A particle is fired into the air with an initial velocity of 60m/s at an angle of 54 degrees from the ground. At time t=5.443, what is the radius of curvature of the path travelled by the particle? I started by coming up with a vector equation for the path travelled by the particle, using the point of launch as the reference point. [tex] \overrightarrow r = (60t\cos 54)\widehat{\underline i } + (60t\sin 54 - 0.5g^2 )\widehat{\underline j } [/tex] The way i thought to approach this problem was to consider a circle which closely approximates the curve at the given instant. For an infinitesimal change in time dt, the velocity along the path is given by: [tex] v = r\frac{{d\theta }}{{dt}} [/tex] I thought that perhaps if i could calculate the speed along the path v and the angular velocity, then i could use those to calculate the radius of curvature at that instant. I am able to calculate the speed easily, but not so sure about the angular velocity. Am i on the right track, or should i be taking a different approach? Thanks in advance, Dan.
Looks okay to me. You don't need speed or angular velocity: curvature is a purely "geometric" property. There are formulas for finding curvature- do you know any? If not what you are looking for is a circle passing through the same point, with the same tangent (first derivative) and same second derivative as this curve. Finding that will give you the radius of curvature and then the curvature.
I managed to come up with [tex] \rho = \frac{{v^3 }}{{v_x g}} = \frac{{(v_x ^2 + v_y ^2 )^{3/2} }}{{v_x g}} [/tex] where v_{x/y} is the initial velocity in the corrosponding direction. Does that look on track for the radius of curvature?
Well, the orginal formula you gave for the trajectory didn't have v_{x} or v_{y} so I don't know. However, one other thing you should think about- did you write the original formula correctly? I didn't notice when I replied before- I was assuming the trajectory must be a parabola- but what you wrote was [tex]\overrightarrow r = (60t\cos 54)\widehat{\underline i } + (60t\sin 54 - 0.5g^2 )\widehat{\underline j }[/tex] That's linear in t: that trajectory is a straight line which has curvature 0 at all points! I assume that "-0.5 g^{2}" was supposed to be "-0.5g t^{2}".
Oops yep youre right. I missed a 't' there. It should be a parabola. Anyway heres my final equation and evaluation for the radius of curvature: [tex] \rho = \frac{{((v_0 \cos \theta )^2 + (v_o \sin \theta - gt)^2 )^{3/2} }}{{gv_0 \cos \theta }} = \frac{{[(60\cos 54)^2 + (60\sin 54 - 9.81(5.443))^2 ]^{3/2} }}{{(9.81)(60\cos 54)}} \approx 130.4m [/tex] Does that look about right?