A) \[\ell =\frac{a}{\sqrt{2}}\]
B) \[\ell =\frac{\sqrt{2}}{3}a\]
C) \[\ell =\frac{4\sqrt{2}}{3}a\]
D) \[\ell =4\sqrt{2}\,s\]
Correct Answer: C
Solution :
When the lamina ABCD is at rest, its centre of gravity G must lie vertically below the comer A by which it is hung. By geometry, \[AG=\sqrt{2}a\] Radius of gyration about this axis is given by \[{{K}^{2}}=\frac{2}{3}{{a}^{2}}\left( {{I}_{G}}=\frac{2}{3}m{{a}^{2}} \right)\] Then length of equivalent simple pendulum is: \[\ell =\sqrt{2}a+\frac{\frac{2}{3}{{a}^{2}}}{\sqrt{2}\,a}=\frac{4\sqrt{2}}{3}a\]You need to login to perform this action.
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