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  • A circular hole of radius is cut out of a circular disc of radius 'a' as shown in figure. The

A circular hole of radius \left ( \frac{a}{2} \right ) is cut out of a circular disc of radius 'a' as shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be :
 
Option: 1 \frac{5}{6}a
Option: 2 \frac{1}{6}a
Option: 3 \frac{10}{11}a  
Option: 4 \frac{2}{3}a

Answers (1)

best_answer

Let \sigma be the surface mass density of the disc

then m=\sigma \mathrm{\pi r}^{2}

As

X_{\text {com }}=\frac{\left(m_{1} x_{1}-m_{2} x_{2}\right)}{\left(m_{1}-m_{2}\right)}

\begin{array}{l} \mathrm{x}_{\mathrm{COM}}=\frac{\left(\sigma \pi \mathrm{a}^{2}\right) \mathrm{a}-\sigma \pi\left(\frac{\mathrm{a}^{2}}{4}\right) \times \frac{3 \mathrm{a}}{2}}{\sigma \pi \mathrm{a}^{2}-\frac{\sigma \pi \mathrm{a}^{2}}{4}} =\frac{\mathrm{a}-\frac{3 \mathrm{a}}{8}}{1-\frac{1}{4}}=\frac{5 \mathrm{a}}{6} \end{array}

Posted by

avinash.dongre

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