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anonymous
 one year ago
The automobile has a speed of 80 ft/s and an
acceleration a of magnitude 10 ft/s2
in the
direction shown when it is at point A. Determine
the radius of curvature of the path at point A, and
the tangential component of acceleration.
(Ans. ρ = 1280 ft)
anonymous
 one year ago
The automobile has a speed of 80 ft/s and an acceleration a of magnitude 10 ft/s2 in the direction shown when it is at point A. Determine the radius of curvature of the path at point A, and the tangential component of acceleration. (Ans. ρ = 1280 ft)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[a = \frac{ v^2 }{ \rho }\] you need to use the component of a in the normal direction dw:1439641571882:dw

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0r=v^2/a*cos(60) r=6400/10*(1/2) =1280

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0as this is a physics board, i am allowed to ask this does anyone know what a radius of curvature actually means in the physical sense?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know that it means anything physically. It's taking a portion of a curve to approximate a circle, and the radius of curvature is the radius of said circle, crudely speaking.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0IrishBoy123, Im not quite sure what you're asking for; however I will do my best. In a physical problem the radius of curvature represents the radius of the circle whose circumferential path is followed during motion. In other words, if I am travelling along an arc (basically anything other than straight line and neglecting for the sake of simplicity constantly varying arcs where the ideas still apply though with more rigor attached) then this arc represents a portion of the circumference of a circle whose radius is precisely the radius of curvature. This concept is more mathematical then physical and it's, in more advanced mathematics, treated as a means to describe curves. If I have a curve with constant curvature, then the idea of a radius of curvature is simple. Just complete the arc (traverse 2 pi radians) and you have a circle whose radius is the radius of curvature of that arc.  **Disclaimer** the rigor police may object to certain phrasing, but I believe that the following is the best (to my ability) way of describing this concept in plain English.  If the arc is constantly varying then the concept of a radius of curvature takes on a calculus like definition in the sense of an instantaneous radius of curvature. Meaning at a given point on the curve, the radius of curvature describes the circle that is tangent to the curve at that point. This may engender the question, what about a straight path? Well to this I would reply, the radius of curvature is meaningless since any circle can be tangent to a line at any point whatsoever. But on a curve that is constantly varying, the idea of tangency must be treated, roughly speaking, as the "largest possible" circle which is tangent at that point. Meaning I can choose many different radii, all of which are < radius of curvature; however if I choose one infinitesimally larger, then the curve will no longer be tangent but will instead intersect the circumference of the great circle in a neighborhood (interval) about the point of tangency.  Sorry if this is difficult to grasp conceptually, but without a rigorous definition it is difficult to explain in words. Hope I was able to describe it in a way that helps.
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