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Question
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 l′ with respect to the ladder and then stops. Neglecting the mass of the rope and the friction in the pulley axle, find the displacement l of the centre of inertia of this system.
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Solution
In the reference frame fixed to the pulley axis the location of C.M. of the given system is described by the radius vector
Δ→rC=MΔ→rM+(M−m)Δ→r(M−m)+mΔ→rm2M
But Δ→rM=−Δ→r(M−m)
and Δ→rm=Δ→rm(M−m)+Δ→r(M−m)
Thus Δ→rC=m→l′2M
Note: one may also solve this problem using momentum conservation.
Δ→rC=MΔ→rM+(M−m)Δ→r(M−m)+mΔ→rm2M
But Δ→rM=−Δ→r(M−m)
and Δ→rm=Δ→rm(M−m)+Δ→r(M−m)
Thus Δ→rC=m→l′2M
Note: one may also solve this problem using momentum conservation.
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