A) \[{{\left( \frac{g}{L} \right)}^{1/2}}\]
B) \[{{\left( \frac{3g}{4L} \right)}^{1/2}}\]
C) \[{{\left( \frac{3\sqrt{3}g}{2L} \right)}^{1/2}}\]
D) \[{{\left( \frac{3g}{2L} \right)}^{1/2}}\]
Correct Answer: D
Solution :
The fall of centre of gravity h \[\frac{\left( \frac{1}{2}-h \right)}{\frac{1}{2}}=\cos \,\,{{60}^{o}}\] \[h=\frac{1}{2}(1-\cos \,{{60}^{o}})\] Decrease in potential energy \[Mgh=Mg\frac{1}{2}(1-\cos {{60}^{o}})\] Kinetic energy of rotation \[=\frac{1}{2}I{{\omega }^{2}}\] \[=\frac{1}{2}\times \frac{M{{I}^{2}}}{3}{{\omega }^{2}}\] \[Mg=\frac{1}{2}(1-\cos \,{{60}^{o}})=\frac{M{{I}^{2}}}{6}{{\omega }^{2}}\] \[\Rightarrow \] \[\omega =\frac{\sqrt{3g}}{2L}\]You need to login to perform this action.
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