Why Does the Minimum Radius of Curvature Occur at the Apex in Parabolic Motion?

F.Turner
Messages
10
Reaction score
0

Homework Statement


A projectile in uniform gravitational field g=-g*j\hat{} with initial location x(t=0)=_{}xo, y(0)=_{}yo show explicitly that the minimum value of radius of curvature \partial of the resulting parabolic trajectory occurs at the apex and equal to (^{}vox)^2/g

Homework Equations


Dynamic vector kinematics equations



The Attempt at a Solution


I have set up two equations that i believe I will need but still I'm stuck i am not to sure if the route I'm taking is correct. Would like to know what should be the initial setup for this problem.
 
Last edited:
Physics news on Phys.org
Start by writing an equation y(x) giving the parabolic trajectory.
 
Okay, I think I'm heading in the right direction now, I am trying to evaluate the radius of curvature as a function of position from there i will attempt to...not sure yet
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top