Question

The densities of two solid spheres A and B of the same radii R vary with radial distance r as ${{\rho }_{a}}(r)=k(\dfrac{r}{R}),{{\rho }_{b}}(r)=k{{(\dfrac{r}{R})}^{5}}$, respectively, where k is a constant. The moments of inertia of the individual spheres about axes passing through their centres are ${{I}_{a}},{{I}_{b}}$ ​, respectively. If $\dfrac{{{I}_{b}}}{{{I}_{a}}}=\dfrac{n}{10}$ ​, the value of n is

Answer 1

Let us first find the mass of a small element of the sphere as the density varies regarding the change in r. Next, the moment of inertia of a hollow sphere is known. Now, as the density is changing, we need to integrate the density term while calculating the moment of inertia of each of the spheres respectively. Finally, we will get the ratio of the moment of inertias of both the spheres.

Formula used:
$I=\dfrac{2}{3}m{{r}^{2}}$

Complete step-by-step answer:
Let us assume a thin shell of width dr at radius r from the centre of the sphere,
The mass then, will be equal to,
$\begin{aligned} & m=d\times V \\\ & \Rightarrow {{m}_{a}}={{\rho }_{a}}(r)\times V \\\ & \Rightarrow {{m}_{a}}={{\rho }_{a}}(r)\times 4\pi {{r}^{2}}dr \\\ & \Rightarrow {{m}_{b}}={{\rho }_{b}}(r)\times 4\pi {{r}^{2}}dr \\\ \end{aligned}$
Now, if we solve for the moment of inertia of the sphere a and b respectively, we get,

& {{I}_{a}}=\int\limits_{0}^{R}{\dfrac{2}{3}\times }{{\rho }_{a}}(r)\times 4\pi {{r}^{2}}dr\times {{r}^{2}} \\\ & \Rightarrow {{I}_{a}}=\dfrac{2}{3}\times \dfrac{4\pi k}{R}\times \dfrac{{{R}^{6}}}{6} \\\ & \Rightarrow {{I}_{b}}=\int\limits_{0}^{R}{\dfrac{2}{3}\times }{{\rho }_{b}}(r)\times 4\pi {{r}^{2}}dr\times {{r}^{2}} \\\ & \Rightarrow {{I}_{b}}=\dfrac{2}{3}\times \dfrac{4\pi k}{{{R}^{5}}}\times \dfrac{{{R}^{10}}}{10} \\\ \end{aligned}$$ As we know the moment of inertias of both the spheres, $\begin{aligned} & \dfrac{{{I}_{b}}}{{{I}_{a}}}=\dfrac{\dfrac{2}{3}\times \dfrac{4\pi k}{{{R}^{5}}}\times \dfrac{{{R}^{10}}}{10}}{\dfrac{2}{3}\times \dfrac{4\pi k}{R}\times \dfrac{{{R}^{6}}}{6}} \\\ & \Rightarrow \dfrac{{{I}_{b}}}{{{I}_{a}}}=\dfrac{6}{10} \\\ & \\\ \end{aligned}$ Now, if we compare the ratio obtained with that of the ratio given in the question, We get n=6. **Additional Information:** Moment of inertia is the name given today rotational inertia, the rotational analogue for the mass in linear motion. It appears in the relationships for the dynamics of rotational motion posters. The moment of inertia must be specified with respect to the extrusion axis of rotation. For play point mass, the moment of inertia is the mass times the square of perpendicular distance to the rotation axes. The point mass relationship becomes the basis for all other moments of inertia Since objects can be built up from a collection of point masses. For a body with indescribable shape or size, or non-uniform mass, the moment of inertia is obtained by assuming a small part and integrating it. **Note:** The moment of inertia of a body is directly proportional to the mass and increases as the mass is moved further away from the axis of rotation, also, the angular momentum increases as the point object is taken away from the axis of rotation. The best example to describe it is the rotating door.