Consider a cylindrical log of radius $R$ which is at rest and a grasshopper is taking a jump from one side of log . Consider the case when the grasshopper is able to jump over it in the situation where :
It just touches the top most part path is symmetrical wrt to circular cross section . 
My question is how do we show that radius of curvature at the time of reaching the highest point is $\geq$ R ? And equality occurs at the time when it reaches top with minimum speed ?
- My main reason for asking is that for the case of touching one i was trying to get the speed at the top , which was given to be √gR. But was not able to show that radius of curvature is exactly same as that of radius of circukar section.
I think it might be related to proving that there is only one circle which can be tangent a parabola at a given point such that all other points of parabola are external to it ? How do we prove this ?
If $a_\perp$ is the projection of the acceleration of the body on a direction perpendicular to its velocity $v$, then we have the kinematical relation: $$ a_\perp={v^2\over r}, $$ where $r$ is the radius of curvature of the trajectory at the point where the body is located.
At the highest point we have $a_\perp=g$, and $v=\sqrt{gR}$ was given: inserting these into the previous equation we thus get: $$ g={gR\over r}, \quad\text{that is:}\quad r=R. $$