Two point masses $M$ each are placed at $(L, 0)$ and $(-L, 0)$. $A$ third point mass $M$ is uniformly rotating on the circle $x^2 + y^2 = L^2$. The equation of the path traced by the $COM$ of the $3$ point masses is:

Four particles of masses $m_1 = 2m$,$m_2 = 4m$,$m_3 = m$,and $m_4$ are placed at the four corners of a square of side $a$. Let the corners be $(0,0)$,$(a,0)$,$(a,a)$,and $(0,a)$ respectively. What should be the value of $m_4$ so that the centre of mass of the system is at the centre of the square,i.e.,at $(\frac{a}{2}, \frac{a}{2})$?

Three point particles of masses $1.0 \; kg$,$1.5 \; kg$,and $2.5 \; kg$ are placed at three corners of a right-angled triangle of sides $4.0 \; cm$,$3.0 \; cm$,and $5.0 \; cm$ as shown in the figure. The center of mass of the system is at a point:

How can a general body be treated in mechanics?

What do you mean by mass element $dm$?

Write a short note on the centre of gravity.