**2 Answers**

written 2.8 years ago by | • modified 2.8 years ago |

![here is the answer for the above question. 100% correct][1]

Density is uniform, so we can use geometrical method.

Area of disc, A1 =πR² Area of cutoff disc, A2 = (π/4)(R/2)² = πR²/16

Xcom = (A1×0-A2X2)/(A1-A2) = ((πR²/16)×(3R/4))/((πR²)- ( πR²/16)) =

-3R/15×4 = -R/20

Hence, center of mass is at R/20 from the center.

written 2.8 years ago by | modified 2.7 years ago by |

Density is uniform, so we can use geometrical method.

Area of disc, $\mathrm{A}_{1}=\pi \mathrm{R}^{2}$

Area of cutoff disc, $\mathrm{A}_{2}=\frac{\pi}{4}\left(\frac{\mathrm{R}}{2}\right)^{2}=\frac{\pi \mathrm{R}^{2}}{16}$ $$ \begin{array}{l} x_{\text {com }}=\frac{A_{1} \times 0-A_{2} x_{2}}{A_{1}-A_{2}}=\frac{-\frac{\pi R^{2}}{16} \times \frac{3 R}{4}}{\pi R^{2}-\frac{\pi R^{2}}{16}}=\frac{-3 R}{15 \times 4}= \\ -\frac{R}{20} \end{array} $$ Hence, center of mass is at $\frac{\mathrm{R}}{20}$ from the center.