F = ma equations for circular movement (roller coaster)

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
 
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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