Solve Vertical Circles Problem: Find Roller Coaster Speed

[chmilne]

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!

 

[Mattara]

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

 

[chmilne]

These are the only forces I see:

UP: F_c, F_N
Down: W

Am I missing any?

 

[chmilne]

CORRECTION:

UP: F_N
Down: W (mg)

 

[HallsofIvy]

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]

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?)

 

[chmilne]

Thanks for the help.

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

Thanks again.