A rod (mass M, length L) is placed vertically on a smooth horizontal surface. Rod is released and after some time velocity of COM is v downwards and at this moment rod makes 60 degrees with horizontal. Find angular momentum of rod about Instantaneous center of rotation.
All equations pertaining to rotational mechanics
The Attempt at a Solution
from energy conservation I wrote (Mg√3L)/4=1/2 Mv^2 + 1/2 ml^2/12 w^2=1/2 I(about ICOR)*w^2
I intuitively guessed ICOR must be on the lower right part of the rod and wrote xw=v where x is distance of ICOR from COM...not sure how to proceed further