Question 63 of 130

Distance of the centre of mass of a solid uniform cone from its vertex is ${z}_{0}$. If the radius of its base is $R$ and its height is $h$ then ${z}_{0}$ is equal to:

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

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Answer: (C) $\dfrac{3h}{4}$

Let the mass of solid cylinder is $M$ and angle with vertical is $\theta$. Now, we take a small element of thickness $dy$ 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,
$\rho =\dfrac{M}{\left(\dfrac{1}{3}\right)\pi {R}^{2}h}=\dfrac{3M}{\pi {R}^{2}h}$.
The mass of the small element is given by,
$dM=\rho dV$
$dM=\dfrac{3M}{\pi {R}^{2}h}\times \pi {r}^{2}dy=\dfrac{3M{y}^{2}{\tan }^{2}\theta }{{R}^{2}h}dy$
The center of mass of the cone is given by,
${z}_{0}=\dfrac{1}{M}{\int }_{0}^{h}ydM$
${z}_{0}=\dfrac{1}{M}{\int }_{0}^{h}y\cdot \dfrac{3M{y}^{2}{\tan }^{2}\theta }{{R}^{2}h}dy$
${z}_{0}=\dfrac{3}{{R}^{2}h}{\left(\dfrac{R}{h}\right)}^{2}{\left[\dfrac{{y}^{4}}{4}\right]}_{0}^{h}$
${z}_{0}=\dfrac{3h}{4}$.

JEE Main Physics practice question on Center of Mass, Momentum & Collision (Centre of mass of continuous mass distribution), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.