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 $x_{cm}=0$ and $z_{cm}=0$
Now we will calculate its $y_{cm}$ which is given by

$$y_{cm}=\frac{\int y\cdot dm}{\int dm}$$
 

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 $dm=\rho dv=\rho \left(\pi r^2\right)dy$
Here $\rho =\frac{M}{V}=\frac{M}{\frac{1}{3}\pi R^{2}H}$
And from similar triangle

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

$$\begin{array}{rl}& y_{cm}=\frac{\int y\cdot dm}{\int dm}\\ & y_{cm}=\frac{1}{M}{\int }_0^{H}y\cdot dm=\frac{1}{M}{\int }_0^{H}y\frac{3M}{\pi R^{2}H}\left(\pi r^2\right)dy=\frac{H}{4}\end{array}$$

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

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