Centre Of Mass Of Hollow Cone - Practice Questions & MCQ

Have a look at the figure of Hollow 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_{c m}=\frac{\int y \cdot d m}{\int d m}
$

So Take a small elemental ring of mass dm of radius r at a vertical distance y from O as shown in figure.

And $r=x \sin \theta$, and $y=x \cos \theta$
Since our element mass is ring so its C.O.M will lie on the $y$-axis.
Now $d m=\sigma d A=\sigma(2 \pi x \sin \theta) d x$

Where

$
\sigma=\frac{M}{\pi R * \sqrt{R^2+H^2}}
$

So

$
\begin{aligned}
& d m=\frac{2 M x d x}{R^2+H^2} \\
& y_{c m}=\frac{1}{M} \int y d m=\frac{1}{M} \int_0^{\sqrt{R^2+H^2}} x \cos \theta * \frac{2 M x d x}{R^2+H^2}=\frac{2 H}{3}
\end{aligned}
$

${ }_{\text {So }} \mathbf{y}_{\mathrm{cm}}=\frac{2 \mathrm{H}}{3}$ from O.