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Displacement of centre of mass 
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#1
Sep1312, 02:15 AM

PF Gold
P: 824

1. The problem statement, all variables and given/known data
A pulley fixed to a rigid support carries a rope whose one end is tied to a ladder with a man and the other end to a counterweight of mass M. The man of mass m climbs up a distance h with respect to the ladder and then stops. If the mass of the rope and the friction in the pulley axle are negligible, find the displacement of the centre of mass of this system. 2. Relevant equations COM[itex]_{y}[/itex]=my[itex]_{1}[/itex]+my[itex]_{2}[/itex]/m+M 3. The attempt at a solution Let the ycoordinate of man be l and that of mass M be 0. Since it goes up a height h ∴ ycoordinate of the man changes to (l+h) but that of mass M remains unchanged. So, if I subtract the final COM from initial COM I get the answer mh/m+M. But the correct answer is mh/2M. I'm unable to bring out mistake in my solution. 


#2
Sep1312, 02:57 AM

HW Helper
P: 4,437

Before man starts climbing the ladder, whether the system is in equilibrium? If yes, you have to take into account the mass of the ladder, which is M m.



#3
Sep1312, 08:04 AM

PF Gold
P: 824

Oh, I forgot that the system is in equilibrium. I must have noticed the word counterweight. This is the key to the answer. Thanks so much as now I have arrived at the correct answer.



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