A) \[\sqrt{\frac{g}{L}}\]
B) \[\frac{1}{2}\sqrt{\frac{g}{L}}\]
C) \[\frac{3}{2}\sqrt{\frac{g}{L}}\]
D) \[2\sqrt{\frac{g}{L}}\]
Correct Answer: C
Solution :
Moment of inertia of the system about the given axis, \[I={{I}_{A}}+{{I}_{B}}+{{I}_{C}}\] Now, as rod is thin, \[{{I}_{A}}=\Sigma m\times {{(0)}^{2}}=0\] Rod B is rotating about one end. \[\therefore \] \[{{I}_{B}}=\frac{M{{L}^{2}}}{3}\] and for rod call points are always at distance L from the axis of rotation, so \[{{I}_{C}}=\Sigma m{{L}^{2}}=M{{L}^{2}}\] \[\therefore \] \[I=0+\frac{M{{L}^{2}}}{3}+M{{L}^{2}}=\frac{4}{3}M{{L}^{2}}\] So, if co is the desired angular speed, gain in kinetic energy due to rotation of \[H\] from horizontal to vertical position, \[{{K}_{R}}=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\,\left[ \frac{4}{3}M{{L}^{2}} \right]{{\omega }^{2}}\] \[=\frac{2}{3}M{{L}^{2}}{{\omega }^{2}}\]and loss in potential energy of the system in doing so \[=0+Mg\frac{L}{2}+MgL=\frac{3}{2}MgL\] So, by conservation of mechanical energy, \[\frac{2}{3}M{{L}^{2}}{{\omega }^{2}}=\frac{3}{2}MgL\] \[\omega =\frac{3}{2}\,\sqrt{\frac{g}{L}}\]You need to login to perform this action.
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