Question

5. If the height of the solid cone is 40 cm.Where is the position of centre of gravity ?

Answer 1

Consider a solid uniform right circular cone of base radius $\sf{R}$ and height $\sf{H.}$ We have to find position of center of mass of this cone, which is along a straight line connecting base center and upper vertex due to symmetry in the cone.

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Consider an elemental cylindrical portion of radius $\sf{r}$ and height $\sf{dx}$ at a perpendicular distance $\sf{x}$ from base center, whose center is at $\sf{O'.}$

We see triangles $\sf{OAB}$ and $\sf{O'A'B}$ are similar. Hence,

$\longrightarrow\sf{\dfrac{OA}{OB}=\dfrac{O'A'}{O'B}}$

$\longrightarrow\sf{\dfrac{R}{H}=\dfrac{r}{H-x}}$

$\longrightarrow\sf{H-x=\dfrac{Hr}{R}}$

$\longrightarrow\sf{x=H-\dfrac{H}{R}\,r\quad\quad\dots(1)}$

Since $\sf{H}$ and $\sf{R}$ are constants,

$\longrightarrow\sf{dx=-\dfrac{H}{R}\,dr\quad\quad\dots(2)}$

The volume of the elemental cylindrical portion is,

$\longrightarrow\sf{dV=\pi r^2\,dx}$

From (2),

$\longrightarrow\sf{dV=-\dfrac{\pi r^2H}{R}\,dr}$

Let $\sf{M}$ be mass of the cone, $\sf{V}$ be its volume, and $\sf{dM}$ be mass of the elemental portion.

Since the cone is uniform, density is same everywhere. Thus,

$\longrightarrow\sf{\dfrac{M}{V}=\dfrac{dM}{dV}}$

$\longrightarrow\sf{\dfrac{M}{\left(\dfrac{1}{3}\pi R^2H\right)}=\dfrac{dM}{\left(-\dfrac{\pi r^2H}{R}\,dr\right)}}$

$\longrightarrow\sf{\dfrac{3M}{\pi R^2H}=-\dfrac{R\ dM}{\pi r^2H\,dr}}$

$\longrightarrow\sf{dM=-\dfrac{3Mr^2\,dr}{R^3}\quad\quad\dots(3)}$

Hence the position of center of mass of the cone is given by,

$\longrightarrow\sf{\bar x=\dfrac{\displaystyle\sf{\int\limits_R^0x\ dM}}{\displaystyle\sf{\int\limits_R^0dM}}}$

The limits are because $\sf{dM}$ is in terms of $\sf{dr.}$

From (1) and (3),

$\longrightarrow\sf{\bar x=\dfrac{\displaystyle\sf{\int\limits_R^0-\left(H-\dfrac{H}{R}\,r\right)\ \dfrac{3Mr^2\,dr}{R^3}}}{\displaystyle\sf{\int\limits_R^0-\dfrac{3Mr^2\,dr}{R^3}}}}$

$\longrightarrow\sf{\bar x=\dfrac{\displaystyle\sf{-\dfrac{3MH}{R^3}\int\limits_R^0\left(1-\dfrac{r}{R}\right)\ r^2\,dr}}{\displaystyle\sf{-\dfrac{3M}{R^3}\int\limits_R^0r^2\,dr}}}$

$\longrightarrow\sf{\bar x=\dfrac{\displaystyle\sf{H\int\limits_R^0\left(r^2-\dfrac{r^3}{R}\right)\,dr}}{\displaystyle\sf{\int\limits_R^0r^2\,dr}}}$

$\longrightarrow\sf{\bar x=\dfrac{\displaystyle\sf{H\int\limits_R^0r^2\,dr-\dfrac{H}{R}\int\limits_R^0r^3\,dr}}{\displaystyle\sf{\int\limits_R^0r^2\,dr}}}$

$\longrightarrow\sf{\bar x=\dfrac{H\left[\dfrac{r^3}{3}\right]_R^0-\dfrac{H}{R}\left[\dfrac{r^4}{4}\right]_R^0}{\left[\dfrac{r^3}{3}\right]_R^0}}$

$\longrightarrow\sf{\bar x=\dfrac{\dfrac{H(0^3-R^3)}{3}-\dfrac{H(0^4-R^4)}{4R}}{\left(\dfrac{0^3-R^3}{3}\right)}}$

$\longrightarrow\sf{\bar x=\dfrac{\dfrac{HR^3}{4}-\dfrac{HR^3}{3}}{-\dfrac{R^3}{3}}}$

$\longrightarrow\sf{\bar x=\dfrac{HR^3\left(\dfrac{1}{4}-\dfrac{1}{3}\right)}{-\dfrac{R^3}{3}}}$

$\longrightarrow\sf{\bar x=\dfrac{H\times-\dfrac{1}{12}}{-\dfrac{4}{12}}}$

$\longrightarrow\sf{\underline{\underline{\bar x=\dfrac{H}{4}}}}$

Hence position of center of mass of the cone is at a perpendicular distance of $\sf{\dfrac{H}{4}}$ from the base center.

According to the question,

• $\sf{H=40\ cm}$

Hence the position of center of mass of the cone is,

$\longrightarrow\sf{\bar x=\dfrac{H}{4}}$

$\longrightarrow\sf{\bar x=\dfrac{40}{4}}$

$\longrightarrow\sf{\underline{\underline{\bar x=10\ cm}}}$

That is, position of center of mass is $\bf{10\ cm}$ vertically from the base center.