- #1

Lavabug

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Need a 2nd opinion on my solution.

A point mass moves frictionlessly in a circle inside a parabolic cup, with the radius at the top being R. The particle's position vector makes an angle theta wrt the center of symmetry (generatrix going from -z to +z).

F=ma

Confined to a certain height implies [tex]mg = N\cos \theta[/tex]

Circular motion at an arbitrary radius r<R gives

[tex]m\frac{v^2}{r}=N\sin\theta[/tex]

which yields

[tex]\frac{v^2}{rg} = tan \theta[/tex]

In my attempt to leave this as a function only of the given parameters (R,g and theta), I use the equation for a paraboloid:

[tex]z=x^2 + y^2[/tex]

and the fact that

[tex]r=sqrt(z)\\

\frac{r}{z}=\tan \theta

[/tex]

to get

[tex]v=\sqrt(g)[/tex]

This doesn't make sense to me on physical grounds. My argument: the speed of the particle in a paraboloid should depend on the angle to enable the requirement that at higher values of r<R, a greater y-component of the normal force is required to remain at that height. Also, units are off by a factor of [L]^(1/2).

Just what am I doing wrong?

## Homework Statement

A point mass moves frictionlessly in a circle inside a parabolic cup, with the radius at the top being R. The particle's position vector makes an angle theta wrt the center of symmetry (generatrix going from -z to +z).

## Homework Equations

F=ma

## The Attempt at a Solution

Confined to a certain height implies [tex]mg = N\cos \theta[/tex]

Circular motion at an arbitrary radius r<R gives

[tex]m\frac{v^2}{r}=N\sin\theta[/tex]

which yields

[tex]\frac{v^2}{rg} = tan \theta[/tex]

In my attempt to leave this as a function only of the given parameters (R,g and theta), I use the equation for a paraboloid:

[tex]z=x^2 + y^2[/tex]

and the fact that

[tex]r=sqrt(z)\\

\frac{r}{z}=\tan \theta

[/tex]

to get

[tex]v=\sqrt(g)[/tex]

This doesn't make sense to me on physical grounds. My argument: the speed of the particle in a paraboloid should depend on the angle to enable the requirement that at higher values of r<R, a greater y-component of the normal force is required to remain at that height. Also, units are off by a factor of [L]^(1/2).

Just what am I doing wrong?

Last edited: