question_answer

A rigid body is made of three identical thin rods, each of length L fastened together in the form of letter H. The body is free to rotate about a horizontal axis that runs along the length of one of the legs of the H shaped body. The body is allowed to fall from rest to position in which the plane of H is horizontal. What is the angular speed of the body when the plane of H is vertical?              

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, \[\frac{2}{3}M{{L}^{2}}{{\omega }^{2}}=\frac{3}{2}MgL\] \[\omega =\frac{3}{2}\,\sqrt{\frac{g}{L}}\]