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Radius of curvature of a Projectile

  1. Aug 30, 2006 #1
    A particle is projeted with a velocity 'u' at an angle [tex]\theta[/tex] with the horizontal. Find the radius of curvature of the parabola traced out by the particle at the point where the velocity makes an angle [tex]\theta/2[/tex] with the horizontal.
    I got [tex]u^2[/tex]/gcos[tex]\theta/2[/tex] but my book gives a different answer.
    Can someone please help me?
  2. jcsd
  3. Aug 30, 2006 #2


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    Well, since you do not claim to have difficulty with the concept of radius of curvature or anything else, I sugest you post your work so we can see where you've gone wrong.
  4. Aug 30, 2006 #3
    radius of curvature = velocity^2/acceleration acting normal to the path towards the centre

    when the angle the velocity makes with the horizontal becomes [tex]\theta/2 , v = ucos\theta/2 + u sin\theta/2[/tex]

    resul v = [tex]u^2[/tex]
    resolving 'g' along the perpendicular towards the centre, I got gcos[tex]\theta/2[/tex]

    so [tex] Radius of c = \frac{u^2}{gcos\theta/2}[/tex]
    Last edited: Aug 30, 2006
  5. Aug 31, 2006 #4
    v_y ^2= u_y ^2+ 2a_ys
    Let the upward direction as positive
    For the y-components and
    0 = u^2{ \sin^2{\theta}} -2gh
    h = \frac {u^2 {\sin^2{\theta}}}{2g}

    For the [tex]\frac {\theta}{2}[/tex]
    [tex]h = \frac {u^2 {\sin^2{\frac {\theta}{2}}}}{2g}[/tex]
    Last edited: Aug 31, 2006
  6. Aug 31, 2006 #5


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    The velocity should include both the x- and the y-velocity components. The normal acceleration is fine since the velocity is tangential to the path of the projectile. The x-component of the velocity will be [tex]u\cos(\theta)[/tex], but the y-component changes, since the projectile is accelerating in the y direction. The problem here is that the angle that the projectile makes with the origin is not [tex]\theta /2[/tex].
    Last edited: Sep 1, 2006
  7. Aug 31, 2006 #6
    so is the velocity vector [tex]ucos\theta + usin\theta/2[/tex] ??
  8. Sep 1, 2006 #7


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    For the y-component of the velocity at the point where its angle with the horizontal is [tex]\theta / 2[/tex] you can use the fact that

    [tex]\tan(\theta /2) = v_y/v_x[/tex]

    This comes for the fact that the velocity vector at any stage in the motion is the resultant of the x-velocity component and y-velocity component of the motion. The angle theta/2 is the angle that the velocity vector is making at that stage with its x-velocity component.

    To be mathematically correct you should rather write you equation in vector notation

    [tex]\vec{v} = \hat{i}v_x + \hat{j}v_y[/tex]

    where the hat vectors are unit vectors in the direction of the x- and y-axes. Just writing a plus between the two components is misleading.
    Last edited: Sep 1, 2006
  9. Sep 1, 2006 #8
    yup i did all that first itself

    [tex]tan\theta/2 = \frac{usin\theta - gt}{ucos\theta}[/tex]

    so do i do this?
    [tex]v = (ucos\theta) + (usin\theta - gt)[/tex]

    so do i take the magnitude of this and divide it by [tex]gcos\theta/2[/tex]
    Last edited: Sep 1, 2006
  10. Sep 1, 2006 #9


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    Close, but you do not know the time. From the first equation in my previous post we have that

    [tex]v_y = v_x \tan(\theta / 2)[/tex]

    which gives us the unknown y-velocity component in terms of the known x-velocity component.
    Last edited: Sep 1, 2006
  11. Sep 2, 2006 #10
    ok thanks

    so [tex] v = utan\theta/2j + ui [/tex]

    [tex] v^2 = u^2tan^2\theta/2 + u^2[/tex]
    [tex] v^2 = u^2sec^2\theta/2[/tex]

    [tex] radius of curvature = \frac{u^2sec^2\theta/2}{gcos\theta/2}[/tex]

    [tex] = \frac{u^2}{gcos^3\theta/2}[/tex]

    is it correct now?
  12. Sep 2, 2006 #11


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    The x-component of the velocity stays the same during all of the motion of the projectile and is given by


    which means that accoding to the formula in my presvious post that the y-component of the velocity at the required position will be

    [tex]u\cos(\theta)\tan(\theta /2)[/tex]
  13. Jul 5, 2011 #12
    could you please help me,how to write an algorithm to calculate the radius is measured in steps of 3 degree of projected objects (irregular shaped objects in an image), then the algorithm will be converted to Matlab code.
  14. Jul 5, 2011 #13


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