# A different type of roller coaster loop problem

Tags:
1. Oct 12, 2016

### JustinT

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
g = 9.8 m/s^2
v^2/r = acentripetal

3. 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

2. Oct 13, 2016

### haruspex

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. Oct 13, 2016

### hmmm27

Might want to check your signs on that one.

4. Oct 13, 2016

### haruspex

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. Oct 14, 2016

### hmmm27

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. Oct 14, 2016

### haruspex

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. Oct 14, 2016

### hmmm27

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".

Last edited: Oct 14, 2016
8. Oct 14, 2016

### haruspex

Maybe, but that does not alter the meaning of acceleration here.

9. Oct 14, 2016

### hmmm27

Perhaps, but "enjoy" implies a tangible experience. One does not experience free fall as an acceleration.

10. Oct 14, 2016

### haruspex

One does not experience standing on firm ground as an acceleration.

11. Oct 14, 2016

### Tom.G

Then what is it that we experience when we are aware of our weight?

12. Oct 14, 2016

### haruspex

We're not discussing GR here. When I stand still, I experience standing still, not accelerating.

13. Oct 14, 2016

### ehild

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. Oct 15, 2016

### haruspex

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. Oct 15, 2016

### hmmm27

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).

Last edited: Oct 15, 2016
16. Oct 15, 2016

### Staff: Mentor

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. Oct 15, 2016

### hmmm27

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.

Last edited: Oct 15, 2016