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Amusement park ride (circular motion) 
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#1
Oct2907, 09:21 AM

#2
Oct2907, 10:52 AM

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Part a) requires a little bit of work but its not too difficult. The max force of static friction ([itex]f_{smax}=\mu_sN[/itex]) for the person to just stay stuck on the wall without moving needs to be equal in magnitude to the persons weight ([itex] W=mg[/itex]). The normal force of the person against the wall is just the centrifugal force.
[tex] F = \frac{mv^2}{r} [/tex] For part b) you've forgotten to take the square root of 6.44. And i'll wait till you've posted your atempts to c) and d) 


#3
Oct2907, 05:41 PM

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how do I relate what you said to get...T= (4 pi^2 R[tex]\mu_s/g)^1/2[/tex] I know that [tex] \sumFx= f_s max = \mu_s N = m(v^2/ r) = mg[/tex] (you said the max force of static friction for the person to stay stuck on the wall without moving needs to be equal in magnitude to the person's weight mg) I looked at what you said again and now I think... [tex] f_s= \mu_s N = \mu_s (mv^2/r)= mg [/tex] I have no idea which is alright or if both are incorrect but I still don't see how I'd show that the maximum period of revolution necessary to keep the person from falling is that equation given when ..generally T= 2 pi r/ v ac= v^2/ r I see from my book in a example that plugging the velocity after rearranging the T equation and plugging into the centripital acceleration equation I can get T= [tex]\sqrt{} 4pi^2 r/ ac[/tex] however I still don't see how I can get the equation given for this particular problem from that.. I really really need help in the equation for a) Thanks 


#4
Oct2907, 05:49 PM

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Amusement park ride (circular motion)



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