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 $M\ne m$.

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

Answer 1

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 $2^{nd}$ law in direction of acceleration of the blocks,

$Mg-T=Ma...\left(i\right)$
$T-mg=ma...\left(ii\right)$
Adding Eq $\left(i\right)\&\left(ii\right)$,
$(M-m)g=(M+m)a$
$\Rightarrow a=\frac{(M-m)g}{(M+m)}....\left(iii\right)$

For the acceleration of centre of mass of the system,
$a_{\text{CM}}=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}$
Considering downward direction as $+ve$, $m_1=M,m_2=m,a_1=+a,a_2=-a$
$\Rightarrow a_{\text{CM}}=\frac{Ma-ma}{(M+m)}$
$\Rightarrow a_{\text{CM}}=\left(\frac{M-m}{M+m}\right)a$
$=\left(\frac{M-m}{M+m}\right)\times \left(\frac{M-m}{M+m}\right)g$
(from (iii))
$\therefore a_{\text{CM}}=\left(\frac{(M-m{)}^2}{(M+m{)}^2}\right)g$
The numerator and denominator both are square terms (always $+ve$, since $M\ne m$).
Hence $+ve$ sign of $a_{\text{CM}}$ represents that acceleration of centre of mass is in vertically downward direction (along $g$), no matter whether $(M>m)$ or $(M<m)$.