question_answer

A thin uniform rod of mass m and length is hinged at the lower end to a level floor and strands vertically. It is now allowed to fall, then its upper end will strike the floor with the velocity

A)  \[\sqrt{2gl}\]                                    

B)  \[\sqrt{5gl}\]

C)   \[\sqrt{3gl}\]                                   

D)  \[\sqrt{mgl}\]

Correct Answer: C

Solution :

                 From law of conservation of energy Potential energy of fall                                 = rotational kinetic energy Potential energy of fall \[=\frac{1}{2}mgl\]                           ... (i)                 Kinetic energy at \[\theta \] is given by \[=\frac{1}{2}\,I{{\omega }^{2}}\] where J is moment of inertia, \[\omega \] the angular velocity. For a rod \[I=\frac{m{{l}^{2}}}{3}\] and \[\omega =\frac{v}{r}\]\[\omega =\frac{v}{r}\]                 \[\therefore \]  \[KE=\frac{1}{2}\frac{m{{l}^{2}}}{3}.\,\frac{{{v}^{2}}}{{{l}^{2}}}\]               ?.. (ii) From Eqs. (i) and (ii), we get                 \[\frac{1}{2}.\frac{m{{l}^{2}}}{3}.\,\frac{{{v}^{2}}}{{{l}^{2}}}=\frac{1}{2}\,mgl\] \[\Rightarrow \]               \[v=\sqrt{3gl}\]