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

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SUMMARY

The center of mass of a cone with its point facing downwards can be derived using calculus, specifically by considering the cone as a stack of disks. The center of mass (Rcm) is calculated using the formula X' = (∫(x Dv) / ∫Dv), which results in a value of 3/4 H. The volume of each disk is expressed as Dv = π[(r / H) X]^2, and integration is performed from 0 to H to find the total volume and center of mass. This method confirms that the center of mass is located at H/4 from the base of the cone.

PREREQUISITES
  • Understanding of calculus, particularly integration techniques
  • Familiarity with the concept of volume and area of geometric shapes
  • Knowledge of axial symmetry in three-dimensional shapes
  • Basic principles of physics related to center of mass
NEXT STEPS
  • Study integration techniques in calculus, focusing on volume calculations
  • Learn about the properties of geometric shapes, specifically cones
  • Explore the concept of center of mass in various shapes
  • Practice problems involving the derivation of center of mass for different solids
USEFUL FOR

Students of physics and mathematics, particularly those studying mechanics and calculus, as well as educators seeking to explain the concept of center of mass in geometric shapes.

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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