A sphere with radius R= 0.200 m has density that decreases with distance rfrom the center of the sphere according to$\mathit{\rho }\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathit{k}\mathit{g}\mathbf{/}{\mathit{m}}^{\mathbf{3}}\mathbf{-}\left(3.00\times {10}^{3}kg/m^4\right)\mathit{r}$ (a) Calculate the total mass of the sphere. (b) Calculate the moment of inertia of the sphere for an axis along a diameter.

Given in the question,

The radius of the sphere $R=0.200\text{\hspace{0.17em}\hspace{0.17em}m}$

The density of the sphere$\rho =3.00\times {10}^3\text{\hspace{0.17em}\hspace{0.17em}kg}/{\text{m}}^{\text{3}}-\left(3.00\times {10}^3\text{\hspace{0.17em}\hspace{0.17em}kg}/{\text{m}}^{\text{4}}\right)r$

$Density=\frac{mass}{\begin{array}{l}volume\\ \rho =\frac{m}{v}\end{array}}$

$I=\frac{2}{5}m{r}^2$

Where m is mass and r is the radius of the sphere

We know the density,

$\begin{array}{l}p=\frac{m}{V}\\ m=\rho V\end{array}$

The volume of the sphere is

$\mathit{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathit{\pi }{\mathit{r}}^{\mathbf{3}}$

Where r is the radius of the the sphere

Therefore

$m=\rho \left(\frac{4}{3}\pi r^3\right)$

Since the density of the sphere is changing with the radius, so total mass can be given as

$\begin{array}{l}{m}_t=\underset{0}{\overset{R}{\int }}\rho \left(\frac{4}{3}\pi r^2\right)dr\\ m_t=\underset{0}{\overset{R}{\int }}\left(3.00\times {10}^3\text{\hspace{0.17em}\hspace{0.17em}kg}/{\text{m}}^{\text{3}}-\left(3.00\times {10}^3\text{\hspace{0.17em}\hspace{0.17em}kg}/{\text{m}}^{\text{4}}\right)r\right)\left(\frac{4}{3}\pi r^2\right)dr\\ m_t=\frac{4}{3}\pi \left[\underset{0}{\overset{R}{\int }}\left(\left(3.00\times {10}^3\text{\hspace{0.17em}\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\right)r^2-\left(3.00\times {10}^3\text{\hspace{0.17em}\hspace{0.17em}kg}/{\text{m}}^{\text{4}}\right)r^3\right)dr\right]\\ m_t=\frac{4}{3}\pi \left[\left(3.00\times {10}^3\text{\hspace{0.17em}\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\right)\left(\frac{R^3}{3}\right)-\left(3.00\times {10}^3\text{\hspace{0.17em}\hspace{0.17em}kg}/{\text{m}}^{\text{4}}\right)\left(\frac{R^4}{4}\right)\right]\end{array}$

Substituting the value of R

$\begin{array}{l}{m}_t=\frac{4}{3}\pi \left[\left(3.00\times {10}^3\text{\hspace{0.17em}\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\right)\left(\frac{{\left(0.200\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^3}{3}\right)-\left(3.00\times {10}^3\text{\hspace{0.17em}\hspace{0.17em}kg}/{\text{m}}^{\text{4}}\right)\left(\frac{{\left(0.200\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^4}{4}\right)\right]\\ m_t=55.3\text{\hspace{0.17em}\hspace{0.17em}kg}\end{array}$

Hence the total mass of the sphere is$m_t=55.3kg$

The moment of inertia of a sphere along its diameter can be given as

$I=\frac{2}{5}m{R}^2$

As we already calculate the mass of the sphere is

$m_t=55.3\text{\hspace{0.17em}\hspace{0.17em}kg}$and $R=0.200\text{\hspace{0.17em}\hspace{0.17em}m}$

The moment of inertia

$\begin{array}{l}I=\frac{2}{5}\left(55.3\text{\hspace{0.17em}\hspace{0.17em}kg}\right){\left(0.200\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^2\\ I={\text{0.80\hspace{0.17em}\hspace{0.17em}kg.m}}^{\text{2}}\end{array}$

The moment of inertia of the sphere is${\text{0.80\hspace{0.17em}\hspace{0.17em}kg.m}}^{\text{2}}$