From a uniform disc of radius $R$, an equilateral triangle of side $\sqrt 3 \,R$ is cut as shown. The new position of centre of mass is :

The distance between the vertex and the centre of mass of a uniform solid planar circular segment of angular size $\theta$ and radius $R$ is given by

A narrow but tall cabin is falling freely near the earth's surface. Inside the cabin, two small stones $A$ and $B$ are released from rest (relative to the cabin). Initially $A$ is much above the centre of mass and $B$ much below the centre of mass of the cabin. A close observation of the motion of $A$ and $B$ will reveal that

Three particles of masses $1\,kg,\,\frac {3}{2}\,kg$ , and $2\,kg$ are located at the vertices of an equilateral triangle of side $a$ . The $x, y$ coordinates of the centre of mass are

A spherical cavity of radius $r$ is curved out of a uniform solid sphere of radius $R$ as shown in the figure below. The distance of the centre of mass of the resulting body from that of the solid sphere is given by

To find the centre of mass of rigid body why it is not possible to know$\sum {{m_i}\overrightarrow {{r_i}} } $ for all the particles ?