Question

Three identical spheres each of radius 'R' are
placed on a horizontal surface touching one
another. If one of the spheres is removed, the
shift in the centre of mass of the system is
​

Answer 1

$\huge\mathfrak\red{Answer}$

$take \: one \: of \: the \: centre \: of \: the \\ \: sphere \: as \: origin \: than \: vector \\ of \: centre \: of \: mass \: (cm) \: is \: \\ as \: follows \\ \\ cm = \frac{2r(m) + 2r(m) + 0(m)}{m + m + m} \\ where \: 2r \: is \: the \: position \: vector \\ of \: the \: centre \: of \: mass \: of \: the \: two \\ spheres \\ \\ cm = \frac{4rm}{3m} \\ cm = \frac{4r}{3} ..........(1) \\ \\ when \: one \: sphere \: is \: removed \: than \: \\ this \: system \: become \: a \: two \: sphere \\ system \: keeping \: the \: origin \: of \: the \\ same \: sphere \: now \: the \: vector \: of \\ cm(2) \: will \: be \\ \\ cm(2) = \frac{2r(m) + 0(m)}{m + m} \\ cm(2) = \frac{2rm}{2m} \\ \\ cm(2) = r \\ \\ change \: in \: cm \: of \: the \: system \\ cm(2) - cm = r - \frac{4r}{3} = \frac{ - r}{3} \\ centre \: of \: mass \: shift \: toward \: the \\ center \: you \: assumed \: as \: origin$