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Circular Roller Coaster Loop Dangers
Name: Garrett
Status: student
Age: N/A
Location: N/A
Country: N/A
Date: N/A
Question:
Can you explain why it would take 8 g's to make it
around a loop on a roller coaster if the loop was perfectly
round? It is hard for me to believe this.
Replies:
Dear Garrett:
You are a bit generous with your 8 g claim. If there is negligible
friction, it works out to be 6 g's of acceleration.
What follows is an excerpt from:
Unterman, Nathan A., Amusement Park Physics: A Teacher's Guide, second
edition; Walch Publishers 2001. Pages 76-77.
ISBM 0-8251-4264-4
Investigating a frictionless circular loop of radius R with the
condition that the train just barely remains on the track at
the top of the loop, we find that the gravitational force must
be equal to the centripetal force due to the circular motion.
Solving, using Newton's second law, we have:
Sum of forces = Fnet = ma = (mv^2 top/r) = Fs-mg
Since the force of support from the track is zero, in this case, we have:
(mv^2 top/r) = mg
Solving for the tangential velocity at the top, we have:
v^2 top = rg
The gravitational potential energy at the top of the circular loop is:
GPE = mg2r
since the diameter of the circle, or 2r, is the height climbed. The
total energy at the top is the gravitational potential energy stored
in Earth's gravitational field due to the change in height and the
kinetic energy just moving the train over the top. Hence:
Etotal = GPE + KE
= mg delta y + (1/2)mv^2
Substituting 2r for the height and rg for v^2
=mg2r + (1/2)m(rg)
= 2.5mgr
In a negligible friction system, this must be equal to the gravitational
potential energy of the train on the lift hill at elevation delta y, before
entering the circular loop; hence:
mg delta y = 2.5 mgr
Therefore, the minimum height of the starting hill that drops the roller
coaster into the circular loop must be 2.5 r of the circular loop. Again,
assuming a friction-free system, by conservation of energy, the gravitational
potential energy at the top of the lift hill must be equal to the kinetic
energy at the bottom of the loop:
mg delta y = 2.5mgr = (1/2)mv^2
(top) (bottom)
Solving for v^2, we have
v^2 bottom = 5rg
At the bottom, the track must not only support the train, but also keep the
train moving with circular motion (mv^2 bottom/r). The forces at the bottom
must be:
Sum F = Fnet = ma = mv^2/r = Fsupport - mg
mv^2/r = F support - mg
F suppport = mv^2/r + mg
Substituting for v^2
Fsupport = m5rg/r + mg = m5g + mg = m6g
Simplifying, a = 6g at the bottom.
In reality, this situation does not exist, since we have ignored frictional
effects. Clearly, an even greater velocity, and thus greater acceleration,
would be necessary; this could be obtained from a taller lift hill or
catapulting the train. This would increase the acceleration at the
bottom, entering the loop at an acceleration of over 6 g's. This large
acceleration is unacceptable, as many riders would be rendered temporarily
unconscious.
The elegance of a clothoid loop is due to its changing radius of curvature,
smaller accelerations are experienced at the bottom that if a circular track
is used.
I hope this helps!
---Nathan A. Unterman
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Update: June 2012
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