Formula for centripetal acceleration

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The formula for centripetal acceleration is a = v²/r, where 'v' is linear speed and 'r' is the radius of the circular path. To achieve an acceleration equal to gravity (9.81 m/s²), the relationship between RPM and radius is explored, leading to the equation v = πωR/30. The discussion highlights the distinction between centripetal and centrifugal forces, noting that the latter is a reaction force felt by riders due to the centripetal force acting on them. The main focus is on how fast a ferris wheel must rotate for riders to feel weightless, with the gravitational force needing to be counteracted by the upward force at the top of the ride. Understanding these forces is essential for analyzing the dynamics of circular motion in amusement rides.
cookiemonster
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The formula for centripetal acceleration might help:

a = \frac{v^2}{r}

toward the center of the circle.

cookiemonster
 
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The point of cookiemonster's formula is that you want to make
a equal to g: 9.81 m/s2. v is, of course, the linear speed. For a circle of radius R, so circumference 2πR, moving with RPM ω, &omega revolutions per minute means covering 2πωR meters per minute or 2πωR/60= πωR/30 meters per second. That is: v= πωR/30 where v is in m/s and &omega is RPM. In terms of RPM, then
a= \frac{v^2}{R}= \frac{\pi^2\omega^2R^2}{900R}= \frac{\pi^2\omega^2R^2}{900}= 9.81\frac{m}{s^2}

Plug in 7.5 for R and solve for ω
 
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Centripetal means "seeking a center." What you really want is the centrifugal force on the riders and, as HallsofIvy says, set that equal to the gravitational force.
 
Strictly speaking, centrifugal force is a reaction force produced by the body on the rider. The only force acting on a body in uniform rotational motion is the centripetal force. However, the reaction force exerted by the ferris wheel on you makes you feel like you experience some kind of a centrifugal force.

Oh, did I just make myself a nitpicker?
 
It's from the viewpoint or perspective of the observer standing on the ground looking at the riders. The problem was, how fast the ferris wheel must rotate to make the riders feel weightless. The gravitational force on the riders is downwards, so at the top the counteracting force must be equal but opposite, i.e. upwards, away from the center of the ferris wheel. Do we call that centripetal? I am not saying that you are wrong, though. Just confusing. Of course from the viewpoint of the riders, the force they feel that is produced by gravity points upward. That's how the pressure on your seat of pants or on the soles of your feet feel. (I am not writing this for your benefit, Sickboy, but for others'). Hence, the force the riders feel that is produced by the ferris wheel is . . . (pause for an answer from the class - centripetal or centrifugal).
 
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