Question

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:
A) $\dfrac{L}{2}$
B) $\dfrac{L}{4}$
C)$\dfrac{L}{8}$
D) none

Answer 1

Hint: First we have to consider the lower end of the rod will end up distance and subtract the distance x away from the lower end when the rod falls. As the question says , the original vertical distance from the lower end (x) has to equal the new (horizontal) distance which is mentioned as x. From this simple calculation we get the correct answer.

Complete step by step solution:
The center of mass we can say the middle point of the rod will directly fall where the lower end used to fall. So, we can tell the lower end of the rod will end up $\dfrac{L}{2}$ from its previous position. Hence 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, it means we can write $\dfrac{L}{2} - x$ away from the middle point. In the order that point should follow a quarter circular path, the original vertical distance from the lower end named as x has to equal the new horizontal distance to the middle point $\dfrac{L}{2} - x$.
Thus,
$\eqalign{
  & x = \dfrac{L}{2} - X \cr
  & \Rightarrow 2x = \dfrac{L}{2} \cr
  & \Rightarrow 4x = L \cr
  & \therefore x = \dfrac{L}{4} \cr} $

The point which will follow a quarter circular path is $\dfrac{L}{4}$ away from the lower end. Hence the correct answer is B.

Note :
In this question we have to understand the question first and name them wisely, then you can solve mathematics, hence you can get a direct answer. If you have good hold on theryotical problems then we can solve very easily.