The sum of moments of all the particles in a system about its center of mass is always . . . . . .

For determining the centre of mass of a body,why is a body considered as being composed of multiple small mass elements?

The distance of the centre of mass from end $A$ of a one-dimensional rod $(AB)$ having mass density $\rho = \rho_{0} \left(1 - \frac{x^{2}}{L^{2}}\right) \text{ kg/m}$ and length $L$ (in meters) is $\frac{3L}{\alpha} \text{ m}$. The value of $\alpha$ is $\ldots \ldots \ldots$ (where $x$ is the distance from end $A$).

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 $(\frac{a}{2}, \frac{a}{2})$?

Mass is distributed uniformly over a thin square plate. If two end points of a diagonal are $(-2, 0)$ and $(2, 2)$,what are the coordinates of the centre of mass of the plate?

The position vectors of two particles of masses $1 \ kg$ and $3 \ kg$ are $\hat{i} + 2\hat{j} + \hat{k}$ and $-3\hat{i} - 2\hat{j} + \hat{k}$ respectively. The position vector of the center of mass of this system is: