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Centripetal Force/Changing direction on ice  Turn length 
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#1
Feb2912, 07:06 PM

P: 68

1. The problem statement, all variables and given/known data
I am running to school at 6m/s when I hit a patch of icy sidealk. The coefficient of friction between the ice and my shoes is 0.16. I wish to turn my 55kg frame to the left. What is the smallest turn I can make? v = 6 m/s μ = 0.16 m = 55 kg 2. Relevant equations Fnet = mv^2/r Ff = μFn Fnet = ma 3. The attempt at a solution Assume that up and left are positive. Fnet = ma Fg + Fn = ma 539 + Fn = 0 Fn = 539 Ff = μFn Ff = 0.16 (539) Ff = 86.24 Fnet = mv^2/r Ff = mv^2/r 86.24 = 55(6)^2/r 86.24 = 1980/r r = 23 m I'm not sure if that's the answer or what to do next, or even if what I did is the right method. The question is kind of confusing me because I don't fully understand how the person is moving. 


#2
Mar112, 12:46 AM

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#3
Mar112, 07:36 AM

P: 68

So you mean like a quarter of the circumference?
C = 2∏r C = 46∏ C = 145 C/4 = 36.25m Therefore the smallest turn the boy can make is 36.25m? 


#4
Mar112, 07:53 AM

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Centripetal Force/Changing direction on ice  Turn length
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