Equation for Centre of Gravity of a Hollow Conical Frustum

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SUMMARY

The discussion centers on deriving a general equation for calculating the center of gravity (COG) of a hollow conical frustum. The established equation for a solid conical frustum is COG = (h/4)[(b^2+2(ba)+3a^2)/(b^2+ba+a^2)], but participants sought a similar formula for hollow frustums. Dave successfully derived an equation and validated it against a numerical method involving hollow cylinders, achieving consistent results. Chet provided additional insights and corrections, enhancing the accuracy of the derived equation.

PREREQUISITES
  • Understanding of hollow conical frustums and their geometric properties
  • Familiarity with calculus, particularly integration techniques
  • Knowledge of center of mass calculations
  • Experience with mathematical modeling and derivation of equations
NEXT STEPS
  • Research the derivation of the center of mass for solid and hollow geometric shapes
  • Study integration techniques for calculating moments and centers of mass
  • Explore applications of the derived COG equation in engineering contexts
  • Investigate the effects of varying wall thickness on the COG of hollow frustums
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Students and professionals in engineering, physics, and mathematics who are involved in structural analysis, particularly those focusing on the mechanics of shapes and their centers of mass.

Dave442
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Homework Statement



I would like to find a general equation for calculating the centre of gravity (COG) of a hollow conical frustum.

Homework Equations



Consider a solid conical frustum as shown below:

6947853_f260.jpg


The COG of this shape may be derived as follows:

COG = (h/4)[(b^2+2(ba)+3a^2)/(b^2+ba+a^2)]

Unfortunately, I can't seem to determine a similar equation for a hollow conical frustum

The Attempt at a Solution



I have been able to find the COG for a hollow conical frustum by splitting the shape up into a large number of hollow cylinders, calculating the total moment about the base axis and then divideding by the total weight. To speed things up however, I was hoping that somebody might help me determine a general equation similar to that shown above. I searched through the previous threads, but everything I found related to a either a solid conical frustum or a hollow cone.

Kind Regards,
Dave
 
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Dave442 said:

Homework Statement



I would like to find a general equation for calculating the centre of gravity (COG) of a hollow conical frustum.

Homework Equations



Consider a solid conical frustum as shown below:

6947853_f260.jpg


The COG of this shape may be derived as follows:

COG = (h/4)[(b^2+2(ba)+3a^2)/(b^2+ba+a^2)]

Unfortunately, I can't seem to determine a similar equation for a hollow conical frustum

The Attempt at a Solution



I have been able to find the COG for a hollow conical frustum by splitting the shape up into a large number of hollow cylinders, calculating the total moment about the base axis and then divideding by the total weight. To speed things up however, I was hoping that somebody might help me determine a general equation similar to that shown above. I searched through the previous threads, but everything I found related to a either a solid conical frustum or a hollow cone.

Kind Regards,
Dave

Hi Dave! Welcome to PF! :smile:

You can find the CM by dividing the frustum into rings. Can you proceed?
 
Last edited:
Hi Pranav,

Thanks for the welcome and the response.

As I mentioned in my original post, I am able to calculate the centre of mass for a hollow conical frustum by dividing into a number of hollow rings. As this approach is quite tedious however, I was hoping that somebody might help me determine a general equation similar to that which I gave for a solid conical frustum in my original post.

Any Ideas?
Dave
 
Dave442 said:
Hi Pranav,

Thanks for the welcome and the response.

As I mentioned in my original post, I am able to calculate the centre of mass for a hollow conical frustum by dividing into a number of hollow rings. As this approach is quite tedious however, I was hoping that somebody might help me determine a general equation similar to that which I gave for a solid conical frustum in my original post.

Any Ideas?
Dave

You can use the same trick of rings to calculate the COG of hollow frustum.

Btw, is your frustum open or closed at both the ends?
 
Start out by expressing the radius r as a function of the distance z measured upward from the base. For example, r = b at z = 0, and r = a at z = h. The differential length dl along the cone is represented by dl=\sqrt{(dr)^2+(dz)^2}. Substitute the equation for r into this equation and obtain dl in terms of dz. The amount of mass between z and z + dz is equal to 2πrdl.

Chet
 
... or you can just treat the frustum as the difference of two cones. You know the CoM and mass of each cone.
 
Hi Guys,

I managed to derive a general equation for calculating the COG of a hollow conical frustum and have attached a pdf with the details. I think it is correct. If I calculate the COG of a hollow conical frustum using this equation I get the same value as when I split the frustum up into several hollow cylinders and calculate the COG by parts. If anyone spots any errors in the proof please let me know!

Cheers,
Dave

Edit: I have updated the attached pdf after Chet spotted a typo.
 

Attachments

Last edited:
Dave442 said:
Hi Guys,

I managed to derive a general equation for calculating the COG of a hollow conical frustum and have attached a pdf with the details. I think it is correct. If I calculate the COG of a hollow conical frustum using this equation I get the same value as when I split the frustum up into several hollow cylinders and calculate the COG by parts. If anyone spots any errors in the proof please let me know!

Cheers,
Dave
Something is wrong with A. The units of the three terms in parenthesis do not match. Also, I recommend using your final result to determine the cg for the case where the wall thickness is thin and constant from top to bottom. This should be very easy to obtain by the method I recommended earlier. It should provide a check on your more general result.

Chet
 
Hi Chet,

Sorry about that, there was a typo in the document I attached in my previous post. I have updated the pdf and you can see that the units of the three terms in parenthesis in A do in fact match. I think the derivation is correct.

For example, consider a hollow conical frustum with the following properties:

Height = 10 m
Outer Diameter at Base = 6 m
Outer Diamter at Top = 2 m
Inner Diameter at Base = 5.5 m
Inner Diameter at Top = 1.5 m
Uniform Wall Thickness = 0.5 m
Uniform Density = 7950 kg/m3

By splitting the frustum up into 100 hollow cylinders and calculating the COG by parts I get COG = 4.111 m above the base. By substituting the properties into the equation that I derived, I get COG = 4.115 m above the base.

What do you think?
Dave
 
  • #10
Dave442 said:
Hi Chet,

Sorry about that, there was a typo in the document I attached in my previous post. I have updated the pdf and you can see that the units of the three terms in parenthesis in A do in fact match. I think the derivation is correct.

For example, consider a hollow conical frustum with the following properties:

Height = 10 m
Outer Diameter at Base = 6 m
Outer Diamter at Top = 2 m
Inner Diameter at Base = 5.5 m
Inner Diameter at Top = 1.5 m
Uniform Wall Thickness = 0.5 m
Uniform Density = 7950 kg/m3

By splitting the frustum up into 100 hollow cylinders and calculating the COG by parts I get COG = 4.111 m above the base. By substituting the properties into the equation that I derived, I get COG = 4.115 m above the base.

What do you think?
Dave
I think, nice job.
I took your results, and manipulated them a little more for the case of constant wall thickness. Here's what I got:

If I let rb=(ri,b+ro,b)/2 and

rt=(ri,t+ro,t)/2 and

δ=(ro,b-ri,b)=(ro,t-ri,t), I get:

A = 4δ(rb+2rt)

B = 3δ(rb+rt)

Com=\frac{h(r_b+2r_t)}{3(r_b+r_t)}

This is the same result I get with my simple integration approach in one of my earlier posts.

Using this result and your data, I get Com = 4.11111m

Chet
 
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  • #11
Hi Chet,

Cheers for that, I wasn't 100% sure what you meant in your earlier post but I get it now - very helpful!

Thanks also for looking over my stuff, much appreciated!

Cheers,
Dave
 

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