Where does g force fit in this centripetal motion question?

AI Thread Summary
In the discussion about centripetal motion and g-force, a scenario is presented where a car travels in a vertical circle with a radius of 6 meters and a mass of 800 kg. The net force equation is analyzed, focusing on the relationship between normal force (fn) and gravitational force (fg) at the bottom of the circle. It is noted that fn and fg cannot be equal, as one must be greater due to the required centripetal acceleration. The question arises about how to incorporate the g-force into the calculations, specifically whether it should be applied to fn or fg. The conversation emphasizes the need to understand radial acceleration to solve the problem effectively.
Henrybar
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a car goes around a vertical circle; determine the period of one cycle if the radius= 6m mass=800kg, gravity=9.8m/s^2 gforce = 6g

my attempt at the bottom of circle:
fnet=ma
fn+fg=(mv^2)/r
fn+fg=(4∏^2mr)/t^2
t=√4∏^2mr/fn+fg
t=√4∏^2(800)(6)/(fg+fg) <-----value of fn and fg will be in opposite directions

fn≠fg so either fn or fg has to be greater. If they both = mg at the bottom, do i multiply the gforce here? to which one? I originally thought that the g force would be multiplied to fn since there is acceleration towards the centre, but would that even make sense?
 
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Henrybar said:
a car goes around a vertical circle; determine the period of one cycle if the radius= 6m mass=800kg, gravity=9.8m/s^2 gforce = 6g

my attempt at the bottom of circle:
fnet=ma
fn+fg=(mv^2)/r
fn+fg=(4∏^2mr)/t^2
t=√4∏^2mr/fn+fg
t=√4∏^2(800)(6)/(fg+fg) <-----value of fn and fg will be in opposite directions

fn≠fg so either fn or fg has to be greater. If they both = mg at the bottom, do i multiply the gforce here? to which one? why?

Lets break this problem into parts, first what's the radial acceleration?
 
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