The center of mass of a body or a system of bodies is a mean position of the total weight of the body where the resultant of the forces applied is considered to be acted upon such that forces, momentum and energy are conserved. The body or system of bodies is balanced around the center of mass and the average of the weighted position coordinates defines its coordinates.

Example: A non-uniform rod having mass per unit length as $\mu =ax$ (a is constant). If its total mass is M and length L.Find position of the centre of mass.

Solution:
Choose a coordinate system with the rod aligned along the x-axis and origin located at the left end of the rod. Choose an infinitesimal mass
element dm located a distance x'. Let the length of the mass element be dx'.
Thus
$dm=\mu (x^{\prime })d{x}^{\prime }$
The total mass is found by integrating the mass element over the length of the rod 
$M={\int }_0^L\mu (x^{\prime })d{x}^{\prime }=a{\int }_0^L{x}^{\prime }d{x}^{\prime }=\frac{a}{2}{x}^{\prime 2}{\mid }_0^L=\frac{a}{2}{L}^2$
or
$a=\frac{2M}{L^2}$
Now center of mass is calculated as

$x_{cm}=\frac{1}{M}{\int }_{body}xdm=\frac{1}{M}{\int }_0^L\mu (x^{\prime })x^{\prime }d{x}^{\prime }=\frac{a}{M}{\int }_0^L{x}^{\prime 2}d{x}^{\prime }$

substituting the value of a
$X_{cm}=\frac{2}{L^2}{\int }_0^L{x}^{\prime 2}d{x}^{\prime }=\frac{2}{3L^2}{x}^{\prime 3}{\mid }_0^L=\frac{2}{3L^2}(L^3-0)=\frac{2}{3}L$

Example: In the figure shown, find out the distance of centre of mass of a system of a uniform circular plate of radius $3R$ from O. On this plate, a hole
of radius $R$ is cut whose centre is at $2R$ distance from centre of the
plate.

Solution: From law of conservation of momentum,
$\bar{x}=\frac{m_1{x}_1+\left(-m_2\right)x_2}{m_1+\left(-m_2\right)}$

$=\frac{A_1{x}_1+\left(-A_2\right)x_2}{A_1+\left(-A_2\right)}$

$A_1=\pi {\left(3R\right)}^2$,
$A_2=\pi R^2$

$x_1=0$,
$x_2=2R$
$\therefore \bar{x}=-R/4$

Definition:
Moment of inertia of a rigid body about an axis of rotation is defined as the sum of product of the mass of each particle and the square of its perpendicular distance from the axis of rotation.

$\therefore I={\sum }_{i=1}^n{m}_i{r}_i^2$

The instantaneous axis of rotation is the axis fixed to a body undergoing planar movement that has zero velocity at a particular instant of time. At this instant, the velocity vectors of the trajectories of other points in the body generate a circular field around this axis which is identical to what is generated by a pure rotation.

For a body performing pure rotation instantaneous axis of rotation lies at the point of contact as given in the above figure.

The perpendicular axis theorem can be used to determine the moment of inertia of a rigid object that lies entirely within a plane, about an axis perpendicular to the plane, given the moments of inertia of the object about two perpendicular axes lying within the plane.
$I_x=I_y+I_z$

The direction of torque is always perpendicular to plane of rotation of body as a cross product is in perpendicular plane to $r$ and $F$ vectors and the torques produced by two forces of couple are in same direction to each other.

Formula to calculate couple is:
$Couple=F\times l$
Where $l$ is perpendicular distance between forces.

Translational Equilibrium: $\sum F=0$ (that is, $F_{net}=0$).
An object may be rotating, even rotating at a changing rate, but may be in translational equilibrium if the acceleration of the center of mass of the object is still zero.
Rotational Equilibrium: $\sum \tau =0$; (${\tau }_{net}=0$). 
An object may be accelerating in a linear fashion (along a straight line and or even turning at a constant rate; an object in rotational equilibrium will NOT be accelerating in a rotational sense (ie. the angular momentum of an object in rotational equilibrium will be constant).

Rolling without slipping is a combination of translation and rotation where the point of contact is instantaneously at rest. When an object experiences pure translational motion, all of its points move with the same velocity as the center of mass; that is in the same direction and with the same speed.

For a ball of radius $r$ moving with translational velocity $v$ and rotating with angular velocity $w$ condition for pure rolling is:
$v=wr$

Example: A sphere has to purely roll upwards. At an instant when the velocity of sphere is $v$, what is the frictional force acting on it?

Solution:
Acceleration of a body rolling (pure)  down the inclined plane ,   $a=\frac{gsin\theta }{1+K}$                            
For  solid sphere     $K=\frac{2}{5}$
Let assume friction force to act upwards.Equation of motion,     
$ma=mgsin\theta -f$    
 $\frac{5}{7}mgsin\theta =mgsin\theta -f$$⟹f=\frac{2}{7}mgsin\theta$ 
As $f$ comes out to be positive thus our assumption is correct.

Example: What will be the work done in rotating a body from angle ${\theta }_1$ to angle ${\theta }_2$ by a constant torque $\tau$ ?

Solution: Work done in angular displacement is given as
${\int }_{{\theta }_1}^{{\theta }_2}dW={\int }_{{\theta }_1}^{{\theta }_2}\tau d\theta$. For constant terms we get the relation as $W=\tau ({\theta }_2-{\theta }_1)$

Rolling motion:
An important case of combined translation and rotation is rolling without slipping, when a body like a sphere, the wheel rolls on a surface, the motion can be treated as the combination of both translational motions of the centre of mass and rotational motion about an axis passing through the centre of mass.

When a torque is applied on a body resisting change of motion (usually due to friction), the instantaneous axis of rotation shifts (usually due to shift of normal). The body topples if the instantaneous axis of rotation falls outside the body. This is shown in the attached diagram.

Example: A cubical block of mass $m$ and side $L$ rests on a rough horizontal surface with coefficient of friction $\mu$. A horizontal force $F$ is applied on the block as shown in the figure. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, find the minimum force required to topple the block.

Solution:
Considering case 1, when minimum force is applied,
$F=\mu N$
$N=mg$
$\therefore F=\mu mg$
Now,
when the force exceeds some value, block topples
Taking $\sum$ moment about the toppling axis,
$\sum M=0$ ................ Still in equilibrium
$\therefore \sum M=\sum F.d$
$\Rightarrow 0=F\times L-mg\times \frac{L}{2}$
$\therefore F=\frac{mg}{2}$