Point mass in a (non-hyperbolic) paraboloid.

In summary: Rg)In summary, the student attempted to find a solution to a problem that required knowledge of a paraboloid, but was unable to get the equation to work. They realized that their equation was incorrect and needed to use the equation for a paraboloid z=Cx2. They then found that v=\sqrt(Rg) and that the normal force in a paraboloid is not necessarily parallel to the position vector from the center.
  • #1
866
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Need a 2nd opinion on my solution.

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?
 
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  • #2
Hi Lavabug! :smile:
Lavabug said:
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]…

… Also, units are off by a factor of [L]^(1/2).

Try [itex]Cz=x^2 + y^2[/itex] :wink:
 
  • #3
Not sure I follow, is this not the proper equation for a paraboloid?
 
  • #5
I did, I suppose my z should really be z/R then, that's the only way I can get v=sqrt(Rg).

However is my physical solution correct? Something about v being independent of the angle makes me worried. Common sense tells me I'd need a greater (and eventually divergent) velocity the higher I wish to make the mass orbit higher and higher (as theta approaches 90º and the normal force starts to vanish). How does my solution account for this?
 
  • #6
i didn't notice before, your [itex]\frac{r}{z}=\tan \theta[/itex] is wrong :redface:
 
  • #7
Why?
 
  • #8
Lavabug said:
Why?

r/z is radius /height

if a point on the rim is P, and if the bottom of the cup is B, then r/z is the slope of the line BP, which goes through the cup: you want the slope of the line that touches the cup :wink:
 
  • #9
Clarify please? I take it my substitution is not right, the x's and y's in my expressions for r and z are not the same. But I don't know how I would leave this independent of anything but R, theta and g. I don't see any convenient substitution that would make it work.
 
  • #10
if z = f(x) = Cx2 is your parabola (that generates the paraboloid z = C(x2 + y2 )),

then the slope is dz/dx :wink:
 
  • #11
Why would I need the slope of this segment? I don't see what you're trying to get at. This is supposed to be a problem that should be done in under a minute...
 
  • #12
oh wait, your θ isn't defined in the usual way, it's …
Lavabug said:
The particle's position vector makes an angle theta wrt the center of symmetry (generatrix going from -z to +z).

so your tanθ = r/z is correct after all (sorry),

but your mg = Ncosθ is wrong, you need cos of the slope (tan-1 dr/dx)
 
  • #13
I should've included a picture:
http://s9.postimg.org/5vouf3fni/IMG_20131015_153139_269.jpg
I just realized that the normal force in a parabolic cup is not necessarily parallel to the position vector from the center. The angle between the normal force and position vector is 90-theta. So my trigonometry is off.
 
  • #14
how do you get 90° - θ ? :confused:

(and i was assuming the "position vector" was measured from the bottom of the cup, not from the centre of the top of the cup)
 
  • #15
Sorry, brainfart.

Have a look at the pic. The position vector starts at the top-center of the paraboloid. A paraboloid of finite length is just like an ellipsoid that was twice as long chopped in half right? All I am given is the top (max) radius R.

The angle the position vector makes wrt the z axis does not coincide with the one the normal force makes at every point, since a paraboloid is not isotropic in theta.

What I need is the angle the normal force makes wrt the vertical, in order to get it's correct projections. But since this is a paraboloid and not a semisphere, it's not the same as the position vector angle, so I'm screwed? There must be a way to do this...
 
  • #16
Lavabug said:
A paraboloid of finite length is just like an ellipsoid that was twice as long chopped in half right?

nooo!

a parabola of finite length is not an ellipse chopped in half
What I need is the angle the normal force makes wrt the vertical, in order to get it's correct projections. But since this is a paraboloid and not a semisphere, it's not the same as the position vector angle, so I'm screwed? There must be a way to do this...

yes, you can use the equation z = Cx2, and then the slope (tan of the angle) is dz/dx

(and the normal is perpendicular to that)
 
  • #17
The normal FORCE from the wall is not perpendicular to the slope you're describing, which is just the slope of the position vector yes?
 
  • #18
Lavabug said:
The normal FORCE from the wall is not perpendicular to the slope you're describing, which is just the slope of the position vector yes?

the normal force is always along the normal, ie the direction perpendicular to the surface

the slope I'm describing is the slope of the cup (the slope of the tangent plane to the cup, ie the slope of the tangent to the generating parabola)
 
  • #19
Ok, but any segment normal to that tangent line (ie the normal force) is NOT parallel to the line from the origin to the mass, in general.

It would be if it was a revolved semicircle/semisphere, but not a paraboloid/egg shaped surface, am I right?
 
  • #20
yes, you're right: the slope of the normal is not θ, the slope of the line from the origin
 

1. What is a point mass in a non-hyperbolic paraboloid?

A point mass in a non-hyperbolic paraboloid refers to a hypothetical object or particle that has a negligible size or volume, but contains all of its mass at a single point. This mass is then placed in a paraboloid shape, which is a three-dimensional geometric figure that resembles a u-shape.

2. How is the motion of a point mass in a non-hyperbolic paraboloid described?

The motion of a point mass in a non-hyperbolic paraboloid is described using the laws of classical mechanics, specifically Newton's laws of motion. These laws explain how the position, velocity, and acceleration of the point mass change over time as it moves within the paraboloid.

3. What factors affect the motion of a point mass in a non-hyperbolic paraboloid?

The motion of a point mass in a non-hyperbolic paraboloid is affected by several factors, including the initial position and velocity of the mass, the shape and orientation of the paraboloid, and any external forces acting on the mass, such as gravity or friction.

4. What is the significance of studying a point mass in a non-hyperbolic paraboloid?

Studying a point mass in a non-hyperbolic paraboloid can help scientists better understand the principles of classical mechanics and the behavior of objects in curved spaces. This knowledge can have applications in various fields, such as astrophysics, where objects often move through curved space.

5. How does the motion of a point mass in a non-hyperbolic paraboloid differ from that of a point mass in a hyperbolic paraboloid?

The motion of a point mass in a non-hyperbolic paraboloid differs from that of a point mass in a hyperbolic paraboloid in terms of the shape of the paraboloid. While a non-hyperbolic paraboloid has a u-shape, a hyperbolic paraboloid has a more complex shape resembling a saddle. This difference results in different trajectories and behaviors for the point mass in each type of paraboloid.

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