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**irspow****Member**- Registered: 2005-11-24
- Posts: 1,055

*edit...sorry, I see the error of my thinking...

I was not considering torque only the mass...what I did is simply split the mass in half without considering the torque applied by those masses about the point c...sorry for wasting the space with my oversight....

The center of mass of a cone of uniform density is (0, h/4) relative to the base center, usually stated as 3h/4 from the vertex, according to everybody. Fine. But follow along for kicks...

If we define the right side of the cone by a line we get:

y=h-hx/r

y=(hr-hx)/r

If we integrate for solid of revolution about the y axis x is the radius...

x=(hr-ry)/h and the volume would be given by the integral...

V(y)=p⌠f(y)^2 dy in this case:

V(y)=p⌠((hr)^2+(ry)^2-2hyr^2)/h^2 dy

V(y)=[(pr^2)/h^2]*⌠h^2+y^2-2hy dy, let k=[pr^2/h^2] for simplicity...

V(y)=k[yh^2+y^3/3-hy^2]

V(y)=(k/3)[3yh^2+y^3-3hy^2] Obviously if the interval is y=[0,h] we get the normal Volume equation for a cone...((pr^2/(3h^2)(h^3)=(phr^2)/3))

But if we find a point c where V(y) for y=[h,c] is equal to V(y) for y=[0,c] then we know that the volume above c and below c are equal.

Considering that the cone is assumed to have uniform density, having equal volume above and below c is the same as having equal mass above and below c, the definition of the center of mass...okay so I say let's find c...

If we use the two intervals to find when they are equal and allow the leading constants to cancel out...the two intervals again are [h,c] and [c,0], you get:

3h^3+h^3-3h^3-3ch^2-c^3+3hc^2=3ch^2+c^3-3hc^2

h^3-3ch^2-c^3+3hc^2=3ch^2+c^3-3hc^2

2c^3-6hc^2+6ch^2-h^3=0

if we let h=1, it does not matter, this is just for ease...and we use an interative process such as Newton's approximation...

2c^3-6c^2+6c-1=0

and the approximation being...

(4c^3-6c^2+1)/(6c^2-12c+6) letting c1=0 we see the the sequence...

0, 1/6, 0.2044..., 0.206295.., 0.206299473.., 0.206299474015900262623.., 0.20629947401590026262414718036385, and then next one does not change anything within the next 20 digits...

Why is point c not h/4? Half of the volume, and thus mass, is above c and below c at this point, how is this not the center of mass? (This is assuming uniform density and using sound solid of revolution integration.)

So, I know, and I have seen, that there are proofs out there that c is h/4, but why does integration of a solid of revolution lead to a different result?

Sorry for being dense, I just want to see the error of my logic...thank you for your time.

*Last edited by irspow (2016-03-10 12:21:25)*

I am at an age where I have forgotten more than I remember, but I still pretend to know it all.

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

If you were to equate the volumes you could do it easier by equating the volume above c.g equal to half of the total volume

However te method itself is wrong. the center of gravity is the point where the moment of total mass equals the sum of the moments of individual masses. In other words}

The concept of c.g. can better be understood if we take the axis of cone horizontally and hang it by a thread at c.g. it is obvious that to keep it in balance the moment of mass/weight of the right hand side(shorter cone) about the thread must be balanced by the moment of the weight/mass of the left hand side which is a frustrum.

*Last edited by thickhead (2016-04-19 19:47:21)*

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