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Analysis of the Ground Function: f(x) with $$f''(\bar{x})=0$$
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[QUOTE="TSny, post: 6854008, member: 229090"] Yes, it's not surprising that the mass cancels out. Here's an outline of the normal force approach. The radius of curvature at a point ##\left(x, f(x)\right)## of the hill is given by $$R =\frac{ \left[1+f’(x)^2\right]^{3/2}}{|f’’(x)|}.$$ At this point, the car may be considered as moving along a circle with this radius. Then Newton’s second law gives $$mg \cos \theta – N = \frac{mv^2}{R}$$ where ##\theta## is the angle satisfying ##\tan \theta = f’(x)##. So, ##\cos \theta = \large \frac 1 {\left[1+f’(x)^2\right]^{1/2}}##. Note that ##m## will cancel when ##N## goes to zero. You have your energy expression for ##v##. Substituting these expressions for ##R, \cos \theta## and ##v## into the 2[SUP]nd[/SUP] law equation, you can get ##N## in terms of ##E_0##, ##f(x)##, ##f’(x)##, and ##f’’(x)##. Setting ##N = 0## gives the relation between ##E_0## and the point ##x## at which the car leaves the surface of the hill. [/QUOTE]
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Analysis of the Ground Function: f(x) with $$f''(\bar{x})=0$$
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