F = ma equations for circular movement (roller coaster)

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SUMMARY

The discussion focuses on deriving expressions for centripetal acceleration at the top of a roller coaster hump and the bottom of a loop using the equation F = ma. At the top of the hump, the equation simplifies to f = mg = m*v^2/Rhump, while at the bottom, it becomes f = F - mg = m*v^2/Rloop. The normal force is critical in these calculations, particularly at the top where it can approach zero if the cart is about to become airborne. The conversation emphasizes the importance of understanding centripetal acceleration over forces in this context.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of centripetal acceleration and its formula (a = v^2/R)
  • Familiarity with roller coaster dynamics and forces acting on objects in circular motion
  • Basic algebra for solving equations involving variables
NEXT STEPS
  • Study the derivation of centripetal acceleration in circular motion scenarios
  • Learn about the role of normal force in roller coaster design and safety
  • Explore the concept of apparent weight in non-inertial reference frames
  • Investigate the physics of roller coasters, focusing on energy conservation and forces
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the dynamics of roller coasters and circular motion analysis.

hsbhsb
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Homework Statement


Come up with expressions for centripetal acceleration at the top of the small hump and the bottom of the loop. (this is one substep in a larger problem)

acQhRmm.png

Homework Equations


f = ma

The Attempt at a Solution



Taking down to be negative, I believe that the f = ma equation at the top of the hump is

f = Fnormal - mg = -m*v^2/Rhump

and at the bottom of the loop it is
f = Fnormal - mg = m*v^2/Rloop

It is simple to solve for v^2/r from there

But the solutions ignore the normal force, saying that the f = ma equations are, respectively

f = mg = m*v^2/Rhump
f = F - mg = m*v^2/Rloop

I don't understand why they do this... In apparent weight circular motion problems like this we always include the normal force in the f = ma equation...
6mVwiIC.png
 
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hsbhsb said:
I don't understand why they do this... In apparent weight circular motion problems like this we always include the normal force in the f = ma equation...
The normal force could be zero at the top of the hump if the cart is about to become airborne. Is there is language in the problem saying that this is the case? The normal force will never be zero at the bottom of the loop. Maybe the "F" in F - mg in the equation for the loop indicates the normal force.
 
hsbhsb said:

Homework Equations


f = ma
That's the start of your difficulties. The question you posted is not concerned with forces, only the centripetal acceleration.
hsbhsb said:
It is simple to solve for v^2/r from there
Not from the point you got to it isn't, since you do not know the normal force.
Forget about forces and think how else you might find the velocities.
kuruman said:
The normal force could be zero at the top of the hump if the cart is about to become airborne
Not that it is relevant to the quoted part of the question, but roller coasters are designed not to come off track, even if the normal force goes negative.
 

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