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Centre Of Mass Of Hollow Cone - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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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 

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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.

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Position of centre of mass for Hollow Cone

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