Q. A straight rod of length `L` is released on a frictionless horizontal floor in a vertical position.As it falls + slips, the distance of a point on the rod from the lower end, which follows a quarter circular locus is:-
The answer is B - L/4.
The center of mass (the middle point) of the rod will fall directly where the lower end used to be. The lower end will end up distance L/2 from it's previous position. All the points between the lower end and the middle will follow a curved path. If the point on the rod is distance x away from the lower end, it will still of course be distance x away from the lower end after the rod falls, which means it will be L/2 - x away from the middle point. In order that the point should follow a quarter circular path, the original (vertical) distance from the lower end (x) has to equal the new (horizontal) distance to the middle point (L/2 - x).
That is, x = L/2 - x
2x = L/2
x = L/4
The point which will follow a quarter circular path is L/4 away from the lower end.