3.0 Centre of mass of the remaining portion

3.1 Motion of mass of the system

Consider a thin square plate of side $L$ and mass $M$. The square plate is divided into 4 equal sections as shown in the figure.

Side length and mass of each square section is $\left( {\frac{L}{2}} \right)$ and $\left( {\frac{M}{4}} \right)$ respectively.

Now, consider if one of the section is removed.

Then the question arises, whether the position of centre of mass will change or remains same.

If changes what will be the new position.

Yes, the position of centre of mass of the remaining portion will change.

As shown in the figure, remaining and removed portion are considered as different individual sections.

Now, from the concept of centre of mass for the system of particles,

$${\overrightarrow r _{COM}} = \frac{{{m_1}{{\overrightarrow r }_1} + {m_2}{{\overrightarrow r }_2} + ... + {m_n}{{\overrightarrow r }_n}}}{{{m_1} + {m_2} + ... + {m_n}}}$$ or $${\overrightarrow r _{OP}} = \frac{{{m_{RP}}{{\overrightarrow r }_{RP}} + {m_{RM}}{{\overrightarrow r }_{RM}}}}{{{m_{RP}} + {m_{RM}}}}\quad ...(iv)$$ where,

$OP$: Original portion

$RP$: Remaining portion

$RM$: Removed portion

From equation $(i)$, $(ii)$, $(iii)$ & $(iv)$ we get, $$\begin{equation} \begin{aligned} \left( {\frac{L}{2}\widehat i + \frac{L}{2}\widehat j} \right) = \frac{{\left( {\frac{{3M}}{4}} \right)\left( {a\widehat i + b\widehat j} \right) + \left( {\frac{M}{4}} \right)\left( {\frac{{3L}}{4}\widehat i + \frac{{3L}}{4}\widehat j} \right)}}{{\frac{{3M}}{4} + \frac{M}{4}}} \\ \left( {\frac{L}{2}\widehat i + \frac{L}{2}\widehat j} \right) = \left( {\frac{{3a}}{4} + \frac{{3L}}{{16}}} \right)\widehat i + \left( {\frac{{3b}}{4} + \frac{{3L}}{{16}}} \right)\widehat j \\ \left( {\frac{{3a}}{4} + \frac{{3L}}{{16}}} \right) = \frac{L}{2}\quad \ \quad \left( {\frac{{3b}}{4} + \frac{{3L}}{{16}}} \right) = \frac{L}{2} \\\end{aligned} \end{equation} $$ or $$a = \frac{{5L}}{{12}}\quad \& \quad b = \frac{{5L}}{{12}}$$