$A$ rod of length $3 \; m$ has a linear mass density that is directly proportional to the distance $x$ from one of its ends. Find the position of its centre of gravity from that end. (in $; m$)

$A$ uniform rod of mass $400 \ g$ and length $1 \ m$ is pivoted at one end. It is displaced by an angle of $60^o$. The increase in its potential energy is ....... $J$.

Centre of mass $(C.M.)$ of three particles of masses $1 \,kg, 2 \,kg$ and $3 \,kg$ lies at the point $(1, 2, 3)$ and $C.M.$ of another system of particles of $3 \,kg$ and $2 \,kg$ lies at the point $(-1, 3, -2)$. Where should we put a particle of mass $5 \,kg$ so that the $C.M.$ of the entire system lies at the $C.M.$ of the first system?

The centre of mass of a system of two particles of masses $m_1$ and $m_2$ separated by a distance $d$ is:

Write the difference between the centre of gravity and the centre of mass of a body.

Find the centre of mass of a uniform $L$-shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is $3 \; kg$.