Equation for Centre of Gravity of a Hollow Conical Frustum

1. Jan 10, 2014

Dave442

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Consider a solid conical frustum as shown below:

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 cant seem to determine a similar equation for a hollow conical frustum

3. 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

2. Jan 10, 2014

Saitama

Hi Dave! Welcome to PF!

You can find the CM by dividing the frustum into rings. Can you proceed?

Last edited: Jan 10, 2014
3. Jan 10, 2014

Dave442

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

4. Jan 10, 2014

Saitama

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?

5. Jan 10, 2014

Staff: Mentor

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

6. Jan 10, 2014

haruspex

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

7. Jan 13, 2014

Dave442

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.

Attached Files:

• COG of Hollow Conical Frustum.pdf
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Last edited: Jan 13, 2014
8. Jan 13, 2014

Staff: Mentor

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

9. Jan 13, 2014

Dave442

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. Jan 13, 2014

Staff: Mentor

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

11. Jan 13, 2014

Dave442

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