The position of the centre of mass of a uniform semi-circular wire of radius $R$ placed in the $x-y$ plane with its centre at the origin and the line joining its ends as the $x$-axis is given by $\left(0, \frac{x R}{\pi}\right)$.
Then,the value of $|x|$ is ...... .

The linear mass density of a rod of length $L$ varies as $\lambda = kx^2$,where $k$ is a constant and $x$ is the distance from one end. The position of the centre of mass of the rod is:

Two bodies $A$ and $B$ have masses $M$ and $m$ respectively,where $M > m$,and they are at a distance $d$ apart. Equal forces are applied to each of them so that they approach each other. The position where they hit each other 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:

The position vectors of three particles of masses $1\, kg, 2\, kg$,and $3\, kg$ are $\vec{r_1} = (\hat{i} + 4\hat{j} + \hat{k})\,m$,$\vec{r_2} = (\hat{i} + \hat{j} + \hat{k})\,m$,and $\vec{r_3} = (2\hat{i} - \hat{j} - 2\hat{k})\,m$ respectively. The position vector of their centre of mass is:

$A$ rigid body consists of a $3 \ kg$ mass and a $2 \ kg$ mass connected by a massless rod. The $3 \ kg$ mass is at $\vec{r}_1 = (2 \hat{i} + 5 \hat{j}) \ m$ and the $2 \ kg$ mass is at $\vec{r}_2 = (4 \hat{i} + 2 \hat{j}) \ m$. Find the length of the rod and the coordinates of the center of mass.