Question

A 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 the same level with the string, taut. Both are then released simultaneously. What is the distance in meters moved by the ring when the string becomes vertical ?

Answer 1

Center of mass in $x$-direction is same

$mx=(1.2-x)\times 2mx=0.8m$

where $x$ is the distance moved by ring from their initial position.