Projectile Motion Graphs: Radius of Curvature as a Function of X and Time

[Warr]

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

 

[Warr]

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

 

[Warr]

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

 

[Warr]

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]

 

[ehild]

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),

You can be sure in your derivation.

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

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]

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

ehild