Centre of mass of a solid cone

shreyaacharya
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im actually bugged of finding a solution for d topic mentioned can any 1 pleasezzz help me
 
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pleasezz help me solve dis problem
 
Do not post homework problems in the tutorials section!

Now, to help you along a bit, let I be the cone's base in the plane z=0, and let (X,Y) denote a point in I. Let the vertex have the coordinates: \vec{r}_{v}=(x_{v},y_{v},z_{v}).

Thus, any point within the cone will lie on some line segment from \vec{r}_{v} to a point (X,Y) in I, so we can therefore represent all points in the cone with the following function:
\vec{r}(X,Y,u)=(\vec{r}_{v}-(X,Y,0))u+(X,Y,0), 0\leq{u}\leq{1},(X,Y)\in{I}
\vec{r}(X,Y,u)\equiv(x(X,Y,u),y(X,Y,u),z(X,Y,u))
This should be useful to you.

In particular, remember that the x-coordinate to any given point in the cone is:
x(X,Y,u)=(x_{v}-X)u+X[/itex]<br /> and similar expressions for the y-and z-coordinates to anyone point in the cone.
 
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Search under my name and using the keyword "cone". I've gone through the entire derivation in an earlier discussion in the HW forum.
 
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this waz a very difficult derivation ...can u be kind enough o explain wid a diagram...thank u
 
Now, remember that the centroid coordinates are gained by averaging the point coordinates over the volume V of the object O.
For example, the horizontal centroid coordinate \hat{x} is given by:
\hat{x}=\frac{\int_{O}xdV}{\int_{O}dV}=\frac{\int_{O}xdV}{V}, dV=dxdydz
Now, just compute the Jacobian to the cone representation I've given, so that you may integrate with respect to the variables (X,Y,u) instead.
 
Since it is Yule, I'll be a bit more charitable:
1. The Jacobian is readily computed to be (1-u)^{2}z_{v}[/tex]<br /> 2. We also have the coordinate representations:<br /> x(X,Y,u)=(x_{v}-X)u+X,y(X,Y,u)=(y_{v}-Y)u+Y, z(X,Y,u)=z_{v}u<br /> <br /> Use these relations to derive the coordinates for the centroid!
 
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