Three equal masses are placed at $(0,0)$,$(a,0)$,and $\left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right)$. Find the coordinates of the center of mass.

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 $A$).

Two particles whose masses are $10\,kg$ and $30\,kg$ and their position vectors are $\hat{i} + \hat{j} + \hat{k}$ and $-\hat{i} - \hat{j} - \hat{k}$ respectively would have the centre of mass at:

In the figure,one-fourth part of a uniform disc of radius $R$ is shown. The distance of the centre of mass of this object from the centre $O$ is:

$A$ slender uniform rod of length $L$ is balanced vertically at a point $P$ on a horizontal surface having some friction. If the top of the rod is displaced slightly to the right,the position of its centre of mass at the time when the rod becomes horizontal is:

What do you mean by mass element $dm$?