Q.
Four particles $A, B, C$ and $D$ with masses $m_A = m, m_B = 2m, m_C = 3m $ and $m_D = 4m$ are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is :
Solution:
$\begin{array}{l} \vec{a}_{ A }=- a \hat{ j } \\ \vec{a}_{ B }= aj \\ \vec{a}_{ C }= ai \\ \vec{a}_{ D }=- aj \\ \vec{a}_{ cm }=\frac{ m _{ a } \vec{a}_{ a }+ m _{ b } \vec{a}_{ b }+ m _{ c } \vec{a}_{ c }+ m _{ d } \vec{a}_{ d }}{ m _{ a }+ m _{ b }+ m _{ c }+ m _{ d }} \\ \vec{a}_{ cm }=\frac{-m a \hat{i}+2 m a \vec{j}+3 m a \hat{i}-4 m a \hat{j}}{10 m} \\ =\frac{3 maĵ -2 ma \hat {j } }{10 m } \\ =\frac{ a _{\hat{A}}}{5}-\frac{ a _{\hat{\widehat{1}}}}{5} \\ =\frac{ a }{5}(\hat{ i }-\hat{ j }) . \\ \end{array}$
