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Q.
Find the distance of the centre of mass of a solid cone of base radius $R$ and height $h$ from its apex.
NTA AbhyasNTA Abhyas 2020
Solution:
Consider an elementary disc of radius r and thickness dy.
$If total mass of cone = M$ $and density = \rho $
Then the mass of elementary disc is $\text{d}\textit{m}=\rho dv=\rho \times \pi r^{2}dy..... (1)$
In similar $\Delta ^{'} \text{s} AOE and AO^{'} \text{C}$
$\frac{y}{h}=\frac{r}{R}\Rightarrow r=\frac{y}{h}R..... (2)$
Put (2) in (1)
$dm=\rho \left(\pi \right)\left(\frac{y}{h} R\right)^{2}\text{d}\textit{y}$
$dm = \rho \times \frac{\pi \text{R}^{2}}{\text{h}^{2}} y^{2} dy$ ∴ The centre of mass of cone lying on the line AO' at a distance $y_{\text{cm}}$ from A can be calculated as
$y_{\text{cm}}=\frac{\displaystyle \int \left(\text{d} \textit{m}\right) y}{\displaystyle \int d m}=\frac{\displaystyle \int \rho \pi \left(\text{R}\right)^{2}}{h^{2}}\frac{y^{3} d y}{\displaystyle \int d m}$ $= \frac{\rho \pi \text{R}^{2}}{\text{h}^{2} M} \displaystyle \int _{0}^{\text{h}} y^{3} dy$
$\because M=\rho \times \frac{1}{3}\pi R^{2}h$
$\Rightarrow y_{\text{cm}}=\frac{\rho \pi R^{2}}{h^{2} \rho \times \frac{\pi R^{2} h}{3}}\times \frac{h^{4}}{4}=\frac{3 h}{4}$
So, the centre of mass lies at $\frac{3 h}{4}$ distance from the vertex.
