The rolling motion is the most commonly observed motion in daily life. The motion of wheel is an example of rolling motion. Round objects like ring, disc, sphere etc. are most suitable for rolling. Let us study the rolling of a disc on a horizontal surface. Consider a point P on the edge of the disc. While rolling, the point undergoes translational motion along with its center of mass and rotational motion with respect to its center of mass.
Combination of Translation and Rotation: We will now see how these translational and rotational motions are related in rolling. If the radius of the rolling object is R, in one full rotation, the center of mass is displaced by 2\(\pi\)R (its circumference). One would agree that not only the center of mass, but all the points on the disc are displaced by the same 2\(\pi\)R after one full rotation. The only difference is that the center of mass takes a straight path; but, all the other points undergo a path which has a combination of the translational and rotational motion. Especially the point on the edge undergoes a path of a cycloid as shown in the figure.

As the center of mass takes only a straight line path, its velocity v_CM is only translational velocity v_TRANS (v_CM = v_TRANS)· All the other points have two velocities. One is the translational velocity v_TRANS, (which is also the velocity of center of mass) and the other is the rotational velocity v_ROT (v_ROT = r\(\omega\)). Here, r is the distance of the point from the center of mass and eo is the angular velocity. The rotational velocity v_ROT is perpendicular to the instantaneous position vector from the center of mass as shown in figure (a). The resultant of these two velocities is v. This resultant velocity v is perpendicular to the position vector from the point of contact of the rolling object with the surface on which it is rolling as shown in figure (b).

We shall now give importance to the point of contact. In pure rolling, the point of the rolling object which comes in contact with the surface is at momentary rest. This is the case with every point that is on the edge of the rolling object. As the rolling proceeds, all the points on the edge, one by one come in contact with the surface; remain at momentary rest at the time of contact and then take the path of the cycloid as already mentioned.
Hence, we can consider the pure rolling in two different ways. (i) The combination of translational motion and rotational motion about the center of mass. (or) (ii) The momentary rotational motion about the point of contact. As the point of contact is at momentary rest in pure rolling, its resultant velocity v is zero (v = 0). For example, in figure, at the point of contact, v_TRANS is forward (to right) and v_ROT is backwards (to the left).

That implies that, v_TRANS and v_ROT are equal in magnitude and opposite in direction (v = v_TRANS -v_ROT = 0). Hence, we conclude that in pure rolling, for all the points on the edge, the magnitudes of v_TRANS and v_ROT are equal (v_TRANS = v_ROT)· As v_TRANS = v_CM and v_ROT = \(R_\omega\), in pure rolling we have,
\(V_{CM}=R_\omega\)

We should remember the special feature of the above equation. In rotational motion, as per the relation v = r\(\omega\), the center point will not have any velocity as r is zero. But in rolling motion, it suggests that the center point has a velocity v_CM given by above equation V_CM - R\(\omega\). For the topmost point, the two velocities v_TRANS and v_ROT are equal in magnitude and in the same direction (to the right). Thus, the resultant velocity v is the sum of these two velocities, v = v_TRANS + v_ROT· In other form,v= 2 v_CM as shown in figure below.

Let us consider a rigid body rotating with angular velocity \(\omega\) about an axis as shown in figure. Every particle of the body will have the same angular velocity \(\omega\) and different tangential velocities v based on its positions from the axis of rotation.
Let us choose a particle of mass mi situated at distance r_i from the axis of rotation. It has a tangential velocity v_i given by the relation, v_i = r_i \(\omega\). The kinetic energy KE_i of the particle is,
KE_i =  \(\frac { 1 }{ 2 } { m }_{ i }{ v }_{ i }^{ 2 }\)
Writing the expression with the angular velocity,
\(KE=\frac { 1 }{ 2 } { m }_{ i }\left( { r }_{ i }\omega \right) ^{ 2 }=\frac { 1 }{ 2 } \left( { m }_{ i }{ r }_{ i }^{ 2 } \right) \omega ^{ 2 }\)

For the kinetic energy of the whole body, which is made up of large number of such particles, the equation is written with summation as,
\(KE=\frac { 1 }{ 2 } \left( \sum { { m }_{ i }{ r }_{ i }^{ 2 } } \right) \omega ^{ 2 }\)
where, the term \(\sum { { m }_{ i }{ r }_{ i }^{ 2 } } \) is the moment of interiaI of the whole body. \(\sum { { m }_{ i }{ r }_{ i }^{ 2 } } \)
Hence, the expression for KE of the rigid body in rotational motion is,
KE = \(\frac{1}{2}\)I\(\omega^2\)
This is analogous to the expression for kinetic energy in translational motion.
KE = \(\frac{1}{2}\)Mv²
Relation between rotational kinetic energy and angular momentum
Let a rigid body of moment of inertia \(\omega\) rotate with angular velocity \(\omega\).
The angular momentum of a rigid body is, L = I\(\omega\)
The rotational kinetic energy of the rigid body is, KE = \(\frac{1}{2}I\omega^2\)
By multiplying the numerator and denominator of the above equation with I, we get a relation between Land KE as,
KE = \(\frac{1}{2}\)\(\frac { I^{ 2 }\omega ^{ 2 } }{ I } =\frac { 1 }{ 2 } \frac { \left( I\omega \right) ^{ 2 } }{ I } \)
\(KE=\frac { { L }^{ 2 } }{ 2I } \)

A point mass is a hypothetical point particle which has non-zero mass and no size or shape. To find the center of mass for a collection ofn point masses, say, m₁, m₂, m₃ ... m_n we have to first choose an origin and an appropriate co-ordinate system as shown in Figure. Let, x₁, x₂, x₃ ... x_n be the X-coordinates of the positions of these point masses in the X direction from the origin.

The equation for the X coordinate of the center of mass is,
\({ x }_{ cm }=\frac { \sum { { m }_{ i }{ x }_{ i } } }{ \sum { { m }_{ i } } } \)
where, \(\sum { { m }_{ i } } \) is the total mass M of all the particles, (\(\sum { { m }_{ i } } =M\)). Hence, 
\({ x }_{ CM }=\frac { \sum { { m }_{ i }{ x }_{ i } } }{ M } \)
Similarly, we can also find y and z coordinates of the center of mass for these distributed point masses as indicated in Figure.
\({ y }_{ CM }=\frac { \sum { { m }_{ i }{ y }_{ i } } }{ M } \)
\({ z }_{ CM }=\frac { \sum { { m }_{ i }{ z }_{ i } } }{ M } \)
Hence, the position of center of mass of these point masses in a Cartesian coordinate system is (x_CM, y_CM z_CM) In general, the position of center of mass can be written in a vector form as,
\(\overset { \rightarrow }{ r_{ CM } } =\frac { \sum { { m }_{ i } } \overset { \rightarrow }{ r } i }{ M } \)
where,  \(\overset { \rightarrow }{ r_{ CM } } ={ x }_{ CM }\hat { i } +{ y }_{ CM }\hat { j } +{ z }_{ CM }\hat { k } \) is the position vector of the center of mass and \(\overset { \rightarrow }{ { r }_{ i } } ={ x }_{ i }\hat { i } +{ y }_{ i }\hat { j } +{ z }_{ i }\hat { k } \) is the position vector of the distributed point mass; where \(\hat { i } ,\hat { j } \) and \(\hat { k } \) are the unit vectors along X, Y and Z-axes respectively.

With point of contact as reference: We can also arrive at the same expression by taking the momentary rotation happening with respect to the point of contact (another approach to rolling). If we take the point of contact as O, then,
\(KE=\frac { 1 }{ 2 } { I }_{ 0 }{ \omega }^{ 2 }\)
Here, I₀ is the moment of inertia of the object about the point of contact. By parallel axis theorem, I₀= I_CM+MR² .Further we can write, I₀= MK² + MR². With V_CM=R\(\omega\) or \(\omega\) =\(\frac { { v }_{ CM } }{ R } \)
\(KE=\frac { 1 }{ 2 } \left( { MK }^{ 2 }+{ MR }^{ 2 } \right) \frac { { v }_{ CM }^{ 2 } }{ R } \)
\(KE=\frac { 1 }{ 2 } { Mv }_{ CM }^{ 2 }\left( 1+\frac { { K }^{ 2 } }{ { R }^{ 2 } } \right) \)

The total kinetic energy (KE) as the sum of kinetic energy due to translational motion (KE_TRANS) and kinetic energy due to rotational motion (KE_ROT).
KE = KE_TRANS+ KE_ROT
If the mass of the rolling object is M, the velocity of center of mass is V_CM, its moment of inertia about center of mass is I_CM and angular velocity is \(\omega\) , then
\(KE=\frac { 1 }{ 2 } { Mv }_{ CM }^{ 2 }+\frac { 1 }{ 2 } { I }_{ CM }{ \omega }^{ 2 }\)
With center of mass as reference: The moment of inertia (I_CM)of a rolling object about the center of mass is, I_CM= MK² and V_CM = R\(\omega\) . Here, K is radius of gyration.
\(KE=\frac { 1 }{ 2 } { Mv }_{ CM }^{ 2 }+\frac { 1 }{ 2 } \left( { MK }^{ 2 } \right) \frac { { v }_{ CM }^{ 2 } }{ { R }^{ 2 } } \)
\(KE=\frac { 1 }{ 2 } { Mv }_{ CM }^{ 2 }+\frac { 1 }{ 2 } { Mv }_{ CM }^{ 2 }\left( \frac { { K }^{ 2 } }{ { R }^{ 2 } } \right) \)
\(\frac { 1 }{ 2 } { Mv }_{ CM }^{ 2 }\left( 1+\frac { { K }^{ 2 } }{ { R }^{ 2 } } \right) \)