(9am to 6pm)

Ask Questions, Get Answers

Want help in doing your homework? We will solve it for you. Click to know more.

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 velocity $V_r$ relative to the ladder. Ignoring the masses of the pulley as well as the friction on the pulley axis, the velocity of the centre of mass of this system is

\[\begin {array} {1 1} (a)\;\frac{m}{M}Vr & \quad (b)\;\frac{m}{2M}Vr \\ (c)\;\frac{M}{m}V_r & \quad  (d)\;\frac{2M}{m}Vr \end {array}\]

1 Answer

Need homework help? Click here.
Mass of Man $=m$
Man of ladder $=M-m$
If man moves up, ladder moves down and counter balancing man moves up.
Velocity of man relative to ladder $=V_r$
Velocity of ladder $= V$
Velocity of man relative to ground $=(V_r-V)$
$V_{cm} =\large\frac{mv_r}{2M}$
answered Nov 22, 2013 by meena.p

Related questions