(i) In pure rolling, the point of the rolling object which comes in contact with the surface is at momentary rest.
(ii) 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. Consider the pure rolling in two different ways.
(a) The combination of translational motion and rotational motion about the center of mass.
(or)
(b) The momentary rotational motion about the point of contact.
(iii) As the point of contact is at momentary rest in pure rolling, its resultant velocity v is zero (v = 0). For example, at the point of contact, v_TRANS is forward (to right) and v_ROT is backwards (to the left).

(iv) 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 = Rω, in pure rolling we have,
v_CM = Rω
(v) 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
Sliding
(i) Sliding is the case when v_CM > Rω (or v_TRANS > v_ROT).The translation is more than the rotation. This kind of motion happens when sudden brake is applied in a moving vehicles or when the vehicle enters into a slippery road. In this case, the point of contact has more of v_TRANS than v_ROT
(ii) Hence, it has a resultant velocity v in the forward direction. The kinetic frictional force (f_k) opposes the relative motion. Hence, it acts in the opposite direction of the relative velocity.
(iii) This frictional force reduces the translational velocity and increases the rotational velocity till they become equal and the object sets on pure rolling. Sliding is also referred as forward slipping.
Slipping
(i) Slipping is the case when v_CM < Rω (or v_TRANS < v_ROT).The rotation is more than the translation ..This kind of motion happens When we. suddenly start the vehicle from rest or the vehicle is stuck in mud.
(ii) In this case, the point of contact has more of v_ROT than V_TRANS. It has a resultant velocity v in the backward direction.
(iii) The kinetic frictional force (f_k) opposes the relative motion. Hence it acts in the opposite direction of the relative velocity.
(iv) This frictional force reduces the rotational velocity and increases the translational velocity till they become equal and the object sets pure rolling. Slipping is sometimes emphasised as backward slipping.

(i) Consider a rigid body rotating about a fixed axis. A point mass m in the body will execute a circular motion about a fixed axis.
(ii) A tangential force \(\vec F\) acting on the point mass produces the necessary torque for this rotation. This force \(\vec F\) is perpendicular to the position vector \(\vec r\) of the point mass.

iii) The torque produced by the force on the point mass m about the axis can be written as,
\(\tau =rFsin90=rF\quad \left[ \because sin90=1 \right] \)
\(\tau =rma\ \left[ \because \left( F=ma \right) \right] \)
\(\tau =rmr\alpha =mr^{ 2 }\alpha \ \left[ \because \left( a=r\alpha \right) \right] \)
\(\tau =\left( mr^{ 2 } \right) \alpha \)
(iv) Hence, the torque of the force acting on the point mass produces an angular acceleration (a) in the point mass about the axis of rotation.
In vector notation,
\(​​\vec \tau =\left( mr^{ 2 } \right) \vec \alpha\)
(v) The directions of \(​​\tau \)  and \(\alpha\) are along the axis of rotation. If the direction of \(\tau\) is in the direction of \(\alpha\), it produces angular acceleration. On the other hand if \(​​\tau \) is opposite to \(\alpha\) angular deceleration or retardation is produced on the point mass.
(vi) The term mr² in equations is called moment of inertia (I) of the point mass. A rigid body is made up of many such point masses. Hence, the moment of inertia of a rigid body is the sum of moments of inertia of all such individual point masses that constitute the body \(\left( I=\sum { { m }_{ i }{ r }_{ i }^{ 2 } } \right) \) Hence, torque for the rigid body can be written as,
\(\vec \tau =\left( \sum { m_{ i }{ r }_{ i }^{ 2 } } \right) \vec a\)
\(\vec \tau = I \vec a\)

(i) The angular momentum of a point mass is defined as the moment of its linear momentum. In other words, the angular momentum Lof a point mass having a linear momentum p at a position r with respect to a point or axis is mathematically written as,
\(\overset { \rightarrow }{ L } =\overset { \rightarrow }{ r } \overset { \rightarrow }{ p } \)
(ii) The magnitude of angular momentum could be written as, L= rp sin\(\theta\) where, ፀ is the angle between \(\overset { \rightarrow }{ r } \) and \(\overset { \rightarrow }{ p } .\overset { \rightarrow }{ L } \) is perpendicular to the plane containing \(\overset { \rightarrow }{ r } \) and \(\overset { \rightarrow }{ p } \)
(iii) As we have written in the case of torque, here also we can associate sinፀ with either  \(\overset { \rightarrow }{ r } \) and \(\overset { \rightarrow }{ p } \)
L=r(p sin\(\theta\)) = r(p丄)
L=(r sin\(\theta\)) p = (r丄)p
where, p丄 is the component of linear momentum p perpendicular to r, and r丄 is the component of position r perpendicular to p.
(iv) The angular momentum is zero (L = 0), if the linear momentum is zero (p = 0) or if the particle is at the origin (\(\overset { \rightarrow }{ r } \)=0) or if  \(\overset { \rightarrow }{ r } \) and \(\overset { \rightarrow }{ p } \) are parallel or antiparallel to each other\(\theta\) =0⁰ or 180⁰)

(i) Torque is defined as the moment of the external applied force about a point or axis of rotation.
(ii) \(\overset { \rightarrow }{ \tau } =\overset { \rightarrow }{ r } \times \overset { \rightarrow }{ F } \)
where, \(\overset { \rightarrow }{ r } \) is the position vector of the point where the force \(\overset { \rightarrow }{ F } \) is acting on the body as shown in Figure.

(iii) Here, the product of \(\overset { \rightarrow }{ r } \) and \(\overset { \rightarrow }{ F } \) is called the vector product or cross product. The vector product of two vectors results in another vector that is perpendicular to both the vectors. Hence, torque (\(\overset { \rightarrow }{ \tau } \)) is a vector quantity.
(iv) Torque has a magnitude (r F sinፀ) and direction perpendicular to \(\overset { \rightarrow }{ r } \) and \(\overset { \rightarrow }{ F } \). Its unit is N m.
\(\overset { \rightarrow }{ \tau } =(r\ F\sin\theta )\hat { n } \)
(v) Here, \(\theta\) is the angle between \(\overset { \rightarrow }{ r } \) and \(\overset { \rightarrow }{ F } \) and \(\hat { n } \) is the unit vector in the direction of \(\overset { \rightarrow }{ \tau } \).

(i) When a rigid body moves, its center of mass will also move along with the body. For kinematic quantities like velocity (v_CM) and acceleration (a_CM) of the center of mass, we can differentiate the expression for position of center of mass with respect to time once and twice respectively. For, simplicity, let us take the motion along, X direction only.
\(\overset { \rightarrow }{ v } _{ CM }=\frac { d\overset { \rightarrow }{ x } _{ CM } }{ dt } =\frac { \sum { { m }_{ i }\left( \frac { d\overset { \rightarrow }{ x } _{ i } }{ dt } \right) } }{ \sum { { m }_{ i } } } =\frac { \sum { { m }_{ i }\overset { \rightarrow }{ v } _{ i } } }{ \sum { { m }_{ i } } } \)
\(\overset { \rightarrow }{ v } _{ CM }=\frac { \sum { { m }_{ i }\overset { \rightarrow }{ { v }_{ i } } } }{ \sum { { m }_{ i } } } \)
\(\overset { \rightarrow }{ a } _{ CM }=\frac { d }{ dt } \left( \frac { d\overset { \rightarrow }{ x_{ CM } } }{ dt } \right) =\left( \frac { d\overset { \rightarrow }{ v } _{ CM } }{ dt } \right) =\frac { \sum { { m }_{ i }\left( \frac { d\overset { \rightarrow }{ v_{ i } } }{ dt } \right) } }{ \sum { { m }_{ i } } } \)
=\(\frac { \sum { { m }_{ i }\overset { \rightarrow }{ a_{ i } } } }{ \sum { { m }_{ i } } } \)
(ii) In the absence of external force, i.e. \(\overset { \rightarrow }{ F_{ ext } } \)= 0 the individual rigid bodies of a system can move or shift only due to the internal forces. 
(iii) This will not affect the position of the center of mass. This means that the center of mass will be in a state of rest or uniform motion. Hence, \(\overset { \rightarrow }{ { v }_{ CM } } \) will be zero when center of mass is at rest and constant when center of mass has-uniform motion (\(\overset { \rightarrow }{ v_{ CM } } =0 \ or \ \overset { \rightarrow }{ v_{ CM } } \)=constant). There will be no acceleration of center of mass, \(\left( \overset { \rightarrow }{ a_{ CM } } =0 \right) \).
From equation
\(0=\frac { \sum { { m }_{ i }\overset { \rightarrow }{ v_{ i } } } }{ \sum { { m }_{ i } } } \)
\(\overset { \rightarrow }{ v_{ CM } } =\frac { \sum { { m }_{ i }\overset { \rightarrow }{ v_{ i } } } }{ \sum { { m }_{ i } } } ;\overset { \rightarrow }{ a_{ CM } } =0\)
(vi) Here, the individual particles may still move with their respective velocities and accelerations due to internal forces