# Virtical Circles

Here's the question.

A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the seat of the car pushing upward on her with a force equal to twice her weight as she goes through the dip. If r=20.0 m, how fast is the roller coaster traveling at the bottom of the dip?

I'm having some real troubles answering this problem. Thus far, here's
what I've come up with.

Fc = ( mv^2 ) / r
FN - 2mg = ( mv^2 ) / r
2mg - mg = ( mv^2 ) / r
(2mg - mg )(r) = ( mv^2 )
(2mg - mg )(r) = ( v^2 )
sqrt( (2mg - mg )(r) ) = v

sqrt( ( 19.6 - 9.8 m/s^2 )( 20.0m ) ) = v
14 m/s = v

I've been informed that 14 m/s is not the correct answer and that to one digit the correct answer is 10 m/s.

It seems to me that I have all the right pieces, but I may be missing something. Will someone help, please? Big thanks!

Create a relationship at the bottom of the coaster. Draw a picture of all of the forces and find the resultant force.

These are the only forces I see:

UP: F_c, F_N
Down: W

Am I missing any?

CORRECTION:

UP: F_N
Down: W (mg)

HallsofIvy
Homework Helper
What do you mean by F_N? The forces felt by the person at the bottom will be the centrifugal force (your F_c?) and the person weight, both downward.

Doc Al
Mentor
chmilne said:
CORRECTION:

UP: F_N
Down: W (mg)
Correct. Those are the only forces acting on the person.

Your analysis--and answer--in post #1 seem correct to me. (It's also true that to one significant digit, 10 m/s is the answer--14 rounds off to 10. But why in the world would you settle for one digit?)

Thanks for the help.

There was something lost in translation when the professor told me 'to one digit.'

Thanks again.