Question

Consider a rectangular plate of dimensions $a\times b$. If the plate is considered to be made up of four rectangles of dimensions $\frac{a}{2}\times \frac{b}{2}$ and we now remove one out of the four rectangles, find the position where the centre of mass of the remaining system will be:

Answer 1

Given,

Mass of each plate is $m$

$m_a=m_b=m_c=m_d=m$

The center of mass of each square in X, Y coordinate $C_1(\frac{a}{4},\frac{b}{4}),C_2(\frac{-a}{4},\frac{b}{4}),C_3(\frac{-a}{4},\frac{-b}{4})and{C}_4(\frac{a}{4},\frac{-b}{4})$ of the square are by,

Formula for center of mass.

$X=\frac{m_a{x}_a+m_b{x}_b+m_c{x}_c-m_d{x}_d}{m_a+m_b+m_c-m_d}$

$Y=\frac{m_a{y}_a+m_b{y}_b+m_c{y}_c-m_d{y}_d}{m_a+m_b+m_c-m_d}$

$X=\frac{[\frac{ma}{4}+\frac{m(-a)}{4}+\frac{m(-a)}{4}-\frac{ma}{4}]}{m+m+m-m}=\frac{-a}{4}$

$X=\frac{[\frac{mb}{4}+\frac{mb}{4}+\frac{m(-b)}{4}-\frac{m(-b)}{4}]}{m+m+m-m}=\frac{-b}{4}$

Hence, position of center of mass $(\frac{-a}{4},\frac{-b}{4})$