Formula for centripetal acceleration

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Homework Help Overview

The discussion revolves around the formula for centripetal acceleration and its application in the context of a ferris wheel. Participants explore the relationship between linear speed, radius, and gravitational force as it pertains to circular motion.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss the formula for centripetal acceleration and its components, questioning the definitions of centripetal and centrifugal forces. There is an exploration of how to relate RPM to linear speed and the implications for the forces experienced by riders on a ferris wheel.

Discussion Status

The discussion is active, with various interpretations of the forces at play being explored. Some participants provide guidance on how to approach the problem, while others raise questions about the definitions and perspectives regarding the forces involved.

Contextual Notes

There is a focus on the conditions under which riders feel weightless and the forces acting on them, with some ambiguity regarding the terminology used to describe these forces. The problem context includes specific values, such as the radius of the ferris wheel, which may influence the calculations discussed.

cookiemonster
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The formula for centripetal acceleration might help:

[tex]a = \frac{v^2}{r}[/tex]

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
[tex]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}[/tex]

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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