JEE Main 2015 — Center of Mass Momentum and Collision Question with Solution

Let the mass of solid cylinder is $M$ and angle with vertical is $\theta$. Now, we take a small element of thickness $\mathit{d}y$ at a distance $y$ from the vertex. The radius of small element is $r$, as shown in the figure below.

Then, $\mathrm{tan\theta }=\frac{R}{h}=\frac{r}{y}$ which will give, $r=y\mathrm{tan\theta }$. The density of the cone is given by,

$\rho =\frac{M}{\left(\frac{1}{3}\right)\pi R^{2}h}=\frac{3M}{\pi R^{2}h}$.

The mass of the small element is given by,

$dM=\rho dV$

$dM=\frac{3M}{\pi R^{2}h}\times \pi r^{2}dy=\frac{3M{y}^2{\tan }^2\theta }{R^{2}h}dy$

The center of mass of the cone is given by,

$z_0=\frac{1}{M}{\int }_0^{h}ydM$

$z_0=\frac{1}{M}{\int }_0^{h}y·\frac{3M{y}^2{\tan }^2\theta }{R^{2}h}dy$

$z_0=\frac{3}{R^{2}h}{\left(\frac{R}{h}\right)}^2{\left[\frac{y^4}{4}\right]}_0^h$

$z_0=\frac{3h}{4}$.