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Homework Help: Radius of Curvature

  1. Sep 6, 2010 #1
    1. The problem statement, all variables and given/known data
    If you paint a dot on the rim of a rolling wheel, the coordinates of the dot may be written as (x,y)=(R[tex]\theta[/tex]+Rsin([tex]\theta[/tex]), R+Rcos([tex]\theta[/tex]) where [tex]\theta[/tex] is measured clockwise from the top. Assume that the wheel is rolling at constant speed, which implies [tex]\theta[/tex] = [tex]\omega[/tex]t.

    a) Find v(t) and a(t) of the dot.

    b) At the instant the dot is at the top of the wheel, what is the radius of curvature of the its path? The radius of curvature is defined to be the radius of the circle that matches up with the path locally at a given point. Hint: You know v and a.

    2. Relevant equations
    x = xo + vt + .5at2

    3. The attempt at a solution
    The thing I am really stuck on is what the book means by radius of curvature. The one sentence definition is the first time it comes up in the book and it has not been mentioned in lecture yet. I think it might be half the path the dot travels on the x axis. My work is the following though, since there could be a mistake there anyway.

    PART a) Taking the derivative a couple times gives
    v = (R[tex]\omega[/tex] + R[tex]\omega[/tex]cos([tex]\omega[/tex]t), -R[tex]\omega[/tex]sin([tex]\omega[/tex]t))
    a = (-R[tex]\omega[/tex]2, -R[tex]\omega[/tex]2cos([tex]\omega[/tex]t))

    Part b) This is the part that is giving me trouble. I initially thought that the only possible radius of curvature would be the original circle. I even tried doing this experimentally by taking a water bottle and rolling it on a piece of paper and noticing how the path matches up with the circumference and so half the circumference could be the radius of curvature.

    Another possible way to do it could be to find the time that passes by seeing how long it takes the y coordinate of position to do one revolution. I think that comes out to t = 2[tex]\pi[/tex]/[tex]\omega[/tex]. When combining that time with x = xo + vt + .5at2 you get x=6[tex]\pi[/tex]R. I think that you need to take half to get when the dot is at the top, so the final answer would be 3[tex]\pi[/tex]. I just don't which one is correct, and really what is meant by radius of curvature.

    Also, I don't know how to get the tex to not look like it is superscript, but the only exponents are a couple things squared.
  2. jcsd
  3. Sep 6, 2010 #2

    I can't give you the definition of radius of curvature ;] but to find it you can use one equation that relates acceleration, speed and radius of motion. If you rewrite Newton's second law for body in circular motion you would get: ma = mv^2/R. Sure, path of the dot isn't circle, but at one instance, when it's at the top, speed and acceleration vectors are perpendicular - perfect conditions for circular motion ;D
    Also i don't know if you wrote answer to part a) as you indented to write it. You can check that when wt = pi, v_y = 0 and a_x = 0.
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