Question

Find the distance of the centre of mass of a solid cone of base radius $R$ and height $h$ from its apex.

• A. $\frac{3h^2}{8R}$
• B. $\frac{h^2}{4R}$
• C. $\frac{3h}{4}$
• D. $\frac{5h}{8}$

Answer 1

Answer: (C)

Consider an elementary disc of radius r and thickness dy.
$Iftotalmassofcone=M$ $anddensity=\rho$
Then the mass of elementary disc is $\text{d}\text{m}=\rho dv=\rho \times \pi r^{2}dy.....(1)$
In similar ${\Delta }^{{}^{\prime }}\text{s}AOEandA{O}^{{}^{\prime }}\text{C}$
$\frac{y}{h}=\frac{r}{R}\Rightarrow r=\frac{y}{h}R.....(2)$
Put (2) in (1)
$dm=\rho \left(\pi \right){\left(\frac{y}{h}R\right)}^2\text{d}\text{y}$
$dm=\rho \times \frac{\pi {\text{R}}^2}{{\text{h}}^2}{y}^{2}dy$
∴ The centre of mass of cone lying on the line AO' at a distance $y_{\text{cm}}$ from A can be calculated as
$y_{\text{cm}}=\frac{\int \left(\text{d}\text{m}\right)y}{\int dm}=\frac{\int \rho \pi {\left(\text{R}\right)}^2}{h^2}\frac{y^{3}dy}{\int dm}$ $=\frac{\rho \pi {\text{R}}^2}{{\text{h}}^{2}M}{\int }_0^{\text{h}}{y}^{3}dy$
$\because M=\rho \times \frac{1}{3}\pi R^{2}h$
$\Rightarrow y_{\text{cm}}=\frac{\rho \pi R^2}{h^2\rho \times \frac{\pi R^{2}h}{3}}\times \frac{h^4}{4}=\frac{3h}{4}$
So, the centre of mass lies at $\frac{3h}{4}$ distance from the vertex.