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Archived Radius of curvature

  1. Sep 14, 2003 #1
    I was helping someone with the following problem...

    A projectile is launched from point A at an angle of 25 degrees relative to the horizontal, with initial velocity = 60 ft/s.
    Determine the speed of the projectile along its trajectory where its radius of curvature is 3/4 of its radius of curvature at point A.
    (Ignore air resistance.)

    We both (independently) worked out a speed = 41.6 m/s, however, the book solution has a speed of 54.5 ft/s.

    Appreciate it if someone is able to verify the book's solution.
    Last edited: Sep 14, 2003
  2. jcsd
  3. Feb 4, 2016 #2


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    In general, the radius of curvature of a path is given by ##R=\frac{|\mathbf v|^3}{|\mathbf v\times \mathbf{\dot v}|}##
    (where ##\mathbf v## is the velocity along the path and ##\mathbf{ \dot v}## is the acceleration)

    In the case of a projectile launched at Earth's surface with speed ##v_0## at an angle θ above the horizontal, (and letting time t=0 be the moment of launch) we have:
    ##\mathbf v = <v_0\cos\theta, v_0\sin\theta-gt, 0>##
    ##\mathbf{ \dot v} = <0, -g, 0>##
    (I include the third, redundant, component just because the cross product is technically only defined in three dimensions.)

    We can thus see that the radius of curvature in this case is proportional to the cube of the speed (specifically ##R=\frac{|\mathbf v|^3}{gv_0\cos\theta}##).
    This means that if the curvature becomes 3/4 of the
    initial curvature, then the speed must become ##\sqrt[3]{3/4}## of the initial speed.

    60*(3/4)1/3≈54.5, so the book is indeed correct.

    (Notice that we didn't actually need to know the exact launch angle; we only need to know that it wasn't launched straight up.)
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