1.

find the COM of the system consisting of particles m1=2kg, m2=1kg, m3=4kg placed at (2,1), (3,2) and (5,4) respectively ​

Answer»

x_{com} \\ = \frac{ m_{1} x _{1} + m_{2} x_{2} + m_{3} x_{3} }{ m_{1} + m_{2} + m_{3} } \\ = \frac{(2)(2) + (1)(3) + (4)(5)}{2 + 1 + 4} = \frac{4 + 3 + 20}{7} \\ = \frac{27}{7} \\ \\ now \\ \\ y_{com} \\ = \frac{ m_{1} y _{1} + m_{2} y_{2} + m_{3} y_{3} }{ m_{1} + m_{2} + m_{3} } \\ = \frac{(2)(1) + (1)(2) + (4)(4)}{2 + 1 + 4} = \frac{2 + 2 + 16}{7} \\ = \frac{20}{7}

HENCE, coordinates of the CENTRE of MASS of the GIVEN SYSTEM are (27/7, 20/7).


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