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Name: Garrett
Status: student
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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.

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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