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Question

The answer scheme tells that C is the critical point and the the inequality $Reaction>0$ should be used over C.

How is this point C decided? Why not B or midpoint of BC? Is there a way to formula a function whose stationary points will indicate that C is the critical point?

Please provide (if possible) both intuitive (using physics) explanation and mathematical explanation.

mathnoob123
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1 Answers1

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A particle in free fall follows a parabolic path. The particle losing contact from the circular track in some point between $B$ and $C$, say $D$, means two things 1) that the trajectory is now parabolic and that 2) the parabola the particle is following is tangent to the circular track(*) in $D$, thus, it will never get contact again with the track. So, you have to assume that the particle is in $C$ and right then loses contact, so is, the particle is in $C$ and the force from the track is zero there.

At $C$, the particle is still in a circular path, so, the acceleration in the direction perpendicular to the track is $v^2/R$, with $v$ the particle's speed and $R$ the track radius. By other side, the force along that direction is the difference between the component of gravitational force along that direction and the normal reaction of the track: $mg\cos40º-N$ The Newton's second law says:

$mg\cos\alpha-N=mv^2/R$ and, with the condition of $N=0$, $v_{max}=\sqrt{gR\cos\alpha}$

You can get $v$ from the energy conservation theorem and with some trigonometry to calculte the heigh $h_{max}$ at wich the particle has to be initially.

(*) I am not sure whether or not you need a proof about this "no return" of a circle tangent to a parabola. As conviction, draw a circle tangent to a parabola and see.

Rafa Budría
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  • I find the no return thing a little confusing. Shouldn't the no return thing depend on the curvature of surface that is ahead of the particle. Let's say for an example there is a hemisphere and the the particle lose contact before the middle of hemisphere. There exists many point on the hemisphere such that the parabolic path of the particle will cause it to return somewhere again on the hemisphere – mathnoob123 May 25 '17 at 18:46
  • And you also didn't clear up one thing. Why at C? Why not B or some other point? – mathnoob123 May 25 '17 at 18:48
  • Yes, the point is about the curvature of a parabola and a circle in the tangency point. The parabola has less curvature than the circle. Draw a circle tangent to a parabola and you'll see why. And all my answer was to explain why not at any point between $B$ and $C$ :( It meant to be a kind of "reductio ad absurdum": "Assume that loses contact in $D$ etc.") – Rafa Budría May 25 '17 at 18:53
  • If the particle is in $C$ is because all along the previous trip the particle never was in free fall. – Rafa Budría May 25 '17 at 19:00
  • But what about a particle that leaves the curvature at a point such that vertical component of velocity is approximately equal to total velocity and the horizontal component is approximately equal to 0?(obviously not talking about the surface used in this question) in such case the particle will fall back to the same point(or a very close one) – mathnoob123 May 25 '17 at 19:03
  • It can happen, of course (e.g. if the particle leaves the track at $C$ almost immediately it will get contact again with the straight part of the track beyond $C$). The trick of the problem is that the track is circular. Another way to see it is thinking that $C$ is the "worst point" in the sense that the slope is maximum and maximum the speed: if the particle is in $C$ "a fortiori" it was in the other points of the circular track. – Rafa Budría May 25 '17 at 19:11