How do I get the Centre of Mass of a cone(standard,right angled)?
I know it must lie on the vertical axis(line joining the vertex and the centre of the base). What element must I take for simplest derivation and then integrate the expression?
Please help!
Thank you.
No Post - my answer was wrong
Solid Cone, hollow cone, center of mass available, not?
Axis through apex, next where weights (volumes) are equal along apex to base.
Hwww... say equal volumes.
vol[cone from apex to center of gravity (a-cog)] minus vol[truncated cone base (
to (cog)] equals zero, or say a delta vol.
Upphs! Curve thrown, center of gravity is throwing into air, spinning, watching the, say, spinning wooden cone, maybe "dumbells", shell or solid, annular.
Would the wooden cone float with the apex up, base down, at some base to height ratio?
When base to height ratio is large, that's a disk, narrow a stick.
So I work on the unstable zone between cone floating on base, or cone on it's side.
This is a job for my notebook, some calculus to integrate volumes from areas incrementally changing, second order, ... zinging through mathematical infinity.
Hope there is a 1/3 to this, makes three dimensional Classic Greek X-ahedrons get the triplet quark/gluons matched to the unitary valence electrons similar three dimensional shapes, that at the level of electron bleachers and pea sized nucleous is located in the center of the playing field.
Got some ideas about matching a nine by nine alpha-numerical pen to paper game to distribute the "numerals" [1, 2, 3, 4, 5, 6, 7, 8, 9] with no repititions for each of nine three by three group. So my wife makes a notation on the left side of a cell for possible connection to other groups, adjoining.
So is this cylinderical, no, since there are edge cases, they have in flat plane only one adjoing.
Back later, best rmuldavin
Axis through apex, next where weights (volumes) are equal along apex to base.
Hwww... say equal volumes.
vol[cone from apex to center of gravity (a-cog)] minus vol[truncated cone base (
to (cog)] equals zero, or say a delta vol.Upphs! Curve thrown, center of gravity is throwing into air, spinning, watching the, say, spinning wooden cone, maybe "dumbells", shell or solid, annular.
Would the wooden cone float with the apex up, base down, at some base to height ratio?
When base to height ratio is large, that's a disk, narrow a stick.
So I work on the unstable zone between cone floating on base, or cone on it's side.
This is a job for my notebook, some calculus to integrate volumes from areas incrementally changing, second order, ... zinging through mathematical infinity.
Hope there is a 1/3 to this, makes three dimensional Classic Greek X-ahedrons get the triplet quark/gluons matched to the unitary valence electrons similar three dimensional shapes, that at the level of electron bleachers and pea sized nucleous is located in the center of the playing field.
Got some ideas about matching a nine by nine alpha-numerical pen to paper game to distribute the "numerals" [1, 2, 3, 4, 5, 6, 7, 8, 9] with no repititions for each of nine three by three group. So my wife makes a notation on the left side of a cell for possible connection to other groups, adjoining.
So is this cylinderical, no, since there are edge cases, they have in flat plane only one adjoing.
Back later, best rmuldavin
Notebook drawing of a thin sheet covered cone, say joined by first "welding" square cells.
Or skewing them into diamond shapes so that the diagonal is equal to the sides.
This sixty degree skewing if rolled may match edge of original flat plane.
Thus one can get those curley drawing displayed by Zenith. Except they are made of triplets tiles joined by the 1/3 charge vertices.
For a solid cone, a pie shaped base that moves along the central axis, say a piece of tort cake.
Rum soaked Tort Cake
What I don't use, I'll power a Brazillian automobile.
Best, rmuldavin
Or skewing them into diamond shapes so that the diagonal is equal to the sides.
This sixty degree skewing if rolled may match edge of original flat plane.
Thus one can get those curley drawing displayed by Zenith. Except they are made of triplets tiles joined by the 1/3 charge vertices.
For a solid cone, a pie shaped base that moves along the central axis, say a piece of tort cake.
Rum soaked Tort Cake
What I don't use, I'll power a Brazillian automobile.
Best, rmuldavin
I would assume you are talking about a solid cone which has volume errr... 1/3*base area*height. Oh I can't be bothered trying to remember how to derive it :
http://www.mathsrevision.net/alevel/pages.php?page=78
quote : The centre of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base
somewhere in there they may explain how they arrive at this.
hope this helps.
http://www.mathsrevision.net/alevel/pages.php?page=78
quote : The centre of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base
somewhere in there they may explain how they arrive at this.
hope this helps.
Hello everybody
Moseley is right. I am talking about a 'normal very obvious cone' with volume = 1/3*pi*height*r^2(where,r = radius of base of the cone). Thax 4 d link Moseley but I still don't know how to derive it. Help please!!!
Moseley is right. I am talking about a 'normal very obvious cone' with volume = 1/3*pi*height*r^2(where,r = radius of base of the cone). Thax 4 d link Moseley but I still don't know how to derive it. Help please!!!
I can't remember how these things are done but essentially you need to sum the mass of the cone from either end until you reach a point where you have accounted for half of the mass.
As far as I can recall you will need to create a volume of revolution of a diagonal line to form a mathematical cone. This can then be integrated with respect to the height or the radius or both maybe. Sorry to be so vague.
In fact these guys seem to know what they are talking about.
http://www.webcalc.net/forum/viewtopic.php?t=166
Good luck.
As far as I can recall you will need to create a volume of revolution of a diagonal line to form a mathematical cone. This can then be integrated with respect to the height or the radius or both maybe. Sorry to be so vague.
In fact these guys seem to know what they are talking about.
http://www.webcalc.net/forum/viewtopic.php?t=166
Good luck.
Break it down into a triangle rotated around one of the non-hypotenuse sides. Call the length of that side H. Then what you want is the distance along that axis from the base (the other non-hypotenuse side) to the point where the area of the triangle above that height is equal to the area from the base to that height. Basic geometry gives us (1- sqrt(2)/2)* H as the height. So for a cone that is obtained from rotating that triangle around the side of length H, the center of mass is on the axis from the vertex to the center of the base, at a distance (1 - sqrt(2)/2) times the height of the cone.
The distance from the base, that is. The analysis assumes a constant density. If not, then you must basically integrate the density thru the volume of the cone to identify the center of mass.