If a body rotates about a fixed axis in such a way that all its particles move in circular paths, the body is said to have the motion of rotation. The particles on the axis of rotation have zero velocity and zero acceleration. (Ref. Fig. 3.2(a))

Refer Fig. 3.2 (b). The angular positions of any two lines 1 and 2 are 91 and e2 from any convenient fixed reference direction.

θ_{1} – θ_{2} =β

The angle β does not change with time in rigid body motion. So differentiating (3.1) with respect to time, we get

θ_{1} – θ_{2} =0 =0

θ_{1} = θ_{2}

Differentiating (3.2) again we get θ_{1} = θ_{2}

Equation (3.1), (3.2) and (3.3) imply that all lines on a rigid body in its plane of motion

have the same change in angular displacement (∆θ_{2} = ∆θ_{1}), same angular velocity, and

the same angular acceleration.

**3.3.1 Angular Motion Relations**

Let w = angular velocity

θ = angular displacement

a = angular acceleration

w = dθ /dt = θ

a = dw /dt = w ,or a = d^{2}θ / dt^{2} =θ

Eleminating dt we get

w = dw =a dθ , or θ dθ = θ dθ

**Special Case**: For rotation with constant a ,

a = dw /dt

or w_{t} –w_{0} =at , or wt = w_{0} +at

w_{t}^{2} = w_{0}^{2} +2a (θ –θ_{0})

θ = θ_{0} +w_{0}t +1/2 at_{2}

**3.3.2 Relationships of Rotation about a Fixed Axis**

Any point P on rotation will move in a circle of radius r. We can apply the idea of circular motion of a particle here

v =wr

a _{n} = w^{2} r =v^{2} /r =vw

a _{t} =ar

Vector Notation:

**SOLVED EXAMPLES**

**Example 3.1.** The square plate rotates about the fixed pivot 0. At the instant represented, its angular velocity is co = 6 rad/s and its angular acceleration is a= 4 rad/s2, in the directions shown in Fig. P-3.1. Calculate the velocity and acceleration of (i) A (ii) B.

(i) For point A :

= – 6k x 270 I +4k x45 j

= – 1620 j – 180 I mm /s

(i) For point B :