Question

Let the solid in Problem 7 have density $\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$.

Show that then ${\mathit{I}}_{\mathbf{z}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{10}}\mathit{M}{\mathit{a}}^{\mathbf{2}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\alpha }$.

Short answer

The required value of $\frac{I_z}{M}is\frac{3}{10}{a}^2{\sin }^2\alpha$.

Step-by-step solution

Given information

The density of the solid is$\cos \theta .$

Concept of spherical coordinates

The spherical coordinates$\left(r,\theta ,\varphi \right)$is related to Cartesian coordinates(x,y,z) by:

$$\begin{gathered} \mathit{r}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}\mathbf{\theta }} \\ \mathbf{}\mathbf{}\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{\right)}\mathit{\varphi } \\ \mathbf{}\mathbf{}\mathbf{=}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{z}}{\mathbf{r}}\mathbf{\right)} \end{gathered}$$

The cylindrical coordinates$\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{,}\mathbf{\varphi }\mathbf{)}$ is related to Cartesian coordinates(x,y,z) by:

$$\begin{gathered} \mathit{r}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{\theta }} \\ \mathbf{}\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{\right)}\mathit{z} \\ \mathbf{}\mathbf{=}\mathit{z} \end{gathered}$$

Evaluate the mass of the solid

Calculate the mass of the solid as follows:

$$\begin{gathered} M=\int \rho dV \\ ={\int }_0^a{r}^{2}dr{\int }_0^a\cos \theta d\theta {\int }_0^{2\tau \tau }d\varphi \\ =\frac{2\tau \tau }{3}{a}^3{\int }_0^a\sin \theta d\theta \left(\cos \theta \right) \\ =\frac{\tau \tau a^3}{3}\sin \alpha \end{gathered}$$

The moment of inertia is as follows:

$$\begin{gathered} I_z=\int \rho \left(x^2+y^2\right)dV \\ =\rho \left(r^2{\sin }^2\theta \right)dV \\ ={\int }_0^{2\tau \tau }d\varphi {\int }_0^{\alpha }\cos \theta {\sin }^3\theta d\theta {\int }_0^a{r}^{4}dr \\ =-\frac{2\tau \tau a^5}{5}{\int }_0^{\alpha }{\sin }^3\theta d\theta \left(\sin \theta \right) \end{gathered}$$

Use the above values to prove .

Write integral as follows:

$$\begin{gathered} I_Z=-\frac{\tau \tau a^5}{10}{\left({\sin }^4\theta \right)}_0^{\alpha } \\ =\frac{\tau \tau a^5}{10}{\sin }^4\alpha \end{gathered}$$

Substitute $\frac{\tau \tau a^5}{10}{\sin }^4\alpha$for $I_z$and $\frac{\tau \tau a^5}{3}\sin \alpha$for M in $\frac{I_z}{M}$.

$\frac{I_z}{M}=\frac{3}{10}{a}^2{\sin }^2\alpha$

Therefore, the required value of $I_z$ is $\frac{3}{10}M{a}^2{\sin }^2\alpha$.