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Simple Projectile in Cylindrical Coordinates

  1. Sep 10, 2010 #1
    1. The problem statement, all variables and given/known data
    A gun can fire shells in any direction with the same speed [tex]v_{0}[/tex]. Ignoring air resistance and using cylindrical polar coordinates with the gun at the origin and [tex]z[/tex] measured vertically up, show that the gun can hit any object inside the surface

    [tex]z = \frac{v^{2}_{0}}{2g} - \frac{g}{2v^{2}_{0}}\rho[/tex]

    Describe this surface and comment on its dimensions.

    2. Relevant equations

    [tex]m\ddot{z} = mg[/tex]
    [tex]\ddot{\rho} = 0[/tex]
    (Angular acceleration is unimportant because the problem has radial symmetry)

    3. The attempt at a solution
    The surface is a sort of parabolic dome (circular paraboloid, I believe would be the technical term?) with its vertex at the maximum height the projectile would go if fired straight up. This is fairly obvious from the given equation.

    Just curious if I could get the solution without using cylindrical coordinates explicitly, I attacked the problem naturally by assigning an angle [tex]\theta[/tex] between the horizontal and the initial velocity of the projectile. Then the equations fall out as follows:

    [tex]v_{0}sin(\theta) = v_{z0}[/tex]

    [tex]z = \frac{v^{2}_{0}sin^{2}(\theta)}{2g}[/tex]

    [tex]\rho = v_{0}cos(\theta)t[/tex]

    [tex]t = \frac{v_{0}sin(\theta)}{g}[/tex]

    [tex]\rho = \frac{v^{2}_{0}cos(\theta)sin(\theta)}{g}[/tex]

    I then made the substitution into the equation for z.

    [tex]z = \rho^{2}(\frac{g}{2v^{2}_{0}cos^{2}(\theta)})[/tex]
    (LaTeX is not displaying this equation for me - it should read:
    z = rho^2*(g/(2v_0^2*cos^2(theta))

    Obviously, the cosine term is not in agreement with the given equation. I may be missing some kind of identity or making some kind of mistake, but I figured I'd move on to trying explicit cylindrical coordinates (since these were kind of fudged) and seeing if the problem resolved itself.

    Integrating the differential equations once we have
    [tex]\dot{z} = -gt + \dot{z_{0}}[/tex]
    [tex]\dot{\rho}= \dot{\rho_{0}}[/tex]
    Integrating the z equation once more we have
    [tex]z = -\frac{1}{2}gt^{2}+\dot{z_{0}}t[/tex]

    I noticed that we can substitute in [tex]t = \frac{\rho}{\dot{\rho_{0}}}[/tex]

    [tex]z = -\frac{1}{2}g(\frac{\rho}{\dot{\rho_{0}}})^{2}+\dot{z_{0}}(\frac{\rho}{\dot{\rho_{0}}})[/tex]

    Furthermore, we can make the substitution [tex]\dot{\rho_{0}} = \sqrt{v^{2}_{0}-\dot{z_{0}}^{2}}[/tex]

    [tex]z = \frac{\dot{z_{0}}\rho\sqrt{v^{2}_{0}-\dot{z^{2}_{0}}-\rho^{2}g}{2(v^{2}_{0}-\dot{z_{0}}^{2}})[/tex]
    (These two equations also aren't showing up. They should read

    dotted rho_0 = sqrt(v_0^2 - dotted z_0^2)

    z = (dot z_0 * rho * sqrt(v_0^2 - dot z_0^2) - rho^2 * g)/(2*(v_0^2 - dot z_0^2)


    Here is where I get stuck, because I don't see how to resolve the problem of having either
    [tex]\dot{z_{0}}[/tex] or [tex]\dot{\rho_{0}}[/tex]
    appear in the equation. Furthermore, on the whole, it seems like I've actually moved further away from the given solution rather than closer to it.
  2. jcsd
  3. Sep 10, 2010 #2


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    Didn't you think it was odd that your [itex]z[/itex] didn't depend on [itex]t[/itex]? :wink:
  4. Sep 10, 2010 #3
    I didn't, actually. Given an angle and an initial velocity, you can determine the maximum height of the parabola. The way I framed the problem was that the surface originally specified [itex](z = \frac{v^{2}_{0}}{2g} - \frac{g}{2v^{2}_{0}}\rho)[/itex]
    would be made up by the vertices of all possible parabolic arcs through which the projectile could travel. I'd been wrestling with this particular part for awhile, because it seems intuitively wrong. I suppose you are saying that I should restrict myself to just the maximum height by allowing t to vary and taking the whole arc into account? I can try that, but I won't be able to give it a shot for a couple days maybe.
  5. Sep 10, 2010 #4


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    You're not looking for the maximum heights, but rather the maximum distance from the starting point(origin) for every possible firing angle....that will give you a curve [itex]r(\theta)[/itex], that when revolved around the z-axis will give you the surface which your projectile cannot escape.
  6. Sep 10, 2010 #5
    Okay, I'll try again with that in mind and come back with my result.
  7. Sep 12, 2010 #6
    Okay, I found a solution online (http://www.physics.umd.edu/courses/Phys410/brill2005/Sol78.html) and was able to use it to guide me. After using your advise I tried to use the distance formula to specify a function s(t, theta), and then tried to maximize it on t for fixed theta. The resulting equation was not solvable for t, so I turned to this solution.

    I substituted my equation from earlier for rho into the equation for z (this time as a function of t!) and then used their method of subtracting the surface from the equation for z. Their solution gets the inequality backward though.
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