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Centre Of Mass Of Hollow Hemisphere - Practice Questions & MCQ

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

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Position of centre of mass for Hollow Hemisphere

Have a look at the figure of Hollow Hemisphere

         

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 height y from origin as shown in figure.

And,    r=Rsin \theta ,\ \ y=Rcos\theta 

\sigma =\frac{M}{2\pi R^2}

So dm=\sigma dA=\sigma (2\pi Rcos\theta )Rd\theta

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

y_{cm}=\int_{0}^{90}Rsin\theta \sigma (2\pi Rcos\theta )Rd\theta =\frac{R}{2}

So \boldsymbol{y_{cm}= \frac{R}{2}\ from\ base}

 

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Position of centre of mass for Hollow Hemisphere

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