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Thank you for taking my question. I am taking an AP Physics course, and I could really use some help with this question. My regular-teacher is out sick this week with the flu, and the substitute wasn't very helpful. I was hoping you could show me what your correct answers are to this question, and (just briefly) explain what process/formula you used to get that answer. Thank you very much for your assistance. Clark

Question:
A roller coaster has a vertical clothoid loop that is 40 m tall. The top of the loop has a radius of 15 m and the entrance and exit at the bottom of the loop have radii of 45 m. For the following questions, assume a frictionless track unless otherwise told different.
1. Find the minimum speed required at the top of the loop for a passenger to feel no force from the seat or shoulder harness?
2. What must the minimum speed be at the bottom of the loop for the coaster to have the speed calculated in the previous question (#1)
3. Assume no other work is done on the train after its ascent of the first hill, what's the minimum height of the hill to make sure it has enough speed at the top of the loop?
4. Finally, accounting for friction, how would the above answers change?

For Part 1, the passenger will not feel a force from the seat (directed upwards since they are upside down) nor the shoulder straps (downward). This means that the forces must balance, i.e., must be equal. The upward force is due to change in direction of the passenger, which is obtained using the equation for acceleration for uniform circular motion

a_circ = m･v^2/R_top

Multiply this by the mass to get a force

F_up = m･a_circ = m･(v_top^2)/R_top

The downward force is due to gravity

F_grav = m･g

Equate these forces and solve for v_top

m･g = m･v^2/R_top  --->  v_top = sqrt(g･R_top)

Part 2: This is answered by using the conservation of energy: The minimum speed at the bottom must give a kinetic energy equal to the kinetic energy at the top,

KE_top = (1/2)･(v_top)^2

PLUS the potential energy at the top obtained by working against gravity to get from the bottom to the top

PE_top = mgh

Symbolically

(1/2)m･(v_bot)^2 = (1/2)･m･(v_top) + mgh

so that

v_bot = sqrt(2gh +v_top^2)

For Part 3, I assume that the train goes to the top of a hill and then rolls down with enough speed to take it through the loop. Thus, it needs enough PE in order to give the train the energy calculated at the top, which we just calculated,

mg･h_hill = (1/2)m(2gh + v_top^2)

or

h_hill = (h + R_top/2).

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#### randy patton

##### Expertise

college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

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26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

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J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

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M.S. MIT Physical Oceanography, B.S. UC Berkeley Applied Math

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