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Projectile motion graphs

  1. Sep 30, 2005 #1
    A projectile is fired at an angle 30 degrees to the horizontal at a muzzle velocity of 450 m/s. Graph the radius of curvature as a function of x, and as a function of time.

    I'm pretty clueless here:

    Here's what I think:

    [tex]\vec a = (v^2/\rho)\hat {e_n} + \dot v\hat {e_t}[/tex]

    where [tex]v[/tex] is the velocity in the direction of motion for any given point in space along the parabola, [tex]\rho[/tex] is the radius of curvature

    so then from this

    [tex]a_n = v^2/\rho[/tex]

    where [tex]a_n[/tex] is the acceleration at any point in space along the parabola in the direction normal to the direction of motion

    so therefore to find a plot of [tex]\rho[/tex] in terms of x, I must find

    [tex]\rho = v(x)^2/a_n(x)^2[/tex]

    so what I need to know is how to find the [tex]v(x)[/tex] and [tex]a_n(x)[/tex]

    or am I going abouit this all wrong..thanks
     
  2. jcsd
  3. Sep 30, 2005 #2
    ok I think I may have gotten [tex]v(x)^2[/tex]

    by setting
    [tex]v_x(x) = v_ocos(30)[/tex] (1)
    and
    [tex]v_y(x) = v_osin(30)-gt[/tex] (2)

    then substituing

    [tex]t = x/{v_ocos(30)}[/tex] into (2)

    then I used

    [tex]v(x)^2 = v_x(x)^2 + v_y(x)^2[/tex]

    after simplifying, got

    [tex]v(x)^2 = v_o^2 - 2gxtan(30) - (gx/{v_ocos(30)})^2[/tex]

    so this is in parabolic form..which would make sense for the magnitude of v with respect to x..

    but if this is right..I'm still confused about how to get [tex]a_n[/tex]

    :S
     
  4. Sep 30, 2005 #3
    I actually just found a mistake on my v(x)..the last term should have a plus in front of it

    still having trouble with [tex]a_n[/tex] though
     
    Last edited: Sep 30, 2005
  5. Sep 30, 2005 #4
    ok, I think I solved it, but I can't be completely sure

    assuming my v(x) is right (with the exception that the sign before the 3rd term should be +ve and not -ve),

    would it make sense that a_n (acceleration normal to the motion) would be given by
    [tex]a_n = gcos(\theta)[/tex], where [tex]\theta[/tex] is the angle of the motion of the projectile relative to the horizontal at any x

    such that I could calculated [tex]cos(\theta)[/tex]

    by trigonometry

    [tex]cos(\theta) = v_x(x)/v(x)[/tex]

    so subbing back into the [tex]a_n[/tex] equation

    [tex]a_n=gv_x(x)/v(x)[/tex]

    and therefore I would be graphing the function
    [tex]\rho = v(x)^2/a_n[/tex]
    [tex]\rho = v(x)^2/(gv_x(x)/v(x)) = v(x)^3/(gv_x(x))[/tex]
     
  6. Sep 30, 2005 #5

    ehild

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    Homework Helper
    Gold Member

    You can be sure in your derivation.

    [tex]v(x)^2 = v_o^2 - 2gxtan(30) + (gx/{v_ocos(30)})^2[/tex]

    It is correct, you solved your problem by yourself, congratulation!!!

    ehild
     
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