Question

Four particles of masses `m_(1) = 1 kg, m_(2) = 2 kg, m_(3) = 3 kg and m_(4) = 4 kg` are located at the corners of a rectangle as shown in Fig. Locate the position of centre of mass.

Answer 1

Here, `m_(1) = 1kg, m_(2) = 2 kg, m_(3) = 3 kg`,
`m_(4) = 4kg`
As is clear form Fig. `m_(1)` is taken at the origin. Therefore, `x_(1) = 0, y_(1) = 0`.
For `m_(2) , x_(2) = a hati, y_(2) = 0`
For `m_(3) , x_(3) = a hati , y_(3) = b hatj`
For `m_(4) , x_(4) = 0, y_(4) = b hatj`
Co-ordinates of centre of mass of the system
are `x = (m_(1)x_(1) + m_(2)x_(2) + m_(3)x_(3) + m_(4)x_(4))/(m_(1) + m_(2) + m_(3) + m_(4))`
`= (1 xx 0 + 2 (a hati) + 3 (a hati) + 4 xx 0)/(1 + 2+ 3+ 4) = 0.5 a hati`
`y = (m_(1)y_(1) + m_(2)y_(2) + m_(3)y_(3) + m_(4)y_(4))/(m_(1) + m_(2) + m_(3) + m_(4))`
`= (1 xx 0 + 2 xx 0 + 3 xx b hatj + 4 xx b hatj)/(1 + 2+ 3+ 4) = 0.7 b hatj`
Position of centre of mass of the system is
`[0.5 a hatj + 0.7 b hatj]`