# Speeds on a roller coaster

## Homework Statement

The radius of curvature of a loop-the-loop for a roller coaster is 14.0 m. At the top of the loop, the force that the seat exerts on a passenger of mass m is 0.35mg. Find the speed of the roller coaster at the top of the loop. (Enter your answers from smallest to largest.)

(This question has two answers, one of which, I found = the speed at the top of the loop. I'm guessing the 2nd answer is the speed at the bottom of the loop?)

## Homework Equations

∑Fr = mar
Fsp + mg = m(v2/R)

a = v2/R

## The Attempt at a Solution

Fsp = m[(v2/R) - g]
.35mg = m[(v2/R) - g]
.35g = (v2/R) - g
1.25g = v2/R
v_top = √(gR1.35) ≈ 13.62 m/s

Now to find the 2nd answer, I think I need to calculate a (since I believe this is constant):

a = v2/R ≈ 13.25 m/s2

But this can't be correct I'll just get the same answer for velocity.

Does anyone know how I am supposed to find the 2nd answer?

PhanthomJay
Homework Helper
Gold Member
1. Acceleration is not constant.
2. At the top , Fsp and mg are in the same direction. Not so at the bottom.
3. Try conservation of energy approach.

We haven't covered conservation of energy yet (we've just finished Newton's laws, friction, uniform circular motion, drag forces, and an intro to work and dot-products).

I sum the forces at the bottom of the roller coaster:

ƩF = Fsp - mg = mv2/R

∴ Fsp = m[(v2/R) + g]

I don't know the mass or the velocity though and I don't have enough information to get another equation to solve for either, do I?

PhanthomJay
Homework Helper
Gold Member
It seems that you assumed what the question was for the 2nd answer. Have you written down the problem statement correctly?

Yes. I copied & pasted the problem:

The radius of curvature of a loop-the-loop for a roller coaster is 14.0 m. At the top of the loop, the force that the seat exerts on a passenger of mass m is 0.35mg. Find the speed of the roller coaster at the top of the loop. (Enter your answers from smallest to largest.)

The "largest" value is the one I found above (≈13.62 m/s).

PhanthomJay
Homework Helper
Gold Member
How do you know it is the largest?

When I entered the answer (we use WebAssign for our homework), it showed as correct in the space that said "largest value" :).

After getting more information from my teacher, here is the solution for the "smallest" value, i.e. the speed at the bottom of the roller coaster:

Assume Fsp = .35mg at the bottom of the roller coaster as well the top.

Fsp - mg = mv2/R
.35g - g = v2/R
√|R(.35g - g)| = v

The question is worded very poorly. Thank you much for all of your assistance, PhanthomJay.

PhanthomJay