Question

Find the position of the CM of uniform lamina $ABCD$ (as shown in figure) from $O$.

• A. $\frac{a}{2\sqrt{3}}$
• B. $\frac{7a}{3\sqrt{3}}$
• C. $\frac{2a}{2\sqrt{3}}$
• D. None

Answer 1

The correct option is B $\frac{7a}{3\sqrt{3}}$

Center of mass of OAD is m and OBC is N.

Now cenetr of mass of ABCD is

$\frac{\sigma (9\times \sqrt{39})\times \frac{2a}{\sqrt{3}}-\sigma (\frac{a}{2}\times \frac{\sqrt{39}}{2})\times \frac{9}{\sqrt{3}}}{\sigma 9\sqrt{39}-\sigma \frac{9}{2}\frac{\sqrt{39}}{2}}=\frac{\sigma a^2(29-\frac{1}{4}a)}{\sigma a^2(\sqrt{3}-\frac{\sqrt{3}}{4})}=\frac{7a}{3\sqrt{3}}$