The $x, y$ coordinates of the centre of mass of a uniform $L-$ shaped lamina of mass $3\, kg$ is
$\left( {\frac{5}{6}\,m,\frac{5}{6}\,m} \right)$
$(1\,m, 1\,m)$
$\left( {\frac{6}{5}\,m,\frac{6}{5}\,m} \right)$
$(2\,m, 2\,m)$
$A$ slender uniform rod of length $\lambda$ 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 :
The position vector of three particles of masses $1\, kg, 2\, kg$ and $3\, kg$ are $\overrightarrow {{r_1}} = (\widehat i + 4\widehat j + \widehat k)\,m,\overrightarrow {{r_2}} = (\widehat i + \widehat j + \widehat k)\,m$ and $\overrightarrow {{r_3}} = (2\widehat i - \widehat j - 2\widehat k)\,m$ respectively. The position vector of their centre of mass is
Mass is distributed uniformly over a thin square plate. If two end points of diagonal are $(-2, 0)$ and $(2, 2)$, what are the coordinates of centre of mass of plate ?
Four particles $A, B, C$ and $D$ with masses $m_A=m, m_B=2m, m_C=3m$ and $m_D=4m$ are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is