From a circular disc of radius R a triangular | vedclass

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Question

$A_{1}=\pi R^{2}, \quad A_{2}=\frac{1}{2} 	imes 2 R 	imes R=R^{2}$

for $\Delta \mathrm{x}_{\mathrm{cm}}=0$ and for $\operatorname{disc} \mathrm{x

Vedclass · Accepted Answer

$\frac{R}{3(\pi - 1)}$

Vedclass · Answer

$A_{1}=\pi R^{2}, \quad A_{2}=\frac{1}{2} 	imes 2 R 	imes R=R^{2}$

for $\Delta \mathrm{x}_{\mathrm{cm}}=0$ and for $\operatorname{disc} \mathrm{x}_{\mathrm{cm}}=0$

so only $\mathrm{y}_{\mathrm{cm}}$ is required

$\begin{aligned} y_{\mathrm{cm}} &=\frac{A_{1} y_{1}-A_{2} y_{2}}{A_{1}-A_{2}}=\frac{\left(\pi R^{2}ight) 0-R^{2} 	imes R / 3}{\pi R^{2}-R^{2}} \ &=\frac{-R}{3(\pi-1)} \end{aligned}$

From a circular disc of radius $R$ a triangular portion is cut (see fig.). The distance of $COM$ of the remaining disc from centre of disc $O$ is:-

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