7.12 Hollow hemisphere

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 uniform hollow hemisphere of mass $M$ and radius $R$

Surface area of the hemisphere is, $(A) = 2\pi {R^2}$

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

For moment of inertia of a uniform hollow hemisphere about an axis passing through its center as shown in the figure,

Consider an infinitesimal small hollow cylinder of mass $dm$ in the shape of a hollow cylinder of radius $r$.

Area of the infinitesimally small section, $\left( {dA} \right) = 2\pi rdy$

Mass of the infinitesimally small section, $\left( {dm} \right) = {\lambda _A}dA$ or $$\begin{equation} \begin{aligned} dm = \frac{M}{{2\pi {R^2}}} \times 2\pi rdy \\ dm = \frac{M}{R}rd\theta \quad (As,y = Rd\theta ) \\\end{aligned} \end{equation} $$

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

Integrating the above equation for the whole hemisphere we get, $$\begin{equation} \begin{aligned} \int\limits_0^{{I_{COM}}} {d{I_{COM}}} = \int\limits_0^{\frac{\pi }{2}} {M{R^2}{{\sin }^3}\theta d\theta } \\ \left[ {{I_{COM}}} \right]_0^{{I_{COM}}} = \int\limits_0^{\frac{\pi }{2}} {M{R^2}\sin \theta \left( {1 - {{\cos }^2}\theta } \right)d\theta } \\\end{aligned} \end{equation} $$ Let, $$\begin{equation} \begin{aligned} \cos \theta = u \\ - \sin \theta d\theta = du \\ d\theta = - \frac{{du}}{{\sin \theta }} \\\end{aligned} \end{equation} $$ Therefore, $$\begin{equation} \begin{aligned} {I_{COM}} = - \int {M{R^2}\left( {1 - {u^2}} \right)du} \\ {I_{COM}} = \int {M{R^2}\left( {{u^2} - 1} \right)du} \\ {I_{COM}} = M{R^2}\left( {\frac{{{u^3}}}{3} - u} \right) \\\end{aligned} \end{equation} $$ or $$\begin{equation} \begin{aligned} {I_{COM}} = M{R^2}\left[ {\frac{{{{\cos }^3}\theta }}{3} - \cos \theta } \right]_0^{\frac{\pi }{2}} \\ {I_{COM}} = M{R^2}\left[ {\left( {0 - 0} \right) - \left( {\frac{1}{3} - 1} \right)} \right] \\ {I_{COM}} = \frac{{2M{R^2}}}{3} \\\end{aligned} \end{equation} $$

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

Consider an infinitesimal small hollow cylinder of mass $dm$ in the shape of a ring of radius $r$.

Area of the infinitesimally small section, $\left( {dA} \right) = 2\pi rdy$

Mass of the infinitesimally small section, $\left( {dm} \right) = {\lambda _A}dA$ or $$\begin{equation} \begin{aligned} dm = \frac{M}{{2\pi {R^2}}} \times 2\pi rdy \\ dm = \frac{M}{R}rd\theta \quad (As,y = Rd\theta ) \\\end{aligned} \end{equation} $$

For moment of inertia for a ring of radius $r$ and mass $dm$ about its diameter is, $$d{I_{COM}} = \frac{{{r^2}dm}}{2}$$

[Diagram]

From parallel axis theorem, $$\begin{equation} \begin{aligned} dI = d{I_{COM}} + {y^2}dm \\ dI = \frac{{{r^2}dm}}{2} + {y^2}dm \\ dI = \left( {\frac{{{r^2} + 2{y^2}}}{2}} \right)dm \\\end{aligned} \end{equation} $$ or $$\begin{equation} \begin{aligned} dI = \left( {\frac{{{r^2} + 2{y^2}}}{2}} \right)\left( {\frac{M}{R}rd\theta } \right) \\ dI = \left( {\frac{{{R^2} + {y^2}}}{2}} \right)\left( {\frac{M}{R}rd\theta } \right)\quad \left\{ {As,\left( {{r^2} = {R^2} - {y^2}} \right)} \right\} \\ dI = \frac{M}{{2R}}\left( {{R^2}r + {y^2}r} \right)d\theta \\\end{aligned} \end{equation} $$ From the diagram, $$\begin{equation} \begin{aligned} \sin \left( {90^\circ - \theta } \right) = \frac{r}{R} \\ \cos \theta = \frac{r}{R} \\ r = R\cos \theta \\ y = R\sin \theta \\\end{aligned} \end{equation} $$ So integrating the above equation with proper limits we get, $$\int\limits_0^I {dI} = \int\limits_0^{\frac{\pi }{2}} {\frac{M}{{2R}}\left[ {{R^3}\cos \theta + {R^3}{{\sin }^2}\theta \cos \theta } \right]d\theta } $$ Let, $$\begin{equation} \begin{aligned} \sin \theta = u \\ \cos \theta d\theta = du \\\end{aligned} \end{equation} $$ So, $$\begin{equation} \begin{aligned} \int {dI} = \int {\frac{{M{R^2}}}{2}\left[ {1 + {u^2}} \right]du} \\ \left[ I \right] = \frac{{M{R^2}}}{2}\left[ {u + \frac{{{u^3}}}{3}} \right] \\\end{aligned} \end{equation} $$ or $$\begin{equation} \begin{aligned} \left[ I \right]_0^I = \frac{{M{R^2}}}{2}\left[ {\sin \theta + \frac{{{{\sin }^3}\theta }}{3}} \right]_0^{\frac{\pi }{2}} \\ I = \frac{{M{R^2}}}{2}\left[ {\left( {1 + \frac{1}{3}} \right) - (0 - 0)} \right] \\ I = \frac{{2M{R^2}}}{3} \\\end{aligned} \end{equation} $$

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

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

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