = 2g

25^2/r = 19.6

r = 31.89m
d = 63.78m
 
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  • #2
JustinT said:
the centripetal acceleration plus the existing 1g
Centripetal force is not an actual applied force. It is that resultant force which produces the centripetal acceleration. Likewise, centripetal acceleration is the radial component of the acceleration which results from all the actual applied forces. Thus, it is not additional to g.
 
  • #3
JustinT said:
I figured at the top of the loop the acceleration acting on the car is all downward
Might want to check your signs on that one.
 
  • #4
hmmm27 said:
Might want to check your signs on that one.
No, it's a reasonable statement, except that the use of "all" applied to accelerations is problematic. Accelerations do not act on masses, forces act on them. The net force acting on the mass produces the resulting acceleration. The acceleration at the top of the loop is downwards, and all forces acting on the mass at that point are also downwards.
 
  • #5
Etymology aside, if the passengers want to "enjoy an acceleration of 3g" at the top of the loop the rotational acceleration is going to have to be 4g, not 2g.
 
  • #6
hmmm27 said:
Etymology aside, if the passengers want to "enjoy an acceleration of 3g" at the top of the loop the rotational acceleration is going to have to be 4g, not 2g.
This is tricky. When I stand still on the floor I am not accelerating (in the Earth frame). In free fall I feel weightless, but I am accelerating at 1g. I read "enjoying an acceleration of 3g" as experiencing an acceleration of 3g and enjoying it. That is not the same as feeling a "force" of 3g.
 
  • #7
haruspex said:
I read "enjoying an acceleration of 3g" as experiencing an acceleration of 3g and enjoying it.

Given what the g-force is going to be at the bottom of the loop, one can only assume that "enjoyment" is used poetically, in the sense of "possibly briefly regaining consciousness".
 
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  • #8
hmmm27 said:
Given what the g-force is going to be at the bottom of the loop, one can only assume that "enjoyment" is used poetically, in the sense of "briefly regaining consciousness".
Maybe, but that does not alter the meaning of acceleration here.
 
  • #9
Perhaps, but "enjoy" implies a tangible experience. One does not experience free fall as an acceleration.
 
  • #10
hmmm27 said:
Perhaps, but "enjoy" implies a tangible experience. One does not experience free fall as an acceleration.
One does not experience standing on firm ground as an acceleration.
 
  • #11
haruspex said:
One does not experience standing on firm ground as an acceleration.
Then what is it that we experience when we are aware of our weight?
 
  • #12
Tom.G said:
Then what is it that we experience when we are aware of our weight?
We're not discussing GR here. When I stand still, I experience standing still, not accelerating.
 
  • #13
Experiencing 3g means that you feel yourself 3 times heavier that normally. In this aspect, 3g is a force 3mg (G-force). Sitting in the roller coaster, it is the sit that acts on your body with that force. You can also say that the sit acts with 3mg normal force on you. Together with the force of gravity, the total force acting at you on the top of the loop is 4mg, and that is the centripetal force at that point.
 
  • #14
ehild said:
Experiencing 3g means that you feel yourself 3 times heavier that normally.
Yes, but that is "experiencing 3g", which, as you say, is taken to mean a net contact force of 3mg. That is fair enough, because we experience contact forces, not accelerations. When I stand still I "experience 1g" (i.e. a contact force of mg, the normal reaction from the floor), but I do not experience an acceleration of 1g. In free fall I experience no contact force, but a downward acceleration of g, etc.
I honestly am not sure whether the question setter has simply blundered, and meant to refer to "enjoying the experience of 3g", or whether this is a trick question, requiring the student to answer with the actual acceleration experienced.
 
  • #15
a large circular loop where cars travel at the top of the loop at 25.0 m/s causing the riders to enjoy an acceleration of 3g.
I would humbly suggest applying the 3g parameter to the bottom of the loop despite the wording, and also read it as force not acceleration (annotating appropriately).
 
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  • #16
hmmm27 said:
I would humbly suggest applying the 3g parameter to the bottom of the loop despite the wording, and also read it as force not acceleration (annotating appropriately).
So the scenario would be that of a loop where the speed at the top is 25 m/s and the speed at the bottom creates the impression of 3x weight for the passengers?
 
  • #17
gneill said:
So the scenario would be that of a loop where the speed at the top is 25 m/s and the speed at the bottom creates the impression of 3x weight for the passengers?

Sounds about right. Alternatively, the original wording using 3g acceleration as a marker at the top (requiring a 2g rotational acceleration + 1g for gravity) yields a 12.5g force at the bottom. Using force as a marker, a 3g net at the top (4g - 1g) yields a not-particularly-survivable 17g's at the bottom.
 
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Related to A different type of roller coaster loop problem

1. What is a "different type of roller coaster loop problem"?

A "different type of roller coaster loop problem" refers to a mathematical problem that involves determining the shape and size of a roller coaster loop that will allow a roller coaster car to safely complete the loop without losing contact with the track.

2. What makes this type of roller coaster loop problem different from traditional loop problems?

This type of roller coaster loop problem differs from traditional loop problems in that it takes into account additional factors such as the speed and acceleration of the roller coaster car, as well as the physical limitations of the human body, in order to design a realistic and safe roller coaster loop.

3. What are some real-life applications of this type of roller coaster loop problem?

This type of roller coaster loop problem is important in the design and construction of real-life roller coasters, as it ensures that the loops are safe and enjoyable for riders. It is also used in amusement park ride inspections to ensure that existing roller coasters are still safe for use.

4. What are some challenges in solving this type of roller coaster loop problem?

One challenge is accurately modeling the forces and movements of a roller coaster car as it travels through the loop, as small variations can greatly affect the design of the loop. Another challenge is balancing safety and excitement, as a loop that is too intense may not be suitable for the general public.

5. How do scientists approach solving this type of roller coaster loop problem?

Scientists approach solving this type of roller coaster loop problem by using mathematical equations and computer simulations to analyze different loop designs and determine the optimal shape and size for the loop. They also take into account safety regulations and rider experience to create a well-designed roller coaster loop.

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