Question

Four point masses are connected by rods of negligible mass and form a square with sides of length 32.2 cm. Three of the masses are 1.5 kg and one is 3.0 kg. How far from the 3.0 kg mass is the center of mass of the system

Answer 1

Placing the
`$3 \ kg$` mass at the
`$\text{origin}$` and line up the square up with the axes.

`$x_(cm) = (\sum_i x_i m_i)/(\sum_i m_i )$`

`$=(1.5 a + 1.5a +0 +0)/(7.5)$`

`$=(3a)/(7.5)$`

`$=(2a)/(5)$`

`$y_(cm) = (\sum_i y_i m_i)/(\sum_i m_i )$`

`$=(1.5 a + 1.5a +0 +0)/(7.5)$`

`$=(3a)/(7.5)$`

`$=(2a)/(5)$`

Therefore, r =
`$\sqrt2 \left((2a)/(5)\right)$`

`$=(2 \sqrt2)/(5)a$`

It s given that the side of the square is a = 32.2 cm

So, r
`$=(2 \sqrt2)/(5)a$`

`$=(2 \sqrt2)/(5)* 32.2$`

= 18.21 cm

So the distance of the 3 kg mass from the center of mass, r= 18.21 cm