Deriving the Center of Mass of a Cone with Point Facing Downwards

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Homework Help Overview

The discussion revolves around deriving the center of mass of a cone with its point facing downwards, characterized by a height H and a radius R. The original poster acknowledges the conventional result of H/4 but seeks to understand the derivation process, particularly struggling with calculus concepts.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the idea of modeling the cone as a stack of disks to facilitate the calculation of the center of mass. The original poster expresses difficulty with calculus, while another participant suggests using a disk element to derive the volume and subsequently the center of mass.

Discussion Status

The discussion is active, with participants providing hints and suggestions for approaching the problem. There is an ongoing exploration of how to set up the integration for the volume of the disks and the center of mass calculation, but no consensus or final method has been established yet.

Contextual Notes

The original poster indicates a lack of confidence in their calculus skills, which may be influencing their ability to engage with the problem effectively. The discussion includes various interpretations of how to calculate the volume of the disks involved.

dowjonez
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I need to find the center of mass of a cone with point facing downwards, of height H and radius R.

Since the density is constant throughout and because of axial symmetry the center must be somewhere on the z-axis.

I know from convention that this is H/4 but i need to derive this.


Rcm = (intregral from 0 to H) of the change in radius

this is where I am stumped
i did really bad in calculus

could anyone help me?
 
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Hint: Consider the cone as a stack of disks.
 
Let Dv Be An Element In The Form Of A Disk That Cuts Through The Cone.

The Radius Of The Disk Is (r / H) X.

The Volume Equals The Area Of The Disk Times The Thickness.

Dv = Pi[(r / H ) X] ^2
Now Intergate From 0 To H

X' = Int (x Dv) / Int Dv = 3/4 H
 
okay so the biggest such disk would have volume pi*R^2*h

what is the volume of the disk under that?
 
the biggest *THIN* disk, at x = H, has radius r = xR/H,
so its Volume = dV = pi R^2 dx.

You need to integrate x from 0 to H .
 

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