Calculate the center of mass of a deformed hollow cone trunk

[Norashii]

Homework Statement

Let a hollow trunk cone as in the figure with density $\mu$, upper $r$, lower radius $R$, height $h$ and one of it's bottom parts pushed to a height $b$ but the upper part is not deformed. Calculate the position of it's center of mass.

Relevant Equations

$r_{cm} =\frac{1}{M} \int \int \int r dm$

I couldn't make progress in this problem, I would appreciate some suggestion on how could I attack this problem.

Thanks in advance!

 

[hutchphd]

Can you find the center of mass of a regular complete cone using the 3D integral? Explicitly what does that look like.

 

[Norashii]

Can you find the center of mass of a regular complete cone using the 3D integral? Explicitly what does that look like.

Yes, for a regular complete cone you can use the symmetry and $x_{cm}=y_{cm}=0$, for the $z$ coordinate you can just use cylindrical coordinates $z_{cm}=\frac{1}{M}\int_{0}^{2\pi}d\theta\int_{0}^{h}\int_{0}^{z\frac{R}{h}}z \rho(z) \mu d\rho dz = \frac{2\pi}{M}\int_{0}^{h}\int_{0}^{z\frac{R}{h}}z \rho(z) \mu d\rho dz = \frac{2\pi}{M}\cdot \frac{R^2}{2h^2}\cdot\frac{h^4}{4}\cdot \frac{3M}{\pi R^2 h} = \frac{3}{4}h$

For the hollow cone : $z_{cm} = \frac{h}{2\pi R \mu \int_{0}^{h}zdz}\cdot \frac{2\pi \mu R}{h}\int_{0}^{h}z^2dz =\frac{2}{h^2}\cdot \frac{h^3}{3} = \frac{2}{3}h$

 

[pasmith]

Are you told how the base of the deformed cone varies with \theta?

 

[Norashii]

Are you told how the base of the deformed cone varies with \theta?

Linearly, but there is no explicit expression given.

 

[hutchphd]

So you need an equation that relates ($z,\theta,\rho$) to define the bottom plane (this will go in the lower limit).
And the upper limit needs to be truncated.
You can also start from a perfect cone and add "negative mass" pieces for the cut out sections. Same difference.

 

[haruspex]

one of it's bottom parts pushed to a height $b$

That's rather unclear.
If it is "pushed up", as though the whole thing is elastic, the angle of the right side will change. E.g. with the origin at bottom left, a point in the cone that was at (x,y,z) is now at (x,y,z') where $z'=z+(h-z)\frac{xb}{2Rh}$.
But is that the only way it might be deformed, consistent with the given description?
And what happens to the density? Is the mass more concentrated in some areas now, as would happen in reality, or is it merely the shape that is changed?
Or does it really mean that part of the base is removed with an angled slice, so no actual deformation?

 

[Norashii]

That's rather unclear.
If it is "pushed up", as though the whole thing is elastic, the angle of the right side will change. E.g. with the origin at bottom left, a point in the cone that was at (x,y,z) is now at (x,y,z') where $z'=z+(h-z)\frac{xb}{2Rh}$.
But is that the only way it might be deformed, consistent with the given description?
And what happens to the density? Is the mass more concentrated in some areas now, as would happen in reality, or is it merely the shape that is changed?
Or does it really mean that part of the base is removed with an angled slice, so no actual deformation?

I'm really sorry for the imprecision, actually you can consider $b << h$ such that the density does not change significantly and can be considered approximately constant.

 

[haruspex]

I'm really sorry for the imprecision, actually you can consider $b << h$ such that the density does not change significantly and can be considered approximately constant.

So are you confirming the mapping of z that I specified in post #7?

Also, merely adding b<<h does not allow us to ignore density change. We would have to do the full analysis, then check whether letting b tend to zero makes the effect of a density change go to zero faster than some other terms. My suspicion is that it would not.

 

[Norashii]

So are you confirming the mapping of z that I specified in post #7?

Also, merely adding b<<h does not allow us to ignore density change. We would have to do the full analysis, then check whether letting b tend to zero makes the effect of a density change go to zero faster than some other terms. My suspicion is that it would not.

Yes, the mapping seems right and about the density, thinking a bit more I believe that it has to change even for a small $b$, because if it stayed constant the $CM$ wouldn't be displaced in the $x-y$ plane which is something that should happen.

 

[haruspex]

about the density, thinking a bit more I believe that it has to change even for a small b, because if it stayed constant the CM wouldn't be displaced in the x−y plane

Isn't it the other way around? If we just deform the shape, but keep the density constant then the CM will shift in the -x direction, whereas if we consider it as mass particles dm shifting vertically then the CM only shifts vertically.

 

[Norashii]

Isn't it the other way around? If we just deform the shape, but keep the density constant then the CM will shift in the -x direction, whereas if we consider it as mass particles dm shifting vertically then the CM only shifts vertically.

Yes, but the CM shifts only vertically if you keep the mass dm constant and just change it's height. However if you deform and the deformed part becomes denser then the CM should move to the denser region, it is like a compression am I wrong?

 

[haruspex]

Yes, but the CM shifts only vertically if you keep the mass dm constant and just change it's height.

That's the same as I wrote in post #11, but that is the case where the density increases. You have pushed the same total mass into a smaller surface area.

If you deform the shape but the density stays the same you have lost mass on the shrunken side.

 

[hutchphd]

Homework Statement:: Let a hollow trunk cone as in the figure with density $\mu$, upper $r$, lower radius $R$, height $h$ and one of it's bottom parts pushed to a height $b$ but the upper part is not deformed. Calculate the position of it's center of mass.
Relevant Equations:: $r_{cm} =\frac{1}{M} \int \int \int r dm$

I couldn't make progress in this problem, I would appreciate some suggestion on how could I attack this problem.

On revisiting this I realize I have no idea what a "hollow trunk cone" means.

\

 

[Norashii]

On revisiting this I realize I have no idea what a "hollow trunk cone" means.

\

It means that it is a cone trunk without thickness

 

[haruspex]

It means that it is a cone trunk without thickness

And a "cone trunk" is what I would call a "frustrated cone" or "frustum".
The one in the question is not just frustrated but quite bent out of shape.

 

[hutchphd]

Oh its an ice cream cone...I see. A soluble form for the problem would involve planar cuts at the top and bottom...no crumpling please.