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 2 years ago
Compare the equation for a line through the origin, y = mx, to the equation Fc=mv^2/r and explain what the slope of the graph of Fc vs. v^2 represents. Remember that the m in the equation for a line represents the slope, and the m in the centripetal force equation represents the mass of an object
 2 years ago
Compare the equation for a line through the origin, y = mx, to the equation Fc=mv^2/r and explain what the slope of the graph of Fc vs. v^2 represents. Remember that the m in the equation for a line represents the slope, and the m in the centripetal force equation represents the mass of an object

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IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0Ouch my eyes. Is this centripetal acceleration or something? Or like Centrifugal force?

erin512
 2 years ago
Best ResponseYou've already chosen the best response.0yes centripetal acceleration :)

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1In the Fc case, rewrite \[F_c=mv^2/r=v^2(m/r)\] What that says to me is because m is constant (ignoring relativistic effects) as v increases, r has to increase very rapidly to maintain a constant force because of the v^2 term in the numerator. If we aren't trying to hold the force constant, but instead hold the radius constant (as well as the mass), the force goes up quite rapidly and the shape of the curve is a parabola opening upward. If we plot F_c vs. v^2, (m/r) is just the slope of the curve. I can't say that I'm positive I've understood what the author of the question is after here...

erin512
 2 years ago
Best ResponseYou've already chosen the best response.0wow you are a genius, thank you so much for your help once again @whpalmer4

erin512
 2 years ago
Best ResponseYou've already chosen the best response.0can u possibly help me with just one more? i'll repost now, this one should be a bit clearer in its description

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1I much prefer solving physics problems to this sort of thing!
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