Center of mass of a cone

  • #1
1. A right circulat cone of constant density (5kg per m^3) is 4 meters from the base to the tip. the diameter of the base is 6 m. find each of the following using inergrals
A) Find the volume and mass of the cone
B) find the center of mass of the cone
C) find the moment of inertia of the cone when rotated about the central axis
d) find the moment of inertia of the cone when rotated about the tip
 

Answers and Replies

  • #2
LowlyPion
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Welcome to PF.

How would you think to go about setting up the integrals?
 
  • #3
I cant figure out what equation to integrate i think for the volume the limits would be from 0-6 but i am not too sure
 
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  • #4
LowlyPion
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Maybe this will help?
http://www.mph.net/coelsner/JSP_applets/cone_ex.htm [Broken]
 
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  • #5
LowlyPion
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You will still need to balance the weight along the height axis.
 
  • #6
i cant figure out how to derive the moment of inertia about the center of the cone
 
  • #7
LowlyPion
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Figure that

[tex]X =\frac{ \int x*m dm}{M} [/tex]

Now the mass at any point is a disk - our dm which is πr²*dx times density that we can just set to 1 because it is uniform.

Observing that the radius at any x can be given by r = R*(1-x/H)

where R is the radius of the base and H is the Height of the cone ... then combine and you get a definite integral in x from 0 to H
 
  • #8
what is m and what is M and is this for C.O.M or the moment of inertia
 
  • #10
LowlyPion
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Forgot to answer your question. M is the Total Mass in the system - in this case the Mass of the cone, and m is incremental mass elements.

Hence the term πr²*dx to replace the m*dm. Because with uniform mass distribution, you can model your incremental m along x as tiny disks of radius r, so you just use the area of that slice πr².

Now r as it turns out is also a function of x along x. You should be able to satisfy yourself that the (1 - x/H) is the relationship that r has with increasing x, as you move from the base along the x axis toward the pointy top.

The rest is substitution and evaluating the integral between 0 and H, and I wouldn't want to spoil your fun at arriving at the answer all together.
 

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