Question

Four particle of masses $m_{1}=2 m , m_{2}$ $=4 m , m_{3}=m$ and $m_{4}$ are placed at four corners of a square. What should be the value of $m_{4}$ so that the centres of mass of all the four particles are exactly at the centre of the square?

• A. 3 m
• B. 5 m
• C. 8 m
• D. none of these

Answer 1

Answer: (D)

Let side of square is $a$.
$x_{c m}=\frac{a}{2}=\frac{m_{1} \times 0+m_{2} \times a+m_{3} \times a+m_{4} \times 0}{m_{1}+m_{2}+m_{3}+m_{4}}$
solve to get $m_{4}=3 m$
$y_{c m}=\frac{a}{2}=\frac{m_{1} \times 0+m_{2} \times 0+m_{3} \times a+m_{4} \times a}{m_{1}+m_{2}+m_{3}+m_{4}}$
solve to get $m_{4}=5 m$
They are contradicting, so it is not possible to have centre of mass at the centre of square.