Roller coaster drop physics problem

[Dragoon]

i don't even know how to derive an equation for this problem i am confused if someone would just show me how to derive the equation that would be appreciated not looking for an answer just how to do it.

A roller coaster is designed so that after a large drop, the cars enter a circular path, radius = 46 meters, which is to provide 2.3 g's for the riders. What is the minimum height the drop can be to achieve this effect?

 

[HallsofIvy]

Do you know a formula for acceleration going around a circle at a constant speed? If I remember correctly, the centripetal acceleration is R\[\omega^2\] where \[\omega\] is the angular velocity in radians per second. 1 radian per second corresponds to a speed of R m/s.

I'm unsure as to whether the "2.3 g's" includes the 1 g they would feel if the car were not moving but I'm going to assume it does not. That means that the motion of the car itself must provide 2.3 g's. That is, we want R\[\omega^2\]= 46\[\omega^2\]= 2.3 g= 2.3*9.8= 22.54 so \[\omega^2\]= 22.54/46= 0.49 and so \[\omega\]= 0.7 radians/sec (the square root). Since 1 radian per second corresponds to 46 meters/sec, the speed of the car, to provide that acceleration, must be 46(0.7)= 32.2 m/s. Now, what height must the car drop from in order to have that speed at the bottom?

 

[Dragoon]

thanks for the help i was working on that problem forever and wasnt sure how to handle it i appreciate it. thanks