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"Masses $M_1$ and $M_2$ are connected to a system of strings and pulleys as shown. The strings are massless and inextensible, and the pulleys are massless and frictionless. Find the acceleration of M1. (Clue if $M_2 = M_1$, A(acceleration)=g/5.)” (The problem is from “An Introduction to Mechanics” – Kleppner & Kolenkow) enter image description here

My try: Let T be the tension of the rope connected to $M_2$. So the tension of the rope connected to $M_1$ will be 2T. The acceleration of both the masses is A. enter image description here

Now, $M_2g – T = M_2A$ … (i)

$2T – M_1g = M_1A$ … (ii)

From (i) and (ii) ,

$$A=\frac{g(2M_2-M_1)}{(2M_2+M_1)}$$ Now if $M_2 = M_1$, I get, A = g/3.

But the answer is A=g/5.

Where am I wrong? Will the accelerations of $M_1$ and $M_2$ not be the same? or, are there anything about the tensions ?

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  • $\begingroup$ The tensions are correct. The problem is that you assume $a_1 = a_2$. If $M_2$ descends 1cm, how far does pulley 2 descend? $\endgroup$ – mmesser314 May 18 '14 at 13:47
  • $\begingroup$ I think, if $M_2$ descends 1cm, then pulley 2(i.e the movable one) also descends 1cm and $M_1$ ascends 1cm as well. Isn't it? But how are the accelerations different? $\endgroup$ – Siddhartha May 18 '14 at 13:58
  • $\begingroup$ The pulley would descend 0.5 cm. Think what happens if you just pull the pulley down 0.5 cm. That alone would cause $M2$ to descend 0.5 cm if the pulley did not rotate. But it does rotate, so $M2# descends farther than that. $\endgroup$ – mmesser314 May 18 '14 at 14:10
  • $\begingroup$ If the accelerations are different let they are $A_1$ and $A_2$ for M1 and M2 respectively. Then the equations are $M_2.g–T=M_2.A_2$ … (i) and $2T–M_1.g=M_1.A_1$ … (ii). Now how to solve for $A_1$? OK it's a different question; pulleys are frictionless, so why do they rotate? $\endgroup$ – Siddhartha May 18 '14 at 14:34
  • $\begingroup$ A frictionless pulley is one where the bearing at the center has no friction. The pulley rotates freely when the rope around it moves. $\endgroup$ – mmesser314 May 18 '14 at 14:51
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Start from the beginning. Why constraint relations? Why are they there? Let me emphasize :

Let's take origin at top pulley which is at rest.

enter image description here

Note that length of top rope is constant : $a+b=k\implies a''+b''=0 \implies a''=-b''$

Also Length of second rope is constant : $(c-b)+(d-b)=k\implies c''+d''=2b''$

Note that $d$ is a constant as the top pulley and ground is rest : $c''=2b''$

Hence, $c''=-2a''$ as stated in comments.

Also, everything we have done is futile and the block $M_2$ will hit the ground very quickly.

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