# Roller coaster change in acceleration?

## Homework Statement

You have taken a job as senior
hench-person with Dr. Caasi Notwen, arch-nemesis of
James Bond. As with all of Bond’s enemies, Dr. Notwen
would like to develop an overly complicated method of
dispatching the British super-spy, so he has tasked you
with designing a “roller coaster of death.” This
frictionless roller coaster is to be designed such that
when the rider (James Bond) reaches the nadir (point B)
on the track, the acceleration he experiences will exceed
the limit of human survival (10 g’s where g = 9.8 m/s2).
At point B the track has a radius of curvature of 25.0 m..

(a) From what height h (Point A) must the cart start (from rest) to ensure Bond’s demise?

ac=v2/R
PE=mgh
KE=(1/2)mv2

## The Attempt at a Solution

10g's = 98.0 m/s2 ≥ ac = v2/R

v = sqrt{acR} = $\sqrt{98*25}$ = 49.5 m/s

PE = KE
mgh = (1/2)mv2
h = v2/2g = 49.52/(2*9.8) = 125.0m

*** I'm rather rusty on my centripetal concepts and am wary of my answer; does it look correct to all of you? ***

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TSny
Homework Helper
Gold Member
There could be some ambiguity in the wording of the question. Your answer is correct if you want the actual acceleration of Bond to be 10 times g.

However, to "experience 10 g's" could be interpreted as experiencing an apparent weight of 10 times your normal weight. That would mean that you would want the normal force acting on Bond to be 10 times his weight. Or, to put it another way, even if bond just sat at rest at point B he would be experiencing 1 g. So, the effect of the motion should be to add another 9 g's if you wanted Bond to experience a normal force of 10 times his weight.

Not sure of the intended interpretation.

There seems to be a second part to the problem, which I'm rather confused on as well; could someone provide any hints?

(b) What force will Bond feel (presuming he survives past Point B) at Point C where the radius of
curvature has been reduced to 15.0 m and is 30 m above point B (i.e., what is the normal force at C)?

Anyone?