Question

The distance between the vertex and the centre of mass of a uniform solid planar circular segment of angular $\sin \theta$ and radius $R$ is given by

• A. $\frac{4}{3}R\frac{\sin\left(\theta /2\right)}{\theta}$
• B. $R\frac{\sin\left(\theta /2\right)}{\theta}$
• C. $\frac{4}{3}R\,\cos\left(\frac{\theta}{2}\right)$
• D. $\frac{4}{3}R\,\cos\theta$

Answer 1

Answer: (A)

Refer to figure below,

Consider a differentiable are of angle $d\theta$
Area of $\Delta O P Q =\frac{1}{2} \times P Q \times O P$
$=\frac{1}{2} R d \theta \times R=\frac{1}{2} R^{2} d \theta$
Centre of mass of this $\Delta\, OPQ$ is at a distance of $\frac{2}{3} R\,cos\,\theta$ from $O$.
So, position of centre of mass of complete segment is
$\bar{y}=\int^{y d m /} \int d m$
$\bar{y}=\frac{\int_{0}^{\theta / 2} \frac{2 R}{3} \cos \theta \cdot \rho \frac{r^{2}}{2} d \theta}{\int_{0}^{\theta / 2} \rho \frac{r^{2}}{2} d \theta}$
where,$\rho= $mass density.
$\therefore \bar{y} =\frac{\frac{2 r}{3} \int_{0}^{\theta / 2} \cos \theta d \theta}{\int_{0}^{\theta / 2} d \theta}=2 \frac{R}{3} \cdot \frac{\sin \theta / 2}{\theta / 2}$
$=\frac{4}{3} R \frac{\sin (\theta / 2)}{\theta}$