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?)
∑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?