A stationary pulley carries a rope whose one end supports a ladder with a man and the other end the counterweight of mass M. The man of mass m climbs up a distance `I^'` with respect to the ladder and then stops. Neglecting the mass of the rope and the friction in the pulley axle, find the displacement I of the centre of inertia of this system.

Explanation

In the reference frame fixed to the pulley axis the location of C.M. of the given system is described by the radius vector
`Deltavecr_C=(Mdeltavecr_M+(M-m)Deltavecr_((M-m))+mDeltavecr_m)/(2M)`
But `Deltavecr_M=-Deltavecr_((M-m))`
and `Deltavecr_m=Deltavecr_(m(M-m))+Deltavecr_((M-m))`
Thus `Deltavecr_C=(m overset(rarr)l)/(2M)`

Note: one may also solve this problem using momentum conservation.