7.6 Cylindrical shell

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 cylinder of mass $M$, length $L$, inner radius $R_1$ and outer radius $R_2$ as shown in the figure.

Volume of the hollow cylinder $V = \pi L\left( {R_2^2 - R_1^2} \right)$

Mass per unit volume $$\left( {{\lambda _V}} \right) = \frac{M}{{\pi L\left( {R_2^2 - R_1^2} \right)}}\quad ...(i)$$

Consider an infinitesimally small section in the shape of a ring of mass $dm$, radius $r$ and thickness $dr$ as shown in the figure.

Area of an infinitesimally small section of a ring, $\left( {dA} \right) = 2\pi rdr$

Volume of an infinitesimally small section of a ring along the cylinder, $(dV)=(dA)L$ $$dV = (2\pi rdr)L\quad ...(ii)$$

Mass of the infinitesimally small section, $\left( {dm} \right) = {\lambda _V}dV$ $$\begin{equation} \begin{aligned} dm = \left( {\frac{M}{{\pi L(R_2^2 - R_1^2)}}} \right)(2\pi Lrdr) \\ dm = \frac{{2Mrdr}}{{(R_2^2 - R_1^2)}}\quad ...(iii) \\\end{aligned} \end{equation} $$

So, moment of inertia of a ring of radius $r$ and mass $dm$ is given by, $$\int {d{I_z}} = \int_m {{r^2}dm} \quad ....(iv)$$

where,

$\int_m : $ Integrating over the entire mass

$r$: Perpendicular distance from an axis of rotation (Radius of the ring)

$dm$: mass of infinitesimally small section

From equation $(iii)$ & $(iv)$ we get, $$\begin{equation} \begin{aligned} \int {d{I_z}} = \int_m {{r^2}\left( {\frac{{2Mrdr}}{{\left( {R_2^2 - R_1^2} \right)}}} \right)} \\ \int {d{I_z} = \frac{{2M}}{{\left( {R_2^2 - R_1^2} \right)}}\int_m {{r^3}dr} } \\\end{aligned} \end{equation} $$

Integrating the above equation for the whole cylindrical shell we get, $$\begin{equation} \begin{aligned} \int\limits_0^{{I_z}} {d{I_z}} = \frac{{2M}}{{\left( {R_2^2 - R_1^2} \right)}}\int\limits_{{R_1}}^{{R_2}} {{r^3}dr} \\ \left[ {{I_z}} \right]_0^{{I_z}} = \frac{{2M}}{{\left( {R_2^2 - R_1^2} \right)}}\left[ {\frac{{{r^4}}}{4}} \right]_{{R_1}}^{{R_2}} \\ \left( {{I_z} - 0} \right) = \frac{{2M}}{{\left( {R_2^2 - R_1^2} \right)}}\left( {\frac{{R_2^4}}{4} - \frac{{R_1^4}}{4}} \right) \\ {I_z} = \frac{{2M}}{{4\left( {R_2^2 - R_1^2} \right)}}\left( {R_2^4 - R_1^4} \right) \\ {I_z} = \frac{{M\left( {R_2^2 - R_1^2} \right)\left( {R_2^2 + R_1^2} \right)}}{{2\left( {R_2^2 - R_1^2} \right)}} \\ {I_z} = \frac{{M\left( {R_2^2 + R_1^2} \right)}}{2}\quad ...(v) \\\end{aligned} \end{equation} $$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} I = M{K^2} \\ \frac{{M\left( {R_2^2 + R_1^2} \right)}}{2} = M{K^2} \\ K = \sqrt {\frac{{\left( {R_2^2 + R_1^2} \right)}}{2}} \\\end{aligned} \end{equation} $$

As we know that the moment of inertia along the axis of a cylindrical shell of inner & outer radius $R_1$ & $R_2$ respectively is given by, $${I_z} = \frac{{M\left( {R_2^2 + R_1^2} \right)}}{2}\quad ...(vi)$$

For hollow cyclinder, $${R_1} = {R_2} = R\quad ...(vii)$$

From equation $(vi)$ & $(vii)$ we get, $${I_z} = M{R^2}$$

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

Consider an infinitesimally small section in the shape of a ring of mass $dM$, radius $R$, width $dr$ and thickness $dy$ at a distance $y$ from the axis of rotation as shown in the figure.

Area of the circular ring $(dA) = 2\pi rdr$

Volume of the infinitesimally small section, $(dV) = dAdy$ $$dV = 2\pi rdrdy\quad ...(viii)$$

Mass of the infinitesimally small section, $(dM) = {\lambda _V}dV$ $$\begin{equation} \begin{aligned} dM = \left( {\frac{M}{{\pi L\left( {R_2^2 - R_1^2} \right)}}} \right)\left( {2\pi rdrdy} \right) \\ dM = \frac{{2M}}{{L\left( {R_2^2 - R_1^2} \right)}}rdrdy\quad ...(ix) \\\end{aligned} \end{equation} $$

Moment of inertia of a ring $r$ and mass $dM$ about its diameter passing through center of mass $(COM)$ is given by, $$d{I_{COM}} = \frac{{(dM){r^2}}}{2}$$

So, moment of inertia of a ring about the axis of rotation of a cylinder is given by,

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

From equation $(ix)$ & $(x)$ we get, $$d{I_x} = \left( {\frac{{{r^2}}}{2} + {y^2}} \right)\left( {\frac{{2M}}{{L\left( {R_2^2 - R_1^2} \right)}}Lrdrdy} \right)$$

Integrating the above equation for the whole cylinder we get, $$\begin{equation} \begin{aligned} \int\limits_0^{{I_x}} {d{I_x}} = \frac{{2M}}{{L\left( {R_2^2 - R_1^2} \right)}}\int\limits_{ - \frac{L}{2}}^{\frac{L}{2}} {\;\int\limits_{{R_1}}^{{R_2}} {\left( {\frac{{{r^3}}}{2} + r{y^2}} \right)} } \\ \left[ {{I_x}} \right]_0^{{I_x}} = \frac{{2M}}{{L\left( {R_2^2 - R_1^2} \right)}}\int\limits_{ - \frac{L}{2}}^{\frac{L}{2}} {\left[ {\frac{{{r^4}}}{8} + \frac{{{r^2}{y^2}}}{2}} \right]_{{R_1}}^{{R_2}}dy} \\ \left( {{I_x} - 0} \right) = \frac{{2M}}{{L\left( {R_2^2 - R_1^2} \right)}}\int\limits_{ - \frac{L}{2}}^{\frac{L}{2}} {\left\{ {\left( {\frac{{R_2^4}}{8} + \frac{{R_2^2{y^2}}}{2}} \right) - \left( {\frac{{R_1^4}}{8} + \frac{{R_1^2{y^2}}}{2}} \right)} \right\}dy} \\ {I_x} = \frac{M}{{L\left( {R_2^2 - R_1^2} \right)}}\int\limits_{ - \frac{L}{2}}^{\frac{L}{2}} {\left( {\frac{{R_2^4 - R_1^4}}{4} + \left( {R_2^2 - R_1^2} \right){y^2}} \right)dy} \\ {I_x} = \frac{M}{{L\left( {R_2^2 - R_1^2} \right)}}\left( {R_2^2 - R_1^2} \right)\int\limits_{ - \frac{L}{2}}^{\frac{L}{2}} {\left( {\frac{{R_2^2 + R_1^2}}{4} + {y^2}} \right)dy} \\ {I_x} = \frac{M}{L}\int\limits_{ - \frac{L}{2}}^{\frac{L}{2}} {\left( {\frac{{R_2^2 + R_1^2}}{4} + {y^2}} \right)dy} \\ {I_x} = \frac{M}{L}\left[ {\left( {\frac{{R_2^2 + R_1^2}}{4}} \right)y + \frac{{{y^3}}}{3}} \right]_{ - \frac{L}{2}}^{\frac{L}{2}} \\ {I_x} = \frac{M}{L}\left\{ {\left[ {\left( {\frac{{R_2^2 + R_1^2}}{4}} \right)\frac{L}{2} + \frac{{{L^3}}}{{24}}} \right] - \left[ { - \left( {\frac{{R_2^2 + R_1^2}}{4}} \right)\frac{L}{2} - \frac{{{L^3}}}{{24}}} \right]} \right\} \\ {I_x} = M\left\{ {\left( {\frac{{R_2^2 + R_1^2}}{4}} \right) + \frac{{{L^2}}}{{12}}} \right\} \\ \\\end{aligned} \end{equation} $$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} I = M{K^2} \\ M\left\{ {\left( {\frac{{R_2^2 + R_1^2}}{4}} \right) + \frac{{{L^2}}}{{12}}} \right\} = M{K^2} \\ K = \sqrt {\left( {\frac{{R_2^2 + R_1^2}}{4}} \right) + \frac{{{L^2}}}{{12}}} \\\end{aligned} \end{equation} $$

As we know that the moment of inertia along the axis of a cylindrical shell of inner & outer radius $R_1$ & $R_2$ respectively is given by, $${I_x} = M\left\{ {\left( {\frac{{R_2^2 + R_1^2}}{4}} \right) + \frac{{{L^2}}}{{12}}} \right\}\quad ...(xi)$$

For hollow cylinder, $${R_1} = {R_2} = R\quad ...(xii)$$

From equation $(xi)$ & $(xii)$ we get, $${I_x} = M\left( {\frac{{{R^2}}}{2} + \frac{{{L^2}}}{{12}}} \right)$$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} I = M{K^2} \\ M\left( {\frac{{{R^2}}}{2} + \frac{{{L^2}}}{{12}}} \right) = M{K^2} \\ K = \sqrt {\left( {\frac{{{R^2}}}{2} + \frac{{{L^2}}}{{12}}} \right)} \\\end{aligned} \end{equation} $$