Question

Two point masses 1 gm each are located at distances 'a' cm and 'b'cm on x and y axes respectively from
origin. Find the position of centre of mass​

Answer 1

Since the first point mass is located at a distance 'a' cm on x axis, the point mass is in the position with coordinates (a, 0).

Then the position vector of this point mass is,

$\mathbf{a}=a\ \mathbf{\hat i}$

Since the second point mass is located at a distance 'b' cm on y axis, the point mass is in the position with coordinates (0, b).

Then the position vector of this point mass is,

$\mathbf{b}=b\ \mathbf{\hat j}$

We have the expression for the center of mass of a system of particles in XY coordinate system.

$(x,\ y)=\left (\dfrac {m_1x_1+m_2x_2}{m_1+m_2},\ \dfrac {m_1y_1+m_2y_2}{m_1+m_2}\right)$

But what about if $m_1=m_2=m\ ?$

$(x,\ y)=\left (\dfrac {mx_1+mx_2}{m+m},\ \dfrac {my_1+my_2}{m+m}\right)\\\\\\(x,\ y)=\left (\dfrac {m(x_1+x_2)}{2m},\ \dfrac {m(y_1+y_2)}{2m}\right)\\\\\\(x,\ y)=\left (\dfrac {x_1+x_2}{2},\ \dfrac {y_1+y_2}{2}\right)$

Thus, about the origin, the center of mass in the system here is,

$(x,\ y)=\left (\dfrac {a+0}{2},\ \dfrac {0+b}{2}\right)\\\\\\\mathbf {(x,\ y)=\left (\dfrac {a}{2},\ \dfrac {b}{2}\right)}$

Then, the center of mass of the system is represented by the position vector,

$\mathbf{p}=\dfrac {a}{2}\ \mathbf{\hat i}\ +\ \dfrac {b}{2}\ \mathbf{\hat j}$