Normal force max min constraints on a roller coaster

In summary: So for N=0 the cart is not in contact with the track at the bottom of the dip. (That's physically impossible for real tracks, but it's a consistent result of your model.) That means the normal force is not an upward force but a downward force, and the equation should reflect that.You should be able to work out what the normal force is for N=0.
  • #1
firewoodwolf
3
0

Homework Statement



passengers cannot have a normal force equal to 0 or greater than 4 times their weight. What are the heights H1 and H2. Coaster starts at H1 and and ends at H2. H1>H2 and there is a low point between the two hills defined as zero height. The radius of the curve at the low point is 15m and the curve of the second hill H2 is 15m. Friction is to be neglected.

Homework Equations



sum of the F= mv2 /r
conservation of mechanical energy k1+U1=k2+U2

The Attempt at a Solution


Fc+N-mg = mv2 / r for the lowest point
Fc+mg-N = mv2 /r

I am unsure how to relate the constraints given to Newtons 2nd law for both hills and the conservation of energy.
Thank you for all insights
 
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  • #2
firewoodwolf said:

Homework Statement



passengers cannot have a normal force equal to 0 or greater than 4 times their weight. What are the heights H1 and H2. Coaster starts at H1 and and ends at H2. H1>H2 and there is a low point between the two hills defined as zero height. The radius of the curve at the low point is 15m and the curve of the second hill H2 is 15m. Friction is to be neglected.

Homework Equations



sum of the F= mv2 /r
conservation of mechanical energy k1+U1=k2+U2

The Attempt at a Solution


Fc+N-mg = mv2 / r for the lowest point
Fc+mg-N = mv2 /r

I am unsure how to relate the constraints given to Newtons 2nd law for both hills and the conservation of energy.
Thank you for all insights

Welcome to Physics Forums.

The "Fc" isn't an actual force, with a known physical cause, acting on the cart. It's simply the net force, which we know must equal mv2/r because the cart can be treated as traveling in a circle at both the lowest and highest points. Thinking that Fc is some extra force to be included in your Newton's 2nd Law equations is a common error of people when they are learning this stuff.

So the only forces are the weight (physical cause: gravity) and the normal force (physical cause: the track is pushing up on the cart, because they are in direct contact). Those two forces must combine (by adding or subtracting them, as appropriate) to give a net force of Fc=mv2/r.

So you really have
[STRIKE]Fc[/STRIKE]+N-mg = mv2 / r for the lowest point
You're also given that N is between 0 and 4x the weight. Try substituting both of those extremes for N in the equation, and see what you can come up with.
 
  • #3
thank you

so taking v[0] as zero at the top of the first hill by cons energy i have mgH[a] = 1/2 mv[2] at the bottom of the hill...

N-mg = mv2 / r yields either v2 =gr or v2 = 3gr based on parameters of question.

using this in the conservation of energy will yield:

h[a] = 1/2 (r) if N=0 or h[a] = 1/2 (3r) if n =4x the weight

i believe these are now correct
 
Last edited:
  • #4
firewoodwolf said:
thank you

so taking v[0] as zero at the top of the first hill by cons energy i have mgH[a] = 1/2 mv[2] at the bottom of the hill...

N-mg = mv2 / r yields either v2 =gr or v2 = 3gr based on parameters of question.
I agree with the v2 = 3gr when N=4mg.
But -- do you see what your equation is missing when you tried N=0?

using this in the conservation of energy will yield:

h[a] = 1/2 (r) if N=0 or h[a] = 1/2 (3r) if n =4x the weight

i believe these are now correct
As I said, you're okay for the N=4mg result.

After you fix up the N=0 part, you can use the same process to look at the 2nd hill of height H2
 
  • #5
thank you again

are you hinting at the negative sign on the mg term?
 
  • #6
Yes :smile:
 

Related to Normal force max min constraints on a roller coaster

What is normal force?

Normal force is the force that a surface exerts on an object that is in contact with it. In the context of a roller coaster, it is the force that the track exerts on the coaster car as it moves along the track.

What is the maximum normal force on a roller coaster?

The maximum normal force on a roller coaster occurs when the coaster car is at the bottom of a loop or at the crest of a hill. This is because the car experiences a change in direction, causing the normal force to increase in order to keep the car on the track.

What is the minimum normal force on a roller coaster?

The minimum normal force on a roller coaster occurs when the car is at the top of a loop or at the bottom of a hill. This is because the car is experiencing a change in direction, causing the normal force to decrease in order to prevent the car from flying off the track.

How do normal force constraints affect the design of a roller coaster?

Normal force constraints play a crucial role in the design of a roller coaster as they determine the maximum and minimum speeds at which the coaster can safely navigate through loops, hills, and other elements. Designers must carefully consider these constraints to ensure the safety and enjoyment of riders.

How does the weight of the coaster car affect the normal force on a roller coaster?

The weight of the coaster car affects the normal force on a roller coaster as it determines the amount of force needed to keep the car on the track. A heavier car will require a greater normal force to prevent it from flying off the track, while a lighter car will require less normal force.

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