Normal force max min constraints on a roller coaster

[firewoodwolf]

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

 

[Redbelly98]

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.

 

[firewoodwolf]

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

 

[Redbelly98]

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

 

[firewoodwolf]

thank you again

are you hinting at the negative sign on the mg term?

 

[Redbelly98]

Yes