Roller coaster loop searching for Height

  • Thread starter gggorillaz
  • Start date
  • #1

Homework Statement


A mass of 100kg is h meters high on a track that extends into a loop that has a radius of 20m. Assume track is frictionless. I need to find minimum height for the mass to make it around the loop without falling off or going backwards.


Homework Equations


Conservation of Energy:
K_1+U_2=K_2+U_2
v=sqrt(2gh)

Force:
F=ma
F=mg

Centripetal Force:
(mv^2)/r=F


The Attempt at a Solution


This may be right, i only need someone to check my work in case im incorrect. So, from K_1+U_1=K_2+U_2 i get K=U
From there i have 1/2mv^2=mgh
masses cancel so v^2=2gh
from there i find centripetal force = (mv^2)/r=F
F=mg=(mv^2)/r
masses again cancel so i find that g=(v^2)/r
I replace v^2 with 2gh so i find that g=2gh/r
g's cancel so i find that h=r/2 plug in for r: h = 20/2=10meters.
Can someone verify that im correct? thanks in advance!
 

Answers and Replies

  • #2
PhanthomJay
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Homework Statement


A mass of 100kg is h meters high on a track that extends into a loop that has a radius of 20m. Assume track is frictionless. I need to find minimum height for the mass to make it around the loop without falling off or going backwards.


Homework Equations


Conservation of Energy:
K_1+U_2=K_2+U_2
v=sqrt(2gh)

Force:
F=ma
F=mg

Centripetal Force:
(mv^2)/r=F


The Attempt at a Solution


This may be right, i only need someone to check my work in case im incorrect. So, from K_1+U_1=K_2+U_2 i get K=U
From there i have 1/2mv^2=mgh
masses cancel so v^2=2gh
from there i find centripetal force = (mv^2)/r=F
F=mg=(mv^2)/r
masses again cancel so i find that g=(v^2)/r
I replace v^2 with 2gh so i find that g=2gh/r
g's cancel so i find that h=r/2 plug in for r: h = 20/2=10meters.
Can someone verify that im correct? thanks in advance!
You are not correct, how can the track be only 10 m high when the loop itself is 40 m high?:wink: Can you spot your error in your equation in the use of the variable 'h' in relation to the height of the top of the loop where the speed is critically at its minimum?
 
  • #3
iIs it that i have a skewed understanding of what centripetal force equals? maybe that ma=mv^2/r but my a isnt = g in this case?

Should i be getting h=2diameter? if so, how do i get there?
 
Last edited:
  • #4
PhanthomJay
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I am not sure of your understanding of the centripetal force, but when you solved for the acceleration of a =g, that is correct, at the top of the loop, since at that point, there is no normal force acting if the coaster is just able to round the top of the loop without leaving the track. You may have assumed that a=g at the bottom of the loop?? Anyway, since v2 = rg is correct at the point at the top of the loop, then when applying conservation of energy , you must apply it between the point h and the point at the top of the loop, to equate velocities and solve for the distance from the top of the track to the top of the loop.
 
  • #5
Ohhh right, I see now, it would now be mgh=1/2mv^2+mg(height of loop) so h=1/2(r)+h =1/2(20)+40 = 50m right?
 
  • #6
PhanthomJay
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Ohhh right, I see now, it would now be mgh=1/2mv^2+mg(height of loop) so htrack=1/2(r)+hloop =1/2(20)+40 = 50m right?
Right:approve:
 
  • #7
Yay! thank you so much for helping!
 

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