7.11 Hollow cone

Physics > Basics of Rotational Motion > 7.0 Moment of inertia of uniform continious rigid bodies

  Basics of Rotational Motion
    1.0 Rigid body
    2.0 Motion of rigid body
    3.0 Kinematics of a plane motion
    4.0 Moment of inertia
    5.0 Radius of gyration $(K)$
    6.0 Theorems of moment of inertia
    7.0 Moment of inertia of uniform continious rigid bodies
7.11 Hollow cone
Consider a uniform hollow cone of mass $M$, radius $R$ and height $H$

Surface area of the cone, $(A) = \pi RL$


Mass per unit area, $({\lambda _A}) = \frac{M}{{\pi RL}}$


Moment of inertia for a uniform hollow right circular cone about an axis through its centre and joining its vertex to centre of base

Consider an infinitesimally small section in the shape of a hollow cylinder of radius $r$ and mass $dm$ as shown in the figure,

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

Mass of the infinitesimally small section, $(dm) = \frac{M}{{\pi RL}}2\pi rdy$

So, moment of inertia of a hollow cylinder of radius $r$ and mass $dm$, $$\begin{equation} \begin{aligned} d{I_{COM}} = {r^2}dm \\ d{I_{COM}} = {r^2}\left( {\frac{M}{{\pi RL}}} \right)2\pi rdy \\ d{I_{COM}} = \frac{{2M}}{{RL}}{r^3}dy \\\end{aligned} \end{equation} $$
From the properties of the similar triangle we get, $$\begin{equation} \begin{aligned} \left( {\frac{{H - y}}{r}} \right) = \frac{H}{R} \\ H - y = \frac{{rH}}{R} \\ y = H\left( {1 - \frac{r}{R}} \right) \\\end{aligned} \end{equation} $$ Also, $$dy = - \frac{H}{R}dr$$ So, $$\begin{equation} \begin{aligned} \int\limits_0^{{I_{COM}}} {d{I_{COM}}} = \int\limits_R^0 { - \frac{{2M}}{{RL}}{r^3}} \left( {\frac{H}{R}} \right)dr \\ \left[ {{I_{COM}}} \right]_0^{{I_{COM}}} = \int\limits_0^R {\frac{{2M}}{{RL}}{r^3}} \left( {\frac{H}{R}} \right)dr \\ \left( {{I_{COM}} - 0} \right) = \frac{{2MH}}{{{R^2}L}}\int\limits_0^R {{r^3}dr} \\ {I_{COM}} = \frac{{2MH}}{{{R^2}L}}\left[ {\frac{{{r^4}}}{4}} \right]_0^R \\ {I_{COM}} = \frac{{2MH}}{{{R^2}L}}\left( {\frac{{{R^4}}}{4}} \right) \\ {I_{COM}} = \frac{{MH{R^2}}}{{2L}} \\\end{aligned} \end{equation} $$
So, moment of inertia of a hollow cone of radius $R$ and passing through its $COM$ is, ${I_{COM}} = \frac{{MH{R^2}}}{{2L}}$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} {I_{COM}} = M{K^2} \\ \frac{{MH{R^2}}}{{2L}} = M{K^2} \\ K = \sqrt {\frac{{H{R^2}}}{{2L}}} \\\end{aligned} \end{equation} $$
So, radius of gyration of a hollow cone of radius $R$ and passing through its $COM$ is, $K = \sqrt {\frac{{H{R^2}}}{{2L}}} $
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