The rotational motion is different from other mechanics concepts because it cannot be studied by assuming the body to be a point body. In rotational motion all particles of the body undergo different linear displacements unlike translational motion, therefore we will assume the body to be rigid body.
Angular displacement: when a body rotates about a fixed axis, each particle travels in circular path. Although different particles will have different linear displacements there angular displacement is same. If the angular displacement is θ, then
where s is the distance traveled while moving in a circular path of radius r.
Angular displacement is a scalar quantity if θ is large and is vector if θ is small. This is so because for large angular displacements it does not satisfy the laws of vector addition. For small angular displacements, its direction is given by right hand curl rule.
Angular velocity: It is rate of change of angular displacement of the body with respect to time. i.e.
As s = θr, therefore we can also write
V = rω
In the vector form
Angular acceleration is an axial vector and its direction is always along the axis of rotation and is given by right hand curl rule. For rigid body linear velocity of particles is different but angular velocity of all bodies is same.
Angular Acceleration: It is rate of change of angular velocity w.r.t. time.
As v = rω, therefore we can also write a = r α
The direction of angular acceleration is same as angular velocity if angular velocity is increasing and opposite to angular velocity if angular velocity is decreasing.
Equation of kinematics: the equation of kinematics can be written analogous to the linear motion as
ω = ω0 + αt
θ = θ0 + ω0t + αt2/2
ω2 = ω02 + 2 α (θ-θ0)
θ = θ0 + t
These equation of motion can be applied if the angular acceleration of the body is constant.
Moment of Inertia: Inertia is the property of the body to resist the changes in the state of motion and moment of inertia is the property, which resists changes in state of rotational motion of the body.
For point mass
I = mr2, where m is the mass and r is the distance from axis of rotation.
For n particles
Where ri is the distance from the axis of rotation of mass mi.
For rigid body:
Where r is the perpendicular distance from the axis of rotation of mass dm.
MI depends upon
[a] Axis of rotation selected
[b] distribution of mass about the axis of rotation.
It does not depend upon the dimension parallel to the axis of rotation. For e.g. if MI of cylinder is calculated about is axis of rotation it does not depend on the length, but only on the mass and radius.
Parallel Axis Theorem: According to parallel axis theorem the MI of the body about given axis is equal to the MI of the body about an axis parallel to given axis and passing through the center of mass of the body IG plus Ma2 where M is the mass of the body and ‘a’ is he distance between the two axis.
I= IG + Ma2
Theorem Of Perpendicular Axis: According to it the sum of MI of the plane lamina about two mutually perpendicular axis lying in its plane is equal to the MI about a axis perpendicular to the plane of lamina and passing through the point of intersection of two axes.
IZ = IX + IY
Important Note regarding MI:  Theorem of parallel axis is applicable for any type of rigid body whereas perpendicular axis theorem is applicable to 2 dimensional bodies only.
 Moment of inertia of the rigid body [symmetrically cut from whole mass] is same as that of the body e.g. a small part of circular disc will have same moment of inertia as that of the disc]
Radius of Gyration: Radius of gyration of the body about an axis is the effective distance from the axis where the whole mass can be assumed to be concentrated so that the moment of inertia remains the same
If K is radius of gyration and M is the mass then I= MK2
K can also be said to be the rms distance from the axis of rotation
Radius of gyration is not a constant quantity it changes with change in axis of rotation of the body.
Angular Momentum: The momentum associated with rotational motion is called angular momentum, it can be associated with a particle or a rigid body.
Suppose a particle of mass m moving with momentum p. Its angular momentum is
For a rigid body of moment of inertia I rotating with angular velocity ω the angular momentum is given by
If a body is in translational as well as rotational motion then its angular momentum is sum of both the components.
Law of Conservation of angular momentum:
Uniform Pure Rolling: Pure rolling means no relative motion at point of contact between the two bodies. For example, consider a disc of radius R moving with linear velocity v and angular velocity ω on a horizontal ground. The disc is said to be moving without slipping if the velocities of point P and Q are equal
Vp = vQ
v = Rω
If Vp >vQ or v > Rω, the motion is said to be forward slipping and if v< Rω is case of backward slipping.
Thus v = Rω is condition for pure rolling on a stationary ground. Sometimes it is simply said rolling. Suppose the base over which the body moves is moving with some velocity say v0, then the condition of pure rolling are different. For example in the above figure vp=vq
v – Rω = v0
Thus in this condition will be different. By uniform pure rolling we mean that v and ω are constant.
Note:  distance moved by the center of mass of the rigid body in one full rotation is 2πR. This can be shown as
S = v T = [ωR] [2π/ω] = 2πR
In forward slipping s > 2πR
In backward slipping s < 2πR
 Instantaneous axis of rotation passes through the bottommost point, as it is a point of zero velocity. Thus the combined motion of rotation and translation can be assumed to be pure rotational motion about the bottommost point with same angular speed ω.
 The speed of point on the circumference of the body at the instant is shown in figure is 2 v sin
 The path of point on the circumference is a cycloid and the distance moved by this point in 1 full rotation is 8R.
Instantaneous center of rotation: Since rolling body does not slip, therefore the point of contact O with the surface is instantaneously at rest and ring momentarily rotates about this point. All the particles appear to describe a circular paths with angular velocity ω about O as center. The point O is then called instantaneous center of rotation.
Velocity of any point: The magnitude of linear velocity of any point is given by the product of angular velocity and the distance of that point from the point of contact the direction of velocity vector is perpendicular to the line joining the point and the point of contact.
Accelerated pure rolling:
So far we were discussing the uniform pure rolling in which v and ω were constants. Now suppose an external force acts on the rigid body, the motion will no longer remain uniform. The condition on a stationary ground is,
v = R ω
Differentiating this equation with respect to time we have
a = R α
Thus in addition to v = R ω at every instant of time, linear acceleration= R α for pure rolling to take place. Here, friction plays an important role in maintaining role in maintaining pure rolling. The friction may sometimes act in forward direction, sometimes in backward direction or under certain conditions it may be zero. Here, we should not forget the basic nature of friction, which is self adjusting force and has tendency to stop the relative motion between the two bodies in contact. Let us take example illustrating the above theory.
Suppose a force F is applied at the topmost point of the rigid body of radius R, mass M and moment of inertia I about an axis passing through the center of the mass. Now the applied force can produce by itself,
[a] linear acceleration ‘a’
[b] an angular acceleration α
If a = R α, then there is no need of friction and force of friction =0. If a < Rα, then to support the linear motion the force of friction f will act in forward direction. Similarly if a > R α then to support the angular motion force of friction will act in backward direction. So, in this case force of friction can be in forward, backward or zero also. Let us assume it in forward direction
For pure rolling to take place
a = R α
solving the above equations we get
From equation following conclusions can be drawn
[a] if I= MR2 then friction force is zero.
[b] if I < MR2, like in the case of solid or hollow sphere, f is positive i.e. force of friction will be forward
[c] If I > MR2, f will be negative.
Here, it should be noted that force of friction F obtained above should be less than μMg for pure rolling to take place. Further, we saw that I < MR2 force of friction acts in forward direction. This is because if I is small α will be more i.e. to support the linear motion friction acts in forward direction.
Rolling on inclined plane:
As we said earlier the force of friction in this case is backward. Equation of motion are
For pure rolling to take, place
a = R α
Solving the equations, we get
Further, the force of friction calculated in equation for pure rolling to take place should be less than or equal to the maximum friction force
Velocity of the body at the bottom of inclined plane is
As here acceleration is constant and body starts from rests so time to taken to reach the bottom of inclined plane is
Important Points for Inclined plane motion:  As factor β= depends only on the shape and not on the mass and radius there same bodies with different mass and radius are allowed to roll down an inclined plane they will reach the bottom at the same time.
 Velocity, acceleration and time of descend depends upon β. Lesser the MI lesser will be β so greater will be velocity and acceleration and lesser will be time of descend.
 The velocity is independent of the inclination of the plane and depends only on the height from which the body descends. Whereas acceleration and time of descend depends upon inclination larger the inclination larger will be acceleration and smaller will be time of descend.
Note: [a] the torque equation can be applied about only two points
 the center of mass of the body
 point about which the body is in pure rotation
[b]In cases where the pulley is having friction, the tension on the two sides of the pulley is different and it is essential to consider the rotational motion of the pulley.
[c] Work done in pure rolling on a stationary ground is zero as the point of application of force is at rest. Therefore, mechanical energy can be conserved if all other dissipative forces are ignored.
[d] the equation ζ = I α holds only in the inertial reference frame. However there exists a special case when the axis of rotation passes through the center of mass otherwise pseudo forces produces pseudo torque on the body.