Centre Of Mass Of Hollow Hemisphere MCQ - Practice Questions & Answers

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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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Centre Of Mass Of Hollow Hemisphere MCQ - Practice Questions with Answers

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