Question

Four particles are placed at the corners of a square as shown in figure. Acceleration of the centre of mass of the particle (in ${\text{m/s}}^2$) is

• A. $\frac{a}{5}(\hat{i}-\hat{j})$
• B. $a(\hat{i}+\hat{j})$
• C. zero
• D. $\frac{a}{5}(\hat{i}+\hat{j})$

Answer 1

The correct option is A $\frac{a}{5}(\hat{i}-\hat{j})$


For a system of particles, the acceleration of the centre of mass is given as,
${\overrightarrow{a}}_{\text{com}}=\frac{m_A{\overrightarrow{a}}_A+m_B{\overrightarrow{a}}_B+m_C{\overrightarrow{a}}_C+m_D{\overrightarrow{a}}_D}{m_A+m_B+m_C+m_D}$
Substituting the acceleration vector as per the given coordinate system,
$\Rightarrow {\overrightarrow{a}}_{\text{com}}=\frac{m(-a\hat{i})+2m\left(a\hat{j}\right)+3m\left(a\hat{i}\right)+4m(-a\hat{j})}{m+2m+3m+4m}$
$\Rightarrow {\overrightarrow{a}}_{\text{com}}=\frac{2ma\hat{i}-2ma\hat{j}}{10m}$
$\Rightarrow {\overrightarrow{a}}_{\text{com}}=\frac{a}{5}(\hat{i}-\hat{j}){\text{m/s}}^2$