Centre Of Mass Of A Solid Cone - Practice Questions & MCQ

Have a look at the figure of a solid cone

Since it is symmetrical about $y$-axis
So we can say that its $x_{c m}=0$ and $z_{c m}=0$
Now we will calculate its $Y \mathrm{~cm}$ which is given by

$
y_{\mathrm{cm}}=\frac{\int y \cdot d m}{\int d m}
$

So Take a small elemental disc of mass $d m$ of radius $r$ at a vertical distance $y$ from the bottom as shown in the figure.

So $\quad d m=\rho d v=\rho\left(\pi r^2\right) d y$

Here

$
\rho=\frac{M}{V}=\frac{M}{\frac{1}{3} \pi R^2 H}
$

And from similar triangle

$
\begin{aligned}
\frac{r}{R} & =\frac{H-y}{H} \\
r & =\left(\frac{H-y}{H}\right) R \\
y_{c m} & =\frac{\int y \cdot d m}{\int d m} \\
y_{c m} & =\frac{1}{M} \int_0^H y \cdot d m=\frac{1}{M} \int_0^H y \frac{3 M}{\pi R^2 H}\left(\pi r^2\right) d y=\frac{H}{4} \\
\text { So, }^{\quad} \mathbf{y}_{\mathrm{cm}} & =\frac{\mathbf{H}}{\mathbf{4}} \text { from bottom O }
\end{aligned}
$

Or, Centre of Mass of a solid cone will lie at distance $\frac{3 h}{4}$ from the tip of the cone.