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Question: Does centrifugal force counteract gravity's force? how much?

(Posted by: Nutria on 2009-06-21 16:02:50)

A rotating mass "wants " to sling away from it's axis of rotation. Thus a 100kg body at the equator is traveling at 1670 kph, or 6012 kps. (Using "Fc = o^2 x r ", I get an ungodly huge number, 203 * 10^18, that must be wrong.) The gravitational force ( F = mg, where g = 9.8 m/ s^2) on that body is only 980 Newtons.

Posted by: Slowfinger on 2009-06-22, 08:08:41

You have problem with units. Period of rotation is 86400s, thus angular velocity is: omega= 2 pi / T = 2 pi / 86400= 7.27 * 10^-5 rad/ s^2 and linear velocity is v = R omega = 6378 * 10^3 * 7.27 * 10^-5 = 463.68 m/ s Thus centrifugal force on a 100kg body is: Fc = m v^2 / R = 100 * 463.68^2 / (6378 * 10^3)= 3.37N

Posted by: D g on 2009-06-21, 16:16:40

Http:/ / en.wikipedia.org/ wiki/ Centripetal_force regentsprep.org/ Regents/ physics/ phys06/ bcentrif/ default.htm centrifugal force is a lay mans force... the only force that physics people talk about in circular motion is CENTRIPETAL FORCE.. it is force going out .. its the force that is generated because the circular object has a tangetial speed component. something travlling 1670 kph isnt travelling 6012 kps its 60 sec / hour so it would be 1670kph/ 60 secph = 27.88 kps = 2.788*10^4 m/ s Ac = v^2/ r = (2.788*10^4)^2/ r Fc = mAc = 100*(2.788*10^4)^2/ r I dont know what number you are using for radius Hope that helps

Posted by: MikeY on 2009-06-21, 16:17:23

Closer to 463 m/ s that's why