Roller coaster minimum speed around loop and effective weight

AI Thread Summary
The discussion focuses on calculating the minimum speed required for a roller coaster to keep passengers from falling out at the top of a 20.0m loop, which is determined to be 14 m/s. At the bottom of the loop, the effective weight of passengers is calculated to be twice their actual weight due to the combined forces of normal force and gravity. For the halfway point in the loop, participants suggest using similar principles to find the net force acting on a passenger, indicating the need to consider both the normal force and gravitational force. The conversation emphasizes the importance of understanding forces in circular motion. Overall, the calculations illustrate the dynamics of roller coaster physics.
and9
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Homework Statement


A roller coaster goes around a vertical loop of 20.0m. a) What is the minimum speed the cars must be moving so that the passengers do not fall out at the top? b) Assuming that the car maintains the same constant speed what is th effective weight of the passengers at the bottom? c) Again assuming car maintains this same constant speed, what is the net force (magnitutde and direction) acting on a 25kg passenger at point half way down the loop?

2. My attempt at solution

a) At top, N -> 0
N + mg = m(v2/r)
mg=m(v2/r)
g=v2/r
vmin=sqrt(rg) = 14 m/s

b) At bottom, N counters mg
N - mg = m(v2/r)
N = m((v2/r) + g)
N = m (9.8 + g) = 19.6m or 2mg

c) I had no idea what to do here?
 
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Welcome to PF!

Hi and9! Welcome to PF! :smile:
and9 said:
c) Again assuming car maintains this same constant speed, what is the net force (magnitutde and direction) acting on a 25kg passenger at point half way down the loop?

c) I had no idea what to do here?

Same as a) and b) … find N, and add it (as a vector) to the weight. :wink:
 
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