Four particles of masses m_1 = 2m,, m_2 = 4m, | vedclass

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Question

Let side of square is $a$

$x_{c n}=\frac{a}{2}=\frac{m_{1} 	imes 0+m_{2} 	imes a+m_{3} 	imes a+m_{4} 	imes a}{m_{1}+m_{2}+m_{3}+m_{4}}$

solv

Vedclass · Accepted Answer

None of these

Vedclass · Answer

Let side of square is $a$

$x_{c n}=\frac{a}{2}=\frac{m_{1} 	imes 0+m_{2} 	imes a+m_{3} 	imes a+m_{4} 	imes a}{m_{1}+m_{2}+m_{3}+m_{4}}$

solve to get $m_{4}=3 m$ $y_{c n}=\frac{a}{2}=\frac{m_{1} 	imes 0+m_{2} 	imes 0+m_{3} 	imes a+m_{4} 	imes a}{m_{1}+m_{2}+m_{3}+m_{4}}$

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.

Four particles of masses $m_1 = 2m,\, m_2 = 4m,\, m_3 = m$ 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?

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