At Which Point Will a Sliding Particle Lose Contact with a Curved Surface?

[carllacan]

Homework Statement

A particle slides on a curved surface. It is restricted to move on a vertcial plane. The intersection of the plane with the surface is described in cartesian coordinates as y + a[cosh(x/a)+1] = 1. At which point will the particle stop having contact with the surface?
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Homework Equations

The section of the surface is described by y + a[cosh(x/a)+1] = 0, with a > 0.

The Attempt at a Solution

I know the particle will leave the surface when the centripetal force is greater to the reaction force. By deriving the equation of the surface I can find the slope of the tangent at every point. That allows me to find a relation between the projection of the gravity force along that tangent and the projection towards the inside of the surface. I think that the centripetal acceleration if equal to the acceleration "along the surface", but I'm not sure of it.
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Another solution I found is to find an expression v(x) from the conservation of energy and then derive it to find the aceleration, but this is the rate of change of the velocity modulus, so I'm not sure that it is related to the centripetal acceleration.

Help please.

 

[TSny]

The Attempt at a Solution

I know the particle will leave the surface when the centripetal force is greater to the reaction force.

I'm not sure what you mean here. "Centripetal force" is usually meant to represent the component of the net force that is directed perpendicular to the velocity vector and in the direction toward the center of curvature of the trajectory. Thus, the centripetal force would include any "reaction force" acting centripetally on the particle. Can you specify more clearly what you mean by "reaction force"? You'll need to reconsider the condition on the forces that holds when the particle leaves the surface.

By deriving the equation of the surface I can find the slope of the tangent at every point. That allows me to find a relation between the projection of the gravity force along that tangent and the projection towards the inside of the surface.

Good.

I think that the centripetal acceleration if equal to the acceleration "along the surface", but I'm not sure of it.

I don't believe there's any relation between the centripetal and tangential components of acceleration that will be directly relevant to this problem. But it is good that you are thinking about the centripetal acceleration.

Another solution I found is to find an expression v(x) from the conservation of energy and then derive it to find the aceleration, but this is the rate of change of the velocity modulus, so I'm not sure that it is related to the centripetal acceleration.

Using energy to get the speed is a good idea. Centripetal acceleration can be expressed in terms of the speed and the radius of curvature of the trajectory.

 

[carllacan]

Using energy to get the speed is a good idea. Centripetal acceleration can be expressed in terms of the speed and the radius of curvature of the trajectory.

Is there any simple way to obtain the radius of curvature? I know how to do it with vectorial calculus, but the vectors are quite complicated in this case.

 

[TSny]

Is there any simple way to obtain the radius of curvature?

Try using the first expression for R shown here .

 

[carllacan]

Try using the first expression for R shown here .

Thanky you very much, that was what I was looking for.

However I'm still stuck. What should I do now? Do I have an expression for the modulus of the component of the gravity force that acts towards the surface and the equate that to the centripetal force?

 

[TSny]

Thanky you very much, that was what I was looking for.

However I'm still stuck. What should I do now? Do I have an expression for the modulus of the component of the gravity force that acts towards the surface and the equate that to the centripetal force?

Yes, that's right. At the moment the object leaves the surface, the centripetal force is due solely to the component of the force of gravity that is perpendicular to the surface. Or, in terms of acceleration, the perpendicular component of the acceleration due to gravity is equal the centripetal acceleration.