# Why changing the mass of forth object does not change the center of mass in the given problem?

The problem is as follows. Four masses of $$1$$ $$kg$$, $$2$$ $$kg$$, $$3$$ $$kg$$, and $$4$$ $$kg$$ are arranged in square shape. The side length of the square is $$1$$ $$m$$. Find the location of the center of mass of this system.

I have found the solution to it to be ($$1/2$$, $$3/10$$) by representing the masses as points on the cortisone plane like here. However in the calculations, $$4$$ $$kg$$ is always multiplied by zero, so that made me wonder if the center of mass is the same even if the fourth object is $$1000$$ $$kg$$. Why is that?

## 2 Answers

No, It would change. Note the expression $$\mathbf{r}_\text{cm}=\frac{\sum_i m_i\mathbf{r}_i}{\sum_im_i}=\frac{m_1\mathbf{r}_1+\sum_im_i\mathbf{r}_i}{m_1+\sum_i m_i}$$ If $$\mathbf{r}_1=0$$, $$\mathbf{r}_\text{cm}=\frac{\sum_{i=2}m_i\mathbf{r}_i}{m_1+\sum_{i=2}m_i}$$ Note $$m_1$$ in the denominator.

In the figure, the origin is taken at the position of the 4kg point mass. Now the coordinates of the origin are $$(0,0)$$ i.e. for the 4kg mass, the position is $$x=0$$ and $$y=0$$. As a result, the 4kg is multiplied with zero in the calculation of the centre of mass of the system.

This basically depends upon the choice of the origin and the coordinate axes. For example, had the origin been selected at the position of the 3kg point mass, we would have to multiply 3kg with zero in that case and the coordinates of the centre of mass would be based on the chosen coordinates system. The coordinates of the centre of mass could have been different from what you have obtained in that case. But it would mean the same location. There is nothing much conceptual about it; it's just about the choice of the coordinate system.

Hope it helps.