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 center 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

Answer: (B)

$\frac{m}{2M}V_r$

Answer 2

The masses of load, ladder and man are M, M-m and m respectively. Their velocities are V(upward), -V and $V_r$ -V respectively $\therefore$ $V_{cm} = \frac{\sum m_1 V_1}{\sum m_1}$ = $\frac{M(\nu) + (M - m )(-\nu) +m(V_r - \nu)}{2M}$ = $\frac{m}{2M } V_r$