Write a short note on the centre of gravity.

(N/A) The point at which the entire weight of a body may be considered as concentrated is known as the centre of gravity $(CG)$.
Take an irregularly shaped cardboard and a narrow-tipped object like a pencil. Locate point $G$ on the cardboard where it can be balanced on the tip of the pencil. This point of balance is the centre of gravity $(CG)$ of the cardboard.
The tip of the pencil provides a vertically upward force due to which the cardboard is in mechanical equilibrium. The reaction of the tip is equal and opposite to $Mg$,the total weight of the cardboard,and hence the cardboard is in translational equilibrium.
There are torques on the cardboard due to the force of gravity. If the total torque on it due to the downward forces is zero,then the cardboard remains in rotational equilibrium.
If $m_{i}$ is the mass of the $i^{\text{th}}$ particle of the cardboard and $\vec{r}_{i}$ is the position vector of the $i^{\text{th}}$ particle with respect to the centre of gravity,then the torque of gravity on the particle is $\vec{\tau}_{i} = \vec{r}_{i} \times (m_{i} \vec{g})$.
The total gravitational torque on the body about the centre of gravity is zero.
$\therefore \vec{\tau}_{g} = \sum \vec{\tau}_{i} = \sum (\vec{r}_{i} \times m_{i} \vec{g}) = (\sum m_{i} \vec{r}_{i}) \times \vec{g} = \vec{0}$.
Since $\sum m_{i} \vec{r}_{i} = 0$ at the centre of gravity,the cardboard remains in rotational equilibrium.