# Center of Mass in 3D object?

How would I find the center of mass in a 3D object (a "spinning top" or "dreidel") that consists of a cylinder welded on top of a box welded on top of an upside down cone? Assume building material is all the same type

Here is a wireframe render of the object:

And here are the dimensions as created:

Top Cylinder: Radius=12, Height=50

Middle Box: Width=70, Height=56, Depth=70

Bottom Cone (inverted after creation): Radius: 35, Height: 28

-
Just integrate over the three parts. What seems to be the problem? – Troy Woo Aug 31 '14 at 9:22
lack of basic math skills, like knowing what "integrate" means :) – davidkomer Aug 31 '14 at 9:22
Ok, at least now we know you want someone to show the details of the calculation. Im sure it will come up, just wait. Or, did you know any CAD software can do this for you? (if only numericals are involved) – Troy Woo Aug 31 '14 at 9:24
Is the design made from solid material, or is it a sheet/plate material (which would mean that the inside is hollow). If the latter, is the material double thickness where the cone and the cylinder are touching the cube? – bubba Aug 31 '14 at 9:39
Why the downvote ??? – bubba Aug 31 '14 at 9:49

Use this facts:

• COM of Cylinder and Cube are at body centre.
• COM of Cone at height $h/4$ from base.

Now do weighted AM of coordinates of COM three figures with volume as their weights.

Let's do Calculation. Simple symmetry says COM will be along the line perpendicular to ground through the point from the cone touching ground. So let's do single calculation: $$\frac{((\pi\times12^2\times50)\times(56+28+25))+((70\times56\times70)\times(28+28))+((\frac13\times\pi\times35^2\times28)\times(\frac34\times28))}{(\pi\times12^2\times50)+(70\times56\times70)+(\frac13\times\pi\times35^2\times28)}\approx 55.82$$ So height above ground must be nearly $55\sim56$ .

-
AM = arithmetic mean – mistermarko Aug 31 '14 at 9:44
And the cone formula is correct only if it's solid. – bubba Aug 31 '14 at 9:47
'Volume' here must refer to the volume of solid material. – mistermarko Aug 31 '14 at 9:52
Volume will be the weight if the density of the three parts are the same and constant (which I now realize that the OP has already specified it). – Dmoreno Aug 31 '14 at 10:18
@Dmoreno If you knew much, for bodies with uniform density their mass is proportional to volume, and the proportionality factor cancels. – ADG Aug 31 '14 at 10:20

Lets assume density is equal to 1. This doesn't affect the location of the center of mass.

Then $$M_c = \text{Mass of cylinder} = \pi \times \text{radius}^2 \times \text{height} = \pi \times 144 \times 50 = 22619.4671$$ $$M_b = \text{Mass of box} = \text{height} \times \text{width} \times \text{depth} = 70\times 56\times 70 = 274400$$ $$M_k = \text{Mass of cone} = \frac{1}{3}\times \pi \times \text{radius}^2 \times \text{height} = \frac{1}{3} \times \pi \times 35^2 \times 28 = 35918.876$$ Let's use $z$ to denote height above the ground plane. So, the tip of the cone is at $z=0$. Then $$z_k = z\text{-coordinate of CoM of cone} = \tfrac34\times 28 = 21$$ $$z_b = z\text{-coordinate of CoM of box} = 28 + \left(\tfrac12 \times 56\right) = 56$$ $$z_c = z\text{-coordinate of CoM of cylinder} = 28 + 56 + \left(\tfrac12 \times 50 \right) = 109$$ Then the height $z$ of the overall center of mass is given by $$z = \frac{z_c*M_c + z_b*M_b + z_k*M_k}{M_c + M_b + M_k}$$ This is a kind of averaging of the three individual $z$-coordinates, weighted by the individual masses, which should make some sense intuitively. If you substitute the numerical values of $z_c$, $M_c$, $z_b$, $M_b$, $z_k$, $M_k$, from above, you get a final answer of $z=55.8248$ (roughly).

-
Am curious, is the downvoter saying this is wrong? If someone enjoys it they'll do it. – mistermarko Aug 31 '14 at 10:28
Well, it was wrong, for a few minutes. I believe it's correct now. – bubba Aug 31 '14 at 10:30
maybe because you haven't provided an explanation to your answer? or maybe not? – ADG Aug 31 '14 at 10:39

The C.M. itself is the mass weighted average of the position vector. For object with uniform density, it is the same as the volume weighted average.

What you need to do is figure out is the Volume and C.M. of individual pieces. For the top cylinder and middle box, the C.M. are at their symmetry center. For the bottom cone, it is at $\frac34$ of its height.

This is because if you cut the bottom cone with a plane at height $h$ from the ground, the area of the cross section will be proportional to $h^2$. So the C.M. of the bottom cone itself is a weighted average.

$$\text{C.M.(bottom cone)} = \frac{\int_0^{28} h^3 dh }{\int_0^{28} h^2 dh} = \frac{\frac14 28^4}{\frac13 28^3} = \frac34 \times 28$$

Combine all these, we have:

$$\newcommand{\mycbox}[2][4pt,border:1px solid;]{\bbox[#1]{\begin{array}{c}#2\end{array}}} \newcommand{\myrlbox}[2][4pt,border:1px solid;]{\bbox[#1]{\begin{array}{rl}#2\end{array}}} \newcommand{\mydnarrow}[2][]{\left\downarrow\rlap{\bbox[#1]{\begin{array}{l}\,\\#2\\\,\end{array}}}\right.} \mycbox[]{ \myrlbox{ {\bf Top}&{\bf Cylinder}\\ \hline \text{Volume} & \pi 12^2 \times 50\\ \text{C.M.} & 50/2 + 56 + 28}\quad \myrlbox{ {\bf Middle}&{\bf Box}\\ \hline \text{Volume} & 70 \times 56 \times 70\\ \text{C.M.} & 56/2 + 28}\\ \myrlbox{ {\bf Bottom}&{\bf Cone}\\ \hline \text{Volume} & 1/3 \times \pi 35^2 \times 28\\ \text{C.M.} & 3/4 \times 28}\\ \mydnarrow{\verb/taking Volume weighed/\\\verb/average of C.M./}\\ \frac{(\pi 12^2 \times 50)(\frac{50}{2}+56+28) + (70 \times 56 \times 70)(\frac{56}{2} + 28) + (\frac13 \times \pi 35^2 \times 28)(\frac34 \times 28)}{ (\pi 12^2 \times 50) + (70 \times 56 \times 70) + (\frac13 \times \pi 35^2 \times 28) }\\ \implies\quad\text{C.M.} = \frac{30747 \pi + 518616}{ 559\pi + 8232} \approx 55.82480568907604 }$$

-
I repeat my above comment. – mistermarko Aug 31 '14 at 10:37
@mistermarko,which one – ADG Aug 31 '14 at 10:37
Inexplicable downvotes. – mistermarko Aug 31 '14 at 10:38
I didn't expected this from you, very bad :( – ADG Aug 31 '14 at 10:38
What's wrong with it then?! – mistermarko Aug 31 '14 at 10:39