An L shaped uniform rod of mass 2M and length 2L (AB = BC = L) is held as shown in Fig. with a string fixed between C and wall so that AB is vertical and BC is horizontal. There is no friction between the hinge and the rod at A. If the string is burnt, find the angle between AB and the vertical at equilibrium position.

An L shaped uniform rod of mass 2M and length 2L (AB = BC = L) is held

1.	An L shaped uniform rod of mass 2M and length 2L (AB = BC = L) is held as shown in Fig. with a string fixed between C and wall so that AB is vertical and BC is horizontal. There is no friction between the hinge and the rod at A. If the string is burnt, find the angle between AB and the vertical at equilibrium position.
Answer» `tan^(-1)(1/3)` `tan^(-1)(1/4)` `tan^(-1)(3)` `tan^-(-1)(1/2)` SOLUTION : `D` is the `CM` of the rod In triangle `ADE`: `tantheta=(L//4)/(3L//4)` `implies tantheta=1/3impliestheta=tan^(-1)(1//3)` here `D` is the CENTRE of MASS. In EQUILIBRIM, `D` will be vertically below `A`. so i equilibrium `AB` will make angle `theta` as calculated above with the vertical.