7.5 Solid cylinder

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 solid cylinder of mass $M$, length $L$ and radius $R$.

Volume of the cylinder $(V) = \pi {R^2}L$

Mass per unit volume $({\lambda _V}) = \frac{M}{{\pi {R^2}L}}\quad ...(i)$

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

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

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

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

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, $$\begin{equation} \begin{aligned} \int {d{I_z} = \int_m {{r^2}\left( {\frac{{2Mrdr}}{{{R^2}}}} \right)} } \\ \int {d{I_z}} = \frac{{2M}}{{{R^2}}}\int_m {{r^3}dr} \\\end{aligned} \end{equation} $$

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

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 disc of mass $dM$, radius $R$ and thickness $dy$ at a distance $y$ from the axis of rotation as shown in the figure.

Area of the circular disc $(A) = \pi {R^2}$

Volume of the infinitesimally small section, $(dV)=Ady$ $$dV = \pi {R^2}dy\quad ...(vi)$$

Mass of the infinitesimally small section of mass $dM = {\lambda _V}dV$ $$\begin{equation} \begin{aligned} dM = \left( {\frac{M}{{\pi {R^2}L}}} \right)\left( {\pi {R^2}dy} \right) \\ dM = \frac{{Mdy}}{L}\quad ...(vii) \\\end{aligned} \end{equation} $$

Moment of inertia of a disc of radius $R$ and mass $dM$ about its diameter passing through centre of mass $(COM)$ is given by, $$d{I_{COM}} = \frac{{\left( {dM} \right){R^2}}}{4}$$

So, moment of inertia of a cylinder is given by,

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

From equation $(vii)$ & $(viii)$ we get, $$d{I_x} = \left( {\frac{{{R^2}}}{4} + {y^2}} \right)\left( {\frac{{Mdy}}{L}} \right)$$

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

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

So, radius of gyration about an axis passing through its centre and perpendicular to its own axis is, $K = \sqrt {\left( {\frac{{{R^2}}}{4} + \frac{{{L^2}}}{{12}}} \right)} $

From parallel axis theorem, $$\begin{equation} \begin{aligned} {I_1} = {I_x} + M{\left( {\frac{L}{2}} \right)^2} \\ {I_1} = M\left( {\frac{{{R^2}}}{4} + \frac{{{L^2}}}{{12}}} \right) + \frac{{M{L^2}}}{4} \\ {I_1} = M\left( {\frac{{{R^2}}}{4} + \frac{{{L^2}}}{3}} \right) \\\end{aligned} \end{equation} $$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} I = M{K^2} \\ M\left( {\frac{{{R^2}}}{4} + \frac{{{L^2}}}{3}} \right) = M{K^2} \\ K = \sqrt {\left( {\frac{{{R^2}}}{4} + \frac{{{L^2}}}{3}} \right)} \\\end{aligned} \end{equation} $$

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

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