Refer to figure below,
Consider a differentiable are of angle $d \theta$.
Area of $\triangle O P Q=\frac{1}{2} \times P Q \times O P$
$=\frac{1}{2} R d \theta \times R=\frac{1}{2} R^2 d \theta$
Centre of mass of this $\triangle O P Q$ is at a distance of $\frac{2}{3} R \cos \theta$ from $O$.
So, position of centre of mass of complete segment is
$\bar{y}=\int y d m / \int d m$
$\bar{y}=\frac{\int \limits_0^{\theta / 2} \frac{2 R}{3} \cos \theta \cdot \rho \frac{r^2}{2} d \theta}{\int \limits_0^{\theta / 2} \rho \frac{r^2}{2} d \theta}$
where, $\rho=$ mass density.
$\therefore \quad \bar{y} =\frac{\frac{2 r}{3} \int \limits_0^{\theta / 2} \cos \theta d \theta}{\int \limits_0^{\theta / 2} d \theta}=2 \frac{R}{3} \cdot \frac{\sin \theta / 2}{\theta / 2}$
$=\frac{4}{3} R \frac{\sin (\theta / 2)}{\theta}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
