Question

Masses $\text{8, 2, 4, 2 }kg\$are placed at the corners $A,\ B,\ C,\ D$ respectively of a square $ABCD$ of diagonal$80cm$. The distance of centre of mass from $A$ will be

• A. $20cm$
• B. $30cm$
• C. $40cm$
• D. $60cm$

Answer 1

Answer: (B)

$30cm$

Answer 2

Let corner A of square ABCD is at the origin and the mass 8 kg is placed at this corner (given in problem) Diagonal of square $d = a\sqrt{2} = 80cm$ ⇒ $a = 40\sqrt{2}cm$

$m _ { 1 } = 8 \mathrm {~kg}$ $m_{2} = 2kg,$ $m_{3} = 4kg,$ $m_{4} = 2kg$

Let ${\overset{\rightarrow}{r}}_{1},{\overset{\rightarrow}{r}}_{2},{\overset{\rightarrow}{r}}_{3},{\overset{\rightarrow}{r}}_{4}$ are the position vectors of respective masses

${\overset{\rightarrow}{r}}_{1} = 0\widehat{i} + 0\widehat{j}$, $\overset{\rightarrow}{r_{2}} = a\widehat{i} + 0\widehat{j}$, $\overset{\rightarrow}{r_{3}} = a\widehat{i} + a\widehat{j}$, $\overset{\rightarrow}{r_{4}} = 0\widehat{i} + a\widehat{j}$

From the formula of centre of mass

$\overset{\rightarrow}{r} = \frac{m_{1}\overset{\rightarrow}{r_{1}} + m_{2}\overset{\rightarrow}{r_{2}} + m_{3}\overset{\rightarrow}{r_{3}} + m_{4}\overset{\rightarrow}{r_{4}}}{m_{1} + m_{2} + m_{3} + m_{4}} = 15\sqrt{2}i + 15\sqrt{2}\widehat{j}$

$\therefore$ co-ordinates of centre of mass $= (15\sqrt{2},15\sqrt{2})$ and co-ordination of the corner $= \lbrack 0,0\rbrack$

From the formula of distance between two points $(x_{1},y_{1})$ and $(x_{2},y_{2})$

distance $= \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

= $\sqrt{(15\sqrt{2} - 0)^{2} + (15\sqrt{2} - 0)^{2}}$ =$\sqrt{900}$ = $30cm$