For the cone in Problem 18, find ${\mathbf{l}}_{\mathbf{x}}\mathbf{/}\mathbf{M}\mathbf{,}\mathbf{}{\mathbf{l}}_{\mathbf{y}}\mathbf{}\mathbf{/}\mathbf{M}\mathbf{,}\mathbf{}{\mathbf{l}}_{\mathbf{z}}\mathbf{/}\mathbf{M}$. Also find $\mathbf{l}\mathbf{/}\mathbf{M}$ about a line through the center of mass parallel to the x axis.

The solid right circular cone is $r^2=z^2,0<z<h$

The center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero. This is the point to which a force may be applied to cause a linear acceleration without an angular acceleration.

It is obvious that$I_y=l_x$

From the diagram, we have the distance of a piece of a cone from the y - axis

$l^2=z^2+r^2{\sin }^2\theta$

So, we can write

$$\begin{gathered} l_y=l_x \\ ={\int }_0^{2\pi }d\theta {\int }_0^{h}dz{\int }_0^z{r}^{2}rdr\left(z^2+r^2{\sin }^2\theta \right) \\ ={\int }_0^{2\pi }d\theta {\int }_0^h\left[\frac{1}{4}{z}^6+\frac{1}{6}{z}^6{\sin }^2\theta \right] \\ ={\int }_0^{2\pi }d\theta {\int }_0^h\left[\frac{1}{28}{h}^7+\frac{1}{6}{h}^7{\sin }^2\theta \right] \\ =\frac{\pi }{14}{h}^7+\frac{h^7}{42}{\int }_0^{2\mathrm{\pi }}{\sin }^2\theta d\theta \\ =\frac{\pi }{14}{h}^7+\frac{\mathrm{\pi }}{42}{h}^7 \\ =\frac{2\mathrm{\pi }}{21}{h}^7 \end{gathered}$$

From the mass found in problem 18, we have

$\frac{{\mathrm{l}}_{\mathrm{y}}}{\mathrm{M}}=\frac{{\mathrm{l}}_{\mathrm{x}}}{\mathrm{M}}=\frac{20}{21}{\mathrm{h}}^2$

Then,

$$\begin{gathered} {\mathrm{l}}_{\mathrm{z}}={\int }_0^{2\mathrm{\pi }}\mathrm{d\theta }{\int }_0^{\mathrm{h}}\mathrm{dz}{\int }_0^{\mathrm{z}}{\mathrm{rdrr}}^2.{\mathrm{r}}^2 \\ =\frac{\mathrm{\pi }}{3}{\int }_0^{\mathrm{h}}{\mathrm{z}}^6\mathrm{dz} \\ =\frac{\mathrm{\pi }}{21}{\mathrm{h}}^7 \\ =\mathrm{Thus},\frac{{\mathrm{l}}_{\mathrm{z}}}{\mathrm{M}}=\frac{10}{21}{\mathrm{h}}^2 \end{gathered}$$

By parallel axis theorem, the following is obtained.

$$\begin{gathered} {\mathrm{l}}_{\mathrm{xm}}={\mathrm{l}}_{\mathrm{x}}-\mathrm{M}{\left(\frac{5}{6}\mathrm{h}\right)}^2 \\ =\frac{65}{252}{\mathrm{h}}^2 \end{gathered}$$

Hence, the value is given the following way.

role="math" localid="1659164022786" $$\begin{gathered} \frac{l_x}{M}=\frac{l_y}{M}=\frac{20}{21}{h}^2 \\ \frac{l_z}{M}=\frac{10}{21}{h}^2 \\ \frac{l_{xm}}{M}=\frac{65}{252}{h}^2 \end{gathered}$$

Thus, the center of mass is obtained as given below.

$$\begin{gathered} \frac{l_x}{M}=\frac{l_y}{M}=\frac{20}{21}{h}^2 \\ \frac{l_z}{M}=\frac{10}{21}{h}^2 \\ \frac{l_{xm}}{M}=\frac{65}{252}{h}^2 \end{gathered}$$