Particle moving down a cone (Newtonian formulation)

[JaJoMarston]

Hi,
This a Classical Mechanics problem I've been trying to solve for a few days now. I cannot use Lagrangian or Hamiltonian formulation, it must be solved using classical Newtonian formulation. One must determine the equations of movement of the particle in cartesian, spherical and cylindrical coordinates, although I'm focusing on cartesian now since I have the feeling that is the hardest part:



1. Homework Statement

So we have a particle of mass m moving in a frictionless inverted cone. We're asked to find the equations of motion of the particle and the forces of constraint (in cartesian coordinates first).

When t=0, the particle is at a distance r₀ from the origin (I guess that's in spherical coordinates), it's at an angle φ(0)=0 and it moves with angular speed ω(0)=w₀ (it's not neccesarily constant throughout the movement). It also moves with an initial speed of v(0)=v₀.

Homework Equations

Second Newton law, that should be it. Also, since the particle is constrained to move inside the cone then:
tan(α)=(x²+y²)/z

We have the weight in the negative Z direction and the normal which should have 3 components.

The Attempt at a Solution

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I figured that we can look at the cone as a tilted plane.

There cannot be any net force in the normal direction of the surface, otherwise the particle would be pulled away from the cone. So from the figure I realized that:
N=mgsin(α)
Now focusing on our set of cartesian axis, the Z component of the normal force must be (by transformation of spherical to cartesian coordinates):
N_z=Ncos(α)=½mgsin(2α)

And so the total acceleration in the z direction must be:
a_z=g(½sin(2α)-1)

This is pretty easy to integrate using the initial conditions so we can get the speed and position in the Z component.

My problem is with the x and y components. Common sense tells me that the net force in x,y should be zero, so there must be a force that balances the normal in those directions. I thought about the centripetal force, but I don't have the slightlest clue of how to express such force. I know that if the particle was moving in a circle of a fixed radius R the centripetal acceleration would be v²/R, but this is not the case at all. We have an angular velocity that might be varying with time, a radius that is not constant and same goes with the speed.

Thanks in advance.

 

[vela]

My problem is with the x and y components. Common sense tells me that the net force in x,y should be zero, so there must be a force that balances the normal in those directions. I thought about the centripetal force, but I don't have the slightlest clue of how to express such force.

Common sense has led you astray. If the net force in the x- and y- directions were zero, there'd be no acceleration in those directions, so the particle would move in a straight line in those directions.

You also seem to have the misconception that centripetal force is another force acting on the particle. There are only two forces acting on the particle, and you've already identified them. (The direction of the normal force as drawn in your figure, however, is wrong.)

 

[JaJoMarston]

Common sense has led you astray. If the net force in the x- and y- directions were zero, there'd be no acceleration in those directions, so the particle would move in a straight line in those directions.

You also seem to have the misconception that centripetal force is another force acting on the particle. There are only two forces acting on the particle, and you've already identified them. (The direction of the normal force as drawn in your figure, however, is wrong.)

Oh I see. So in the X direction we only have the X component of the normal, which should be:
a_x=Nsinαcos(φ) where φ is a function of time.
And in the Y component:
a_y=Nsin(α)sin(φ)

My problem is with the normal now then, I made the assumption that the normal has the direction shown in the figure so I can relate it to a component of the weight.

Thanks a lot.

 

[vela]

Consider the extreme case where $\alpha = 90^\circ$. You'd expect the normal to point straight up, so the x- and y- components should equal 0. Do your expressions give you that result?

The normal force has its name because it's perpendicular to the surface the object is on. The geometry of the problem dictates what direction it points. Because the particle accelerates in all three directions, there's no simple relationship between N and mg.

 

[JaJoMarston]

Consider the extreme case where $\alpha = 90^\circ$. You'd expect the normal to point straight up, so the x- and y- components should equal 0. Do your expressions give you that result?

The normal force has its name because it's perpendicular to the surface the object is on. The geometry of the problem dictates what direction it points. Because the particle accelerates in all three directions, there's no simple relationship between N and mg.

No they don't, but I don't see the flaw either. I would expect to be able to express N in terms of something, not just leave it there. I feel like the problem is too general to be solved by Newton laws.