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

Have a look at the figure of Hollow Hemisphere

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 origin as shown in figure.

$
\begin{aligned}
&\text { And, } r=R \sin \theta, \quad y=R \cos \theta\\
&\begin{aligned}
& \sigma=\frac{M}{2 \pi R^2} \\
& \text { So }_0 d m=\sigma d A=\sigma(2 \pi R \cos \theta) R d \theta \\
& y_{c m}=\frac{\int y \cdot d m}{\int d m} \\
& \text { So }_0 \\
& y_{c m}=\int_0^{90} R \sin \theta \sigma(2 \pi R \cos \theta) R d \theta=\frac{R}{2} \\
& \boldsymbol{y}_{c m}=\frac{\boldsymbol{R}}{2} \text { from base }
\end{aligned}
\end{aligned}
$