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Question

Consider the pulley-block system shown in the figure below. If the system is released from rest, then identify the correct statement regarding acceleration of the centre of mass of the system. Assume Mm.



A
It is in vertically upward direction.
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B
It is in vertically downward direction.
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C
The direction of acceleration of centre of mass depends on which mass is greater.
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D
The centre of mass does not accelerate.
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Solution

The correct option is B It is in vertically downward direction.

Let M>m. Then the direction of the accelerations of the blocks will be as shown in the figure. To validate string constraint, both blocks will have same magnitude of acceleration i.e a.

Applying Newton's 2nd law in direction of acceleration of the blocks,

MgT=Ma ...(i)
Tmg=ma ...(ii)
Adding Eq (i) & (ii),
(Mm)g=(M+m)a
a=(Mm)g(M+m) ....(iii)

For the acceleration of centre of mass of the system,
aCM=m1a1+m2a2m1+m2
Considering downward direction as +ve, m1=M, m2=m, a1=+a, a2=a
aCM=Mama(M+m)
aCM=(MmM+m)a
=(MmM+m)×(MmM+m)g
(from (iii))
aCM=((Mm)2(M+m)2)g
The numerator and denominator both are square terms (always +ve, since Mm).
Hence +ve sign of aCM represents that acceleration of centre of mass is in vertically downward direction (along g), no matter whether (M>m) or (M<m).

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