Radius of curvature of a Projectile

[konichiwa2x]

Hi,
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?

 

[quasar987]

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.

 

[konichiwa2x]

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]

 

[data1217]

[tex]
v_y ^2= u_y ^2+ 2a_ys
[/tex]
Let the upward direction as positive
For the y-components and
[tex]\theta[/tex]
[tex]
0 = u^2{ \sin^2{\theta}} -2gh
[/tex]
[tex]
h = \frac {u^2 {\sin^2{\theta}}}{2g}
[/tex]

For the [tex]\frac {\theta}{2}[/tex]
[tex]h = \frac {u^2 {\sin^2{\frac {\theta}{2}}}}{2g}[/tex]

 

[andrevdh]

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].

 

[konichiwa2x]

so is the velocity vector [tex]ucos\theta + usin\theta/2[/tex] ??

 

[andrevdh]

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.

 

[konichiwa2x]

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]

 

[andrevdh]

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.

 

[konichiwa2x]

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?

 

[andrevdh]

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

[tex]u\cos(\theta)[/tex]

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]

 

[kerimai]

hi,
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.

 

[SteamKing]

The radius of curvature is defined in the attached link:
http://mathworld.wolfram.com/RadiusofCurvature.html