7.7 Solid sphere

2.1 Pure translational motion
2.2 Pure rotational motion
2.3 Plane motion

3.1 Angular velocity $\omega $
3.2 Angular acceleration $\left( \alpha \right)$
3.3 Kinematics equation for rotational motion
3.4 Analogy between translational motion & rotational motion

6.1 Parallel-axis theorem
6.2 Perpendicular axis theorem

7.1 Thin rod
7.2 Rectangular lamina
7.3 Circular ring
7.4 Circular disc
7.5 Solid cylinder
7.6 Cylindrical shell
7.7 Solid sphere
7.8 Hollow sphere
7.9 Spherical shell
7.10 Solid cone
7.11 Hollow cone
7.12 Hollow hemisphere
7.13 Parallelopiped
7.14 List of moment of inertia $(I)$ and radius of gyration $(K)$ of different bodies

Consider a solid sphere of radius $R$ and mass $M$.

Volume of the solid sphere is, $(V) = \frac{4}{3}\pi {R^3}$

Mass per unit volume, $\left( {{\lambda _V}} \right) = \frac{M}{V} = \frac{{3M}}{{4\pi {R^3}}}$

Consider an infinitesimally small section in the shape of a disc of radius $r$, thickness $dy$ and mass $dm$ at a distance $x$ from an axis of rotation.

Area of disc of radius $r$, $A = \pi {r^2}$

Volume of the infinitesimally small section of mass, $dV = \pi {r^2}dy$

Mass of an infinitesimally small section, $dm = {\lambda _V}dV$ $$\begin{equation} \begin{aligned} dm = \left( {\frac{{3M}}{{4\pi {R^3}}}} \right)\left( {\pi {r^2}dy} \right) \\ dm = \frac{{3M}}{{4{R^3}}}{r^2}dy\quad ...(i) \\\end{aligned} \end{equation} $$

Moment of inertia of an axis about the diameter of a disc of radius $r$ and mass $dm$ is, $$d{I_{COM}} = \frac{{\left( {dm} \right){r^2}}}{4}$$

From parallel axis theorem, $$\begin{equation} \begin{aligned} d{I_x} = d{I_{COM}} + {y^2}dm \\ d{I_x} = \frac{{{r^2}dm}}{4} + {y^2}dm \\ d{I_x} = \left( {\frac{{{r^2}}}{4} + {y^2}} \right)dm\quad ...(ii) \\\end{aligned} \end{equation} $$

From equation $(i)$ & $(ii)$ we get, $$d{I_x} = \left( {\frac{{{r^2}}}{4} + {y^2}} \right)\left( {\frac{{3M}}{{4{R^3}}}{r^2}dy} \right)\quad ...(iii)$$

As we know, $$\begin{equation} \begin{aligned} {y^2} + {r^2} = {R^2} \\ {r^2} = {R^2} - {y^2}\quad ...(iv) \\\end{aligned} \end{equation} $$

From eqution $(iii)$ & $(iv)$ we get, $$d{I_x} = \frac{{3M}}{{4{R^3}}}\left\{ {\frac{{{{\left( {{R^2} - {y^2}} \right)}^2}}}{4} + {x^2}\left( {{R^2} - {y^2}} \right)} \right\}dy$$

Integrating the above equation with proper limits we get, $$\begin{equation} \begin{aligned} \int\limits_0^{{I_x}} {d{I_x}} = \int\limits_{ - R}^R {\frac{{3M}}{{4{R^3}}}\left\{ {\left( {\frac{{{R^4} + {y^4} - 2{R^2}{y^2}}}{4}} \right) + \left( {{R^2}{y^2} - {y^4}} \right)} \right\}dy} \\ \int\limits_0^{{I_x}} {d{I_x}} = \int\limits_{ - R}^R {\frac{{3M}}{{16{R^3}}}\left( {{R^4} - 3{y^4} + 2{R^2}{y^2}} \right)dy} \\ \left[ I \right]_0^{{I_x}} = \frac{{3M}}{{16{R^3}}}\left[ {{R^4}x - \frac{{3{x^5}}}{5} + \frac{{2{R^2}{x^3}}}{3}} \right]_{ - R}^R \\ \left( {{I_x} - 0} \right) = \frac{{3M}}{{16{R^3}}}\left\{ {\left( {{R^5} - \frac{{3{R^5}}}{5} + \frac{{2{R^5}}}{3}} \right) - \left( { - {R^5} + \frac{{3{R^5}}}{5} - \frac{{2{R^5}}}{3}} \right)} \right\} \\ {I_x} = \frac{{3M}}{{16{R^3}}}\left( {2{R^5} - \frac{{6{R^5}}}{5} + \frac{{4{R^5}}}{3}} \right) \\ {I_x} = \frac{{3M}}{{16{R^3}}}\left( {\frac{{32{R^5}}}{{15}}} \right) \\ {I_x} = \frac{{2M{R^2}}}{5} \\\end{aligned} \end{equation} $$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} {I_{x}} = MK_{x}^2 \\ \frac{{2M{R^2}}}{5} = MK_{x}^2 \\ {K_{x}} = \sqrt {\frac{2}{5}} R \\\end{aligned} \end{equation} $$

From parallel axis theorem, $$\begin{equation} \begin{aligned} I = {I_{COM}} + M{R^2} \\ I = \frac{{2M{R^2}}}{5} + M{R^2} \\ I = \frac{{7M{R^2}}}{5} \\\end{aligned} \end{equation} $$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} I = M{K^2} \\ \frac{{7M{R^2}}}{5} = M{K^2} \\ K = \sqrt {\frac{7}{5}} \\\end{aligned} \end{equation} $$