7.8 Hollow 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 hollow sphere of radius $R$ and mass $M$

Surface area of the uniform hollow sphere is, $(A) = 4\pi {R^2}$

Mass per unit area, $\left( {{\lambda _A}} \right) = \frac{M}{A} = \frac{{M}}{{4\pi {R^2}}}$

Consider an infinitesimally small section in the shape of a ring of radius $r$, thickness $Rd\theta $ and mass $dm$ along the axis of rotation.

Area of ring of radius $r$, $dA = 2\pi rRd\theta $

Mass of the infinitesimally small section, $\left( {dm} \right) = {\lambda _A}dA$ $$\begin{equation} \begin{aligned} dm = \frac{M}{{4\pi {R^2}}}(2\pi rRd\theta ) \\ dm = \frac{{Mrd\theta }}{{2R}} \\\end{aligned} \end{equation} $$

Moment of inertia of ring of radius $r$ and mass $dm$ is,$$d{I_{COM}} = {r^2}dm$$ So, $$\begin{equation} \begin{aligned} d{I_{COM}} = {r^2}dm \\ d{I_{COM}} = {r^2}\left( {\frac{{Mrd\theta }}{{2R}}} \right) \\ d{I_{COM}} = \frac{{M{r^3}d\theta }}{{2R}} \\\end{aligned} \end{equation} $$ As we know, $$r = R\sin \theta $$ So, we can write, $$\begin{equation} \begin{aligned} d{I_{COM}} = \frac{{M{R^3}{{\sin }^3}\theta d\theta }}{{2R}} \\ d{I_{COM}} = \frac{{M{R^2}}}{2}{\sin ^3}\theta d\theta \\\end{aligned} \end{equation} $$

Integrating the above equation with proper limits we get, $$\begin{equation} \begin{aligned} \int\limits_0^{{I_{COM}}} {d{I_{COM}}} = \frac{{M{R^2}}}{2}\int\limits_0^\pi {{{\sin }^3}\theta d\theta } \\ \left[ {{I_{COM}}} \right]_0^{{I_{COM}}} = \frac{{M{R^2}}}{2}\int\limits_0^\pi {{{\sin }^3}\theta d\theta } \\ \left( {{I_{COM}} - 0} \right) = \frac{{M{R^2}}}{2}\int\limits_0^\pi {\sin \theta \left( {1 - {{\cos }^2}\theta } \right)d\theta } \\\end{aligned} \end{equation} $$ Let, $$\begin{equation} \begin{aligned} \cos \theta = u \\ - \sin \theta d\theta = du \\ \cos \pi = - 1\quad \ \quad \cos 0 = 1 \\\end{aligned} \end{equation} $$ So, $$\begin{equation} \begin{aligned} {I_{COM}} = \frac{{M{R^2}}}{2}\int\limits_1^{ - 1} {\left( {{u^2} - 1} \right)du} \\ {I_{COM}} = \frac{{M{R^2}}}{2}\left[ {\frac{{{u^3}}}{3} - u} \right]_1^{ - 1} \\ {I_{COM}} = \frac{{M{R^2}}}{2}\left[ {\left( {\frac{{ - 1}}{3} + 1} \right) - \left( {\frac{1}{3} - 1} \right)} \right] \\ {I_{COM}} = \frac{{M{R^2}}}{2}\left[ {\frac{4}{3}} \right] \\ {I_{COM}} = \frac{{2M{R^2}}}{3} \\\end{aligned} \end{equation} $$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} {I_{COM}} = MK_{COM}^2 \\ \frac{{2M{R^2}}}{3} = MK_{COM}^2 \\ {K_{COM}} = \sqrt {\frac{2}{3}} R \\\end{aligned} \end{equation} $$

From parallel axis theorem, $$\begin{equation} \begin{aligned} I = {I_{COM}} + M{R^2} \\ I = \frac{{2M{R^2}}}{3} + M{R^2} \\ I = \frac{{5M{R^2}}}{3} \\\end{aligned} \end{equation} $$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} I = M{K^2} \\ \frac{{5M{R^2}}}{3} = M{K^2} \\ K = \sqrt {\frac{5}{3}} R \\\end{aligned} \end{equation} $$