Distance of the centre of mass of a solid uniform cone from its...

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

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

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

The mass of the small element is given by,

$d M = ρ d V$

$d M = \dfrac{3M}{\pi{R}^{2}h} \times \pi{r}^{2} d y = \dfrac{3M{y}^{2}\tan^{2}\theta}{{R}^{2}h} d y$

The center of mass of the cone is given by,

${z}_{0} = \dfrac{1}{M}{∫}_{0}^{h}y d M$

${z}_{0} = \dfrac{1}{M}{∫}_{0}^{h}y \cdot \dfrac{3M{y}^{2}\tan^{2}\theta}{{R}^{2}h} d y$

${z}_{0} = \dfrac{3}{{R}^{2}h}{\left(\dfrac{R}{h}\right)}^{2}{\left[\dfrac{{y}^{4}}{4}\right]}_{0}^{h}$

${z}_{0} = \dfrac{3h}{4}$.