Roller coaster minimum speed around loop and effective weight

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SUMMARY

The discussion focuses on calculating the minimum speed required for a roller coaster to prevent passengers from falling out at the top of a 20.0m vertical loop, which is determined to be 14 m/s using the formula vmin = sqrt(rg). Additionally, the effective weight of passengers at the bottom of the loop is calculated as 2mg, where N counters mg. For the net force acting on a 25kg passenger halfway down the loop, participants suggest using similar principles to find the normal force and adding it vectorially to the weight.

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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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