0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
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 M≠m.

A
It is in vertically upward direction.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
It is in vertically downward direction.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
The direction of acceleration of centre of mass depends on which mass is greater.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
The centre of mass does not accelerate.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
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, Mg−T=Ma ...(i) T−mg=ma ...(ii) Adding Eq (i) & (ii), (M−m)g=(M+m)a ⇒a=(M−m)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=Ma−ma(M+m) ⇒aCM=(M−mM+m)a =(M−mM+m)×(M−mM+m)g (from (iii)) ∴aCM=((M−m)2(M+m)2)g The numerator and denominator both are square terms (always +ve, since M≠m). 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).

Suggest Corrections
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program