Centre Of Mass Of A Solid Cone MCQ - Practice Questions with Answers

Have a look at the figure of 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 $y_{c m}$ which is given by

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

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

So $d m=\rho d v=\rho\left(\pi r^2\right) d y$
Here $\rho=\frac{M}{V}=\frac{M}{\frac{1}{3} \pi R^2 H}$
And from similar triangle

$$
\begin{gathered}
\frac{r}{R}=\frac{H-y}{H} \\
r=\left(\frac{H-y}{H}\right) R
\end{gathered}
$$

$$
\begin{aligned}
& y_{c m}=\frac{\int y \cdot d m}{\int d m} \\
& y_{c m}=\frac{1}{M} \int_0^H y \cdot d m=\frac{1}{M} \int_0^H y \frac{3 M}{\pi R^2 H}\left(\pi r^2\right) d y=\frac{H}{4}
\end{aligned}
$$

$\mathrm{So}_, \mathbf{y}_{\mathbf{c m}}=\frac{\mathbf{H}}{\mathbf{4}}$ from battom 0

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