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Normal force inside a hollow cone

  • #1

Homework Statement


A hollow cone is put upside-down with its symmetry axis vertical. The surface of it makes an angle of theta with the vertical direction as shown in the figure . A small puck of mass m slides without friction on the inner side of this cone and remains within a horizontal plane as it rotates about the axis of the cone.

EDIT: The question: What is the normal force exerted on the puck by the inner surface of the cone in terms of m, g, and theta?

Homework Equations




The Attempt at a Solution



1) Resolve X and Y coordinate so that the surface of the inside of the cone is x and y is perpendicular to that surface.

2) Resolve force of gravity into x and y components.

3) Fy = Fn - mgsin(theta) = 0

Fn = mgsin(theta) (WRONG)

The actual solution

1) Set X and Y coordinate normally

2) Resolve the NORMAL force

3) Fy = Nsin(theta) = mg
N = mg/sin(theta)


So I'm always used to resolving gravity into components when doing force problems, and so I did that.
However, to get the correct solution in this problem
help.png
you must resolve the normal force into components. I'm not sure why one way is wrong and the other way is correct.
 
Last edited:

Answers and Replies

  • #2
Nathanael
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A hollow cone is put upside-down with its symmetry axis vertical. The surface of it makes an angle of theta with the vertical direction as shown in the figure . A small puck of mass m slides without friction on the inner side of this cone and remains within a horizontal plane as it rotates about the axis of the cone.
What's the problem?
 
  • #3
What's the problem?
Whoops. Sorry for my incompetence.

I added the question

"What is the normal force exerted on the puck by the inner surface of the cone in terms of m, g, and theta?"
 
  • #4
rock.freak667
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From my understanding, both are the same thing.

Isn't
Resolve X and Y coordinate so that the surface of the inside of the cone is x and y is perpendicular to that surface.
the same as

Set X and Y coordinate normally
and then you resolve your forces based on your coordinate system or how you oriented the axes.
 
  • #5
Nathanael
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You can use either coordinates system, both will give the same answer as long as you know which direction the forces are pointing in the coordinate system you choose.

I think what you did was to take the normal force to be equal to the component of gravity perpendicular to the surface. If you do it this way then there will still be a component of gravity parallel to the surface and so the puck would be sliding down the cone.
 
  • #6
Nathanael
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1) Resolve X and Y coordinate so that the surface of the inside of the cone is x and y is perpendicular to that surface.

2) Resolve force of gravity into x and y components.

3) Fy = Fn - mgsin(theta) = 0

Fn = mgsin(theta) (WRONG)
In your coordinate system the equation Fy=0 would not be true.
 
  • #7
You can use either coordinates system, both will give the same answer as long as you know which direction the forces are pointing in the coordinate system you choose.

I think what you did was to take the normal force to be equal to the component of gravity perpendicular to the surface. If you do it this way then there will still be a component of gravity parallel to the surface and so the puck would be sliding down the cone.
That's exactly what I did.
You can use either coordinates system, both will give the same answer as long as you know which direction the forces are pointing in the coordinate system you choose.

I think what you did was to take the normal force to be equal to the component of gravity perpendicular to the surface. If you do it this way then there will still be a component of gravity parallel to the surface and so the puck would be sliding down the cone.

That's exactly what I did.

Fy = 0 in my coordinate system but Fx != 0 since there will be mgcos(theta).
This would mean that the mass will be sinking down into the cone and since it's stated that in the question it goes around a horizontal circle, I can't resolve the gravity because then the x component of it will be unaccounted for?
 
  • #8
Nathanael
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The "Fy=0" was the mistake of your solution. It would be simpler to use the normal coordinate system. Since the puck is going in a circle, there is a centripetal force acting on it (which comes from the normal force) which acts in the horizontal direction (the x-direction of the normal coordinate system).
In your coordinate system, part of the centripetal force will be in your y direction and therefore Fy≠0.

If you use the normal coordinate system it is simpler because the centripetal force acts along the x-direction, so you are justified in writing Fy=0
 
  • #9
The "Fy=0" was the mistake of your solution. It would be simpler to use the normal coordinate system. Since the puck is going in a circle, there is a centripetal force acting on it (which comes from the normal force) which acts in the horizontal direction (the x-direction of the normal coordinate system). In your coordinate system, this means that part of the centripetal force will be in your y direction and therefore Fy≠0.

If you use the normal coordinate system it is simpler because the centripetal force acts along the x-direction. Thus you are justified in writing Fy=0

Interesting. I see that since the centrifugal force wouldn't be in the x or y direction in my coordinate system.
Guess I'm just used to blindly resolving the force of gravity because of all those ramp questions.


Much thanks for the insights.
 
  • #10
Nathanael
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Every problem is a new one :)
 
  • #11
TSny
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Just to emphasize Nathanael's important observation: If you know the direction of acceleration of an object, then it is often a good idea to set up your coordinate system so that one of your coordinate axes is parallel to the acceleration.
 

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