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 $\Delta OPQ=\frac{1}{2}\times PQ\times OP$
$=\frac{1}{2}Rd\theta \times R=\frac{1}{2}{R}^{2}d\theta$
Centre of mass of this $\Delta OPQ$ is at a distance of $\frac{2}{3}Rcos\theta$ from $O$.
So, position of centre of mass of complete segment is
$\bar{y}={\int }^{ydm/}\int dm$
$\bar{y}=\frac{{\int }_0^{\theta /2}\frac{2R}{3}\cos \theta \cdot \rho \frac{r^2}{2}d\theta }{{\int }_0^{\theta /2}\rho \frac{r^2}{2}d\theta }$
where,$\rho =$mass density.
$\therefore \bar{y}=\frac{\frac{2r}{3}{\int }_0^{\theta /2}\cos \theta d\theta }{{\int }_0^{\theta /2}d\theta }=2\frac{R}{3}\cdot \frac{\sin \theta /2}{\theta /2}$
$=\frac{4}{3}R\frac{\sin (\theta /2)}{\theta }$