Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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

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

Can you answer this question?

1 Answer

0 votes
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

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App