The distance between the vertex and the centr | vedclass

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Question

(A) Consider a uniform solid circular sector of radius $R$ and total angular size $	heta$. Let the sector be symmetric about the $y$-axis.Consider an

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$\frac{4}{3} R \frac{\sin (	heta / 2)}{	heta}$

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(A) Consider a uniform solid circular sector of radius $R$ and total angular size $	heta$. Let the sector be symmetric about the $y$-axis.Consider an infinitesimal triangular element of the sector with an angular width $d\alpha$ at an angle $\alpha$ from the $y$-axis.The area of this infinitesimal triangle is $dA = \frac{1}{2} R^2 d\alpha$.The centre of mass of a triangle is at a distance of $\frac{2}{3} R$ from the vertex $O$. For this element,the distance of its centre of mass from the $x$-axis is $y = \frac{2}{3} R \cos \alpha$.The $y$-coordinate of the centre of mass of the entire sector is given by:$\bar{y} = \frac{\int y dA}{\int dA} = \frac{\int_{-	heta/2}^{	heta/2} (\frac{2}{3} R \cos \alpha) (\frac{1}{2} R^2 d\alpha)}{\int_{-	heta/2}^{	heta/2} \frac{1}{2} R^2 d\alpha}$Simplifying the expression:$\bar{y} = \frac{\frac{1}{3} R^3 \int_{-	heta/2}^{	heta/2} \cos \alpha d\alpha}{\frac{1}{2} R^2 [\alpha]_{-	heta/2}^{	heta/2}} = \frac{\frac{1}{3} R^3 [\sin \alpha]_{-	heta/2}^{	heta/2}}{\frac{1}{2} R^2 	heta} = \frac{\frac{1}{3} R^3 (2 \sin(	heta/2))}{\frac{1}{2} R^2 	heta}$$\bar{y} = \frac{2}{3} R \frac{2 \sin(	heta/2)}{	heta} = \frac{4}{3} R \frac{\sin(	heta/2)}{	heta}$

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

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