A thin uniform rod of mass 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 :
The kinetic energy at the point Q is given by \[=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\frac{m{{l}^{2}}}{3}\frac{{{\upsilon }^{2}}}{{{l}^{2}}}\] \[=\frac{1}{2}\times \frac{1}{3}m{{\upsilon }^{2}}\] ??(i) The potential energy at G \[=\frac{1}{2}mgl\] .....(ii) From equations (i) and (ii), we get \[\frac{1}{2}\frac{m{{\upsilon }^{2}}}{3}=\frac{1}{2}mgl\] \[\upsilon =\sqrt{3gl}\]