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 his given by \[\frac{\left( \frac{L}{2}-h \right)}{\left( \frac{L}{2} \right)}=\cos \,60{}^\circ \] or \[h=\frac{L}{2}(1-cos\,60{}^\circ )\] \[\therefore \] Decrease in potential energy \[=\,\,Mgh=Mg\frac{L}{2}\,\left( 1-cos\,60{}^\circ \right)\] Kinetic energy of rotation = \[=\,\,\frac{1}{2}1{{\omega }^{2}}=\frac{1}{2}\times \frac{M{{L}^{2}}}{3}\omega \] \[l=\frac{M{{L}^{2}}}{3}\] (because rod is rotating about an axis passing through its one end)] According to law of conservation of energy \[Mg\frac{L}{2}(1-cos\,60{}^\circ )=\frac{M{{L}^{2}}}{6}{{\omega }^{2}}\Rightarrow \omega =\sqrt{\frac{3g}{2L}}\]You need to login to perform this action.
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