(adsbygoogle = window.adsbygoogle || []).push({}); [SOLVED] Riding a Loop the Loop (velocity at the bottom of the loop)

1. The problem statement, all variables and given/known data

A car in an amusement park ride rolls without friction around the track shown in the figure. It starts from rest at point A at a height h above the bottom of the loop. Treat the car as a particle.

part A - What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B)?

found this to be 5/2R, since:

sigma Fy = Fc - N - mg = 0

N = 0 and Fc = mv^2/R

so Fy = Fc = mg and mv^2/R = mg and v^2 = GR

h = (1/2mgR + 2mgR)/mg = 1/2R + 2R = 5/2R

now I'm stuck on the velocity at the point C.

part B - If the car starts at height h = 4.50 R and the radius is R = 22.0m , compute the speed of the passengers when the car is at point C, which is at the end of a horizontal diameter.

2. Relevant equations

mgR = 1/2mv^2

v^2 = [tex]\sqrt{2mgR}[/tex]

3. The attempt at a solution

got v = 29 which was wrong, I might be using the wrong formulas or the wrong radius, or the formulas in the wrong way; this is an energy conservation problem.

any help is appreciated.

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# Homework Help: Riding a Loop the Loop (velocity at the bottom of the loop)

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