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Suppose I have a smooth parabolic surface on the xy plane, $y=-x^2$, and there is a ball on kept on the surface at $(0,0)$. What minimum velocity should I give to the ball, tangentially, so that it loses contact with the surface the next instant? [Assuming the surface and the ball are kept normally on the Earth.]

I thought of approximating the parabola with a circle as then we can use $\frac{mv^2}{R} > mg$, by taking $R=\frac12$ as the radius of curvature. However, I am not sure that this is correct, because for example, we could have taken the approximation to be the tangent line, which would make $v$ infinite.

So my question is, can we use the radius of curvature here? If yes, then why does it give the exact answer? And if no, how should I solve this?

Added: If we take the parabola as a circle locally, then it would have a radius of $R$, and the forces on the ball would be $mg$ downward and $N$ (normal) upward. Now, $v$ the tangential velocity must be sufficient so that the resultant force is greater than $\frac{mv^2}{R}$, so as to ensure the ball does not continue on a circular path the next instant. Thus we have $mg - N < \frac{mv^2}{R}$. Thus, $mg < \frac{mv^2}{R} $, even if normal were to become zero.

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  • $\begingroup$ May you explain by which method, you reached to $\frac{mv^2}{R} \geq mg$? Add it (calculations) to your question. I think it will be better! $\endgroup$ – lucas Jul 9 '16 at 14:05
  • $\begingroup$ @lucas I have added my logic :) $\endgroup$ – Sawarnik Jul 9 '16 at 15:39
  • $\begingroup$ Your logic is true and it is not an approximation! Because the equation you have used is valid for each instance and R is the radius of curvature at that instance. (As you know at each point of the path there is a circle tangent to the path and you can assume that the particle is moving on this circle at that instance) $\endgroup$ – lucas Jul 9 '16 at 15:45
  • $\begingroup$ @lucas But the radius would changing at every instant? And by using $\frac{mv^2}{R}$ we are assuming its travelling in a circle of constant radius? $\endgroup$ – Sawarnik Jul 9 '16 at 15:59
  • $\begingroup$ I said the equation is valid for any instance and loosing contact occurs at one of those any instance. $\endgroup$ – lucas Jul 9 '16 at 16:02
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I think you intended that the velocity you want to apply to the ball is horizontal(direction of the x-axes). In this case, the ball have to travel in a parabolic trajectory with the coefficient of x^2 between -1 and 0(in this way this trajectory is above the parabolic surface y=-x^2). The equation of the parabolic trajectory of a ball falling down are: (v is the velocity, t is time)

y=-1/2*g*t^2

enter image description here

combining these equations you can obtain:

y= -1/2 *g *1/v^2 *x^2

So the solution came from:
[![enter image description here][4]][4]

Finnally you can solve it and find that if:

v > (g/2)^(1/2)
the ball loses the contact.

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  • $\begingroup$ Thank you! This was the easiest method and yet I didn't think of it. However, I would still like to clarify my doubt on using the radius of curvature, which that is the essence of the question. $\endgroup$ – Sawarnik Jul 9 '16 at 16:01
  • $\begingroup$ You are right, I have forgotten to answer this point: $\endgroup$ – blu potatos Jul 12 '16 at 11:32
  • $\begingroup$ Th approximation you used is valid. The circle with radius 1/2 is the circle that stay exactly under the parabola. (y+1/2)^2+x^2=1/4 is the circle equation. So if you use mv^2/R>mg you can find the same solution of the answer I've posted some time ago. $\endgroup$ – blu potatos Jul 12 '16 at 11:37
  • $\begingroup$ s3.amazonaws.com/grapher/exports/cwtwqvxkrs.png This is the graphic that show that your approximation is valid. It also quite easy using a bit of algebra to show that only the circle with radius 1/2 can be used for this approximation! $\endgroup$ – blu potatos Jul 12 '16 at 12:03

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