Question

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.
A. `tan^(-1)(1/3)`
B. `tan^(-1)(1/4)`
C. `tan^(-1)(3)`
D. `tan^-(-1)(1/2)`

Answer 1

Correct Answer - A

`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.