Conservation of energy on loop. Nearly done

AI Thread Summary
The discussion centers on calculating the downward force of the rails on a roller coaster car at the top of a loop, using conservation of energy principles. The initial potential energy is converted into kinetic energy at the top of the loop, leading to the equation for centripetal force. The calculated centripetal force was initially found to be 31,392 N, but the correct force exerted by the tracks is 16.7 kN after accounting for gravitational force. The key realization is that the net inward force (centripetal force) must subtract the weight of the car to find the force from the tracks. This understanding aligns with Newton's second law, confirming the approach to solving the problem.
irishbob
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Homework Statement


A 1500-kg roller coaster car starts from rest at a height H=23.0m above the bottom of a 15.0-m-diameter loop. If friction is negligible, determine the downward force of the rails on the car when the upside-down car is at the top of the loop.

Homework Equations


Conservation of energy: U+K=Esys
U=mgh
K=0.5mv2
F=ma
acentripetal=v2/r

The Attempt at a Solution


Uinitial=Esys because starts from rest. Uinitial=mghinitial
at the top of the loop: Esys=U+K=mghtop+0.5mv2
meaning mghinitial=mghtop+0.5mv2
simplifying: v2=2(ghinitial-ghtop).
so:
F=macentripetal=mv2/r=
1500kg*2(9.81m/s2*23m-9.81m/s2*15)/7.5mSo that gives an answer of 31,392N. I know the answer is 16.7 kN (back of book), and I know you get there by subtracting mg. So why am I subtracting mg? I would think that would be the TOTAL force and not solely the force of the tracks on the car.

Any and all help is much appreciated!

Edit: You know, I think I get it. I solved for centripetal force, which is the total inward force toward the center of the circle. Because mg is in that direction, I subtract it and get the force from the tracks. Is this right?
 
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irishbob said:
Edit: You know, I think I get it. I solved for centripetal force, which is the total inward force toward the center of the circle. Because mg is in that direction, I subtract it and get the force from the tracks. Is this right?
Yes, that's right. The 'centripetal force' is just the net force. To find the contribution from the tracks, you must subtract mg. It's always best to think in terms of Newton's 2nd law:
ΣF = ma
N + mg = mv2/r
N = mv2/r - mg
 
irishbob said:

Homework Statement


A 1500-kg roller coaster car starts from rest at a height H=23.0m above the bottom of a 15.0-m-diameter loop. If friction is negligible, determine the downward force of the rails on the car when the upside-down car is at the top of the loop.


Homework Equations


Conservation of energy: U+K=Esys
U=mgh
K=0.5mv2
F=ma
acentripetal=v2/r

The Attempt at a Solution


Uinitial=Esys because starts from rest. Uinitial=mghinitial
at the top of the loop: Esys=U+K=mghtop+0.5mv2
meaning mghinitial=mghtop+0.5mv2
simplifying: v2=2(ghinitial-ghtop).
so:
F=macentripetal=mv2/r=
1500kg*2(9.81m/s2*23m-9.81m/s2*15)/7.5m


So that gives an answer of 31,392N. I know the answer is 16.7 kN (back of book), and I know you get there by subtracting mg. So why am I subtracting mg? I would think that would be the TOTAL force and not solely the force of the tracks on the car.

Any and all help is much appreciated!

Edit: You know, I think I get it. I solved for centripetal force, which is the total inward force toward the center of the circle. Because mg is in that direction, I subtract it and get the force from the tracks. Is this right?

Glad you worked it out - I highlighted some wording in the question, and your pondering which re-inforces what you realized.
 
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