Centre Of Mass Of Hollow Hemisphere - Practice Questions & MCQ

Have a look at the figure of Hollow Hemisphere

Since it is symmetrical about $y$-axis
So we can say that its $x_{c m}=0$ and $z_{c m}=0$
Now we will calculate its $y_{c m}$ which is given by

$
y_{c m}=\frac{\int y \cdot d m}{\int d m}
$

So, Take a small elemental ring of mass dm of radius r at a height y from the origin as shown in the figure.

And, $r=R \sin \theta, \quad y=R \cos \theta$

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

$\mathrm{So}_{\mathrm{o}} d m=\sigma d A=\sigma(2 \pi R \cos \theta) R d \theta$
$\mathrm{So}_{\mathrm{cm}}=\frac{\int y \cdot d m}{\int d m}$
$y_{c m}=\int_0^{\frac{\pi}{2}} R \sin \theta \sigma(2 \pi R \cos \theta) R d \theta=\frac{R}{2}$
${ }_{\mathrm{So}} \boldsymbol{y}_{\mathrm{cm}}=\frac{\boldsymbol{R}}{2}$ from base