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Centre Of Mass Of Hollow Cone - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

4 Questions around this concept.

Solve by difficulty

A hollow cone and a hollow semicircular shell are placed as shown in the diagram. Each has mass M . What is the y-coordinate of COM of system

A

$\frac{10}{13}R$

B

$\frac{5}{7}R$

C

$\frac{3}{7}R$

D

$\frac{13}{12}R$

Concepts Covered - 1

Position of centre of mass for Hollow Cone

Have a look at the figure of Hollow Cone

Since it is symmetrical about y-axis

So we can say that its $x_{cm} = 0$ and $z_{cm} = 0$

Now we will calculate its ${y_{cm}$ which is given by

$y_{cm} = \frac{\int y.dm}{\int dm}$

So Take a small elemental ring of mass dm of radius r at a vertical distance y from O as shown in figure.

And $r=xsin\theta , \ and \ y=xcos\theta$

Since our element mass is ring so its C.O.M will lie on the y-axis.

Now $dm=\sigma dA=\sigma (2\pi xsin\theta )dx$

Where $\sigma = \frac{M}{\pi R*\sqrt{R^2+H^2}}$

So $dm=\frac{2Mxdx}{R^2+H^2}$

$y_{cm}=\frac{1}{M} \int ydm=\frac{1}{M}\int_{0}^{\sqrt{R^2+H^2}}xcos\theta *\frac{2Mxdx}{R^2+H^2}=\frac{2H}{3}$

So $\mathbf{y_{cm}= \frac{2H}{3}}$ from O.

Study it with Videos

Position of centre of mass for Hollow Cone

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