The following question is given in my textbook
Two uniform cones with base radius $r$ are joined together by their plane faces.Their lines of symmetry are aligned, the height of one cone is $6r$ and the other is $2r$. Given that the smaller cone is $50\%$ denser than the larger, find the distance of the centre of mass from their joint plane face.
The way I worked on it is by first drawing a simple diagram
Then, finding the total volume and mass of the individual cones, for the larger cone:
$$v_1=\frac{1}{3} \pi r^2 (6r)=2\pi r^3$$
$$m_1=2\pi r^3 \rho$$
For the smaller cone:
$$v_2= \frac{1}{3} \pi r^2 (2r)=\frac{2}{3} \pi r^3$$
$$m_2=\pi r^3 \rho$$
Then proceeding to find the $\bar{x}$ of each using the result that:
For a right cone the centre of mass lies $(\frac{1}{4}h,0)$ measured from the vertex and $(\frac{3}{4}h,0)$ from the base
For the larger cone:
$$\bar{x}=\frac{1}{4}(6r)=\frac{3}{2}r$$
For the smaller cone:
$$\bar{x}=\frac{1}{4}(2r)=\frac{1}{2}r$$
Next calculating the center of mass for the whole shape:
$$\bar{x}=\frac{(2\pi r^3 \rho)(\frac{3}{2}r)+ (\pi r^3 \rho)(\frac{1}{2}r)}{\pi r^3 \rho + 2\pi r^3 \rho}=\frac{7}{6}r$$
However the answer given in the book is that the distance is $\frac{5}{6}r$ I do not know how they obtained this answer, is there a problem in my working and how do I proceed with solving this correctly?
Best Answer
I think you have to use a better notation since you do not distinguish the 3 centers of mass.
I'll resume what you have:
Supposing you place your coordinate system on the heigths of the two cones with the origin on the base of the two cones. Then
$$\begin{cases} v_1 = 2\pi r^3 \\ m_1=2\pi r^3 \rho_1 \\ \bar{x_1} = -\frac{3}{2}r \end{cases} \qquad \begin{cases} v_2= \frac{2}{3} \pi r^3 \\ m_2= \pi r^3 \rho_1 \\ \bar{x}_2 = \frac{1}{2}r \end{cases}$$
Now as you wrote:
$$\bar{x}=\frac{m_1 \bar{x_1} + m_2 \bar{x_2}}{m_1 + m_2} = \frac{(2\pi r^3 \rho_1)(-\frac{3}{2}r)+ (\pi r^3 \rho_1)(\frac{1}{2}r)}{\pi r^3 \rho_1 + 2\pi r^3 \rho_1}= \frac{(-3 + \frac{1}{2}) r}{1 + 2} = -\frac{5}{6}r$$
Please also note that