Question

A rope thrown over a pulley has a ladder with a man of mass $m$ on one of its ends and a counter balancing mass $M$ on its other end. The man climbs with a velocity $V_r$ relative to ladder. Ignoring the masses of the pulley and the rope as well as the friction on the pulley axis, the velocity of the centre of mass of this system is :

• A. $\frac{m}{M}{V}_r$
• B. $\frac{m}{2M}{V}_r$
• C. $\frac{M}{m}{V}_r$
• D. $\frac{2M}{m}{V}_r$

Answer 1

The correct option is B $\frac{m}{2M}{V}_r$

Mass of Man $=m$

=m

Man of ladder $M=m$

=M−m

If a man moves up, the ladder moves down and counterbalancing man moves up.

Velocity of man relative to ladder $=V_r$

=Vr

Velocity of ladder $=V$

=V

Velocity of man relative to ground $(V_r-V)$

$V_{cm}=\frac{MV+m\left(V_r-V\right)-\left(M-m\right)V}{2M}$

$V_{cm}=\frac{m{V}_r}{2M}$

Hence,

option $B$ is correct answer.
=(

Vr−V)

Vcm