Question

Particle of mass $2m$ is connected by an inextensible string of length $1.2m$ to a ring of mass M which is free to slide on a horizontal smooth rod initially the ring and the particle are at same level with the string taught what are generally simultaneously the distance in meter moved by the ring when the string become verticle is

Answer 1

first of all, one should consider that there is no outer force acting on the system in a horizontal direction. Indeed, there is no friction acting on the ring, and the gravitational force acting on the particle is acting vertically. The force of tension in the string is the inner force in the system. In this case, the centre of mass of the system cannot move horizontally. Initially, the centre can be found as

$mx=(1.2-x)+2m$

$mx=2.4m-2mx$

$x=\frac{2.4}{3}=0.8m$