What is the speed of the roller coaster at the bottom of the dip?

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To determine the speed of a roller coaster at the bottom of a dip with a radius of 22.0 m, the forces acting on a passenger must be analyzed. The net force equation is given by Fn - mg = m(v^2/r), where Fn is the normal force and mg is the gravitational force. The normal force is three times the passenger's weight, so it should be expressed as 3mg. Clarification was provided that using 3m instead of 3mg was incorrect, leading to confusion. The correct application of forces confirms the calculations for determining the roller coaster's speed at the dip's bottom.
Amber430
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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 three times her weight as she goes through the dip. If r = 22.0 m, how fast is the roller coaster traveling at the bottom of the dip?

At the bottom of the dip, Fn -mg= m(v^2/r). I plugged in 3m for Fn since it's 3 times her weight, and I'm not sure if I'm doing this right.
 
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Amber430 said:
At the bottom of the dip, Fn -mg= m(v^2/r).
Good.
I plugged in 3m for Fn since it's 3 times her weight, and I'm not sure if I'm doing this right.
Careful. 3m is a mass, not a force. What's her weight? What's 3 times her weight?
 
It looks like everything you have done is correct. When you say that you plugged in 3m for Fn you mean (3m)*g, right? If so, it looks correct.
 
No, I was not using 3mg, just 3m. That's where I messed it up. It makes sense now, thank you so much!
 
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