Rotating spheres of variable densities!

Consider the above diagram for both spheres $A$ and $B$ . Take a sphere of radius $r$ ( $r < R$ ) and at a further distance $dr$ draw a small spherical shell of radius $r+dr$ from the center. The thickness $dr$ of this shell is very small, so we can consider the surface area to remain the same as of the sphere of radius $r$ .

Therefore, the volume of the shell will be

$V = 4\pi r^2 \cdot dr$

• For sphere A-

$\begin{aligned} & dm_A = V \cdot \rho_A (r) \\ \implies & dm_A = 4\pi k \left( \dfrac{r^3}{R} \right) \cdot dr \end{aligned}$

So,

$\begin{aligned} I_A &=& \int r^2 \,dm_A \\ &=& \dfrac{4\pi k}{R} \int_0^R r^5 \,dr \\ &=& \dfrac{4\pi k}{R} \cdot \dfrac{R^6}{6} \\ &=& \dfrac23 \pi k R^5 \end{aligned}$

• For sphere B-

$\begin{aligned} & dm_B = V \cdot \rho_B (r) \\ \implies & dm_B = 4\pi k \left( \dfrac{r^7}{R^5} \right) \cdot dr \end{aligned}$

So,

$\begin{aligned} I_B &=& \int r^2 \,dm_B \\ &=& \dfrac{4\pi k}{R^5} \int_0^R r^9 \,dr \\ &=& \dfrac{4\pi k}{R^5} \cdot \dfrac{R^{10}}{10} \\ &=& \dfrac25 \pi k R^5 \end{aligned}$

Thus,

$\dfrac{I_B}{I_A} = \dfrac{\frac25 \pi k R^5}{\frac23 \pi k R^5} = \dfrac35 = \dfrac{6}{10}$

So, $\boxed{n=6}$ .

Suppose the density is

$\rho_n(r) = k \left ( \frac{r}{R} \right )^{n}$

Then, for any of the spheres, the moment of inertia may be written as

$I_n = \int r^{2}\rho_n(r)dV = 4\pi \int_{0}^{R} r^{4}k\left ( \frac{r}{R} \right )^{n}dr = \frac{4\pi k}{n+5}R^{5}$

What we want is

$\frac{I_5}{I_1} = \frac{6}{10}$

The moment of inertia of a solid is given by the formula,

$I = \displaystyle \int \int \int \delta(x,y,z) \| \mathbf{r} \| dV$

where $\delta(x, y, x)$ is the density at any point $(x, y, z)$ and $\| \mathbf{r} \|$ is a vector from the axis of rotation to that point $(x, y, z)$ . Since we have a formula for density in terms of a radial distance, it is convenient to integrate using spherical coordinates.

In spherical coordinates, $\rho$ is distance from the origin(radius), $\phi$ is the smaller angle made by our radius with the positive z-axis, and $\theta$ is the smaller angle made with radius and the positive x-axis.

So, we can change our integral to

$I = \displaystyle \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} \delta(\rho\sin{\phi}\cos{\theta}, \rho\sin{\phi}sin{\theta}, \rho\cos{\phi})\rho^2\rho^2\sin{\phi} d\rho d\phi d\theta$

The extra $\rho^2\sin{\phi}$ comes from the Jacobian when converting between rectangular and spherical integrals.

In the case of $A$ and $B$ the integrals become, respectively,

$I_A = \frac{k}{R}\displaystyle \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} \rho^5\sin{\phi} d\rho d\phi d\theta = \frac{2}{3}\pi R^5k$

and

$I_B = \frac{k}{R^5}\displaystyle \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} \rho^9\sin{\phi} d\rho d\phi d\theta = \frac{2}{5}\pi R^5k$ .

Thus, $\frac{I_B}{I_A} = \frac{6}{10}$ , so $\boxed{n = 6}$ .

haha.. i made it level 5 ! i guessed 2 times i.e. 1 and 3 and it's level jumped from 4 to 5 then i thought i should do it , but i think level 5 is too much for this problem !