Consider a rigid body rotating about z-axis with an Angular and Linear Velocities ω as shown in Fig. 5.4(a).

rigid-body-rotating

Fig. 5.4 (a)

Imagine a point P in the rigid body at a perpendicular distance r from the axis of rotation. OP represents the reference line of the rigid body. As the body rotates, the point P moves along a circle of radius r with a linear velocity v whereas the line OP rotates with angular velocity ω as shown in Fig. 5.4 (b).

we are interested in finding out the relation between ω and v. as the axis of rotation is fixed, so the direction of ω always remains the same and ω can be manipulated as a scalar. As regards the linear velocity of the point P, we consider its magnitude only which can also be treated as a scalar.

Suppose during the course of its motion, the point P moves through a distance P1P2 = ∆S in a time interval ∆t during which reference line OP has an angular displacement ∆Ө radian during this interval. ∆S and ∆Ө are related by Eq. 5.1.

∆S = r∆Ө

Dividing both sides by ∆t

dividing-both-sides

In the limit when ∆t ⇾ 0 the ratio ∆S/∆t represents v, the magnitude of the velocity with which point P is moving on the circumference of the circle. Similarly ∆ϴ/∆t represents the angular velocity ω of the reference line OP. So equation 5.6 becomes

V = rω                   …………….                              (5.7)

In Fig 5.4 (b), it can be seen that the point P is moving along the arc P1P2. In the limit when ∆t ⇾ 0, the length f arc P1P2 becomes very small and its direction represents the direction of tangent to the circle at point P1. Thus the velocity with which point P is moving on the circumference or the circle has a magnitude v and its direction is always along the tangent to the circle at that point. That is why the linear velocity of the point P is also known as tangential velocity.

moving-along-the-arc-P1P2

Fig. 5.4 (b)

Similarly Eq 5.7 shows that if the reference line OP is rotating with an angular acceleration a, the point P will also have a linear or tangential acceleration at. Using Eq 5.7 it can be shown that the two accelerations are related by

at = ra                   …………….                              (5.8)

Eqs 5.7 and 5.8 show that on a rotating body, points that are at different distances from the axis do not have the same speed or acceleration, but all points on a rigid body rotating about a fixed axis do have the same angular displacement, angular speed and angular acceleration at any instant. Thus by the use of angular variables we can describe the motion of the entire body in a simple way.

Point to ponder

amusement-parks

You may feel scared at the top of roller coaster ride in the amusement parks but you never fall down even when you are upside down. Why?

 Equations of Angular Motion

The equations (5.2, 5.3, 5.4 and 5.5) of angular motion are exactly analogous to those in linear motion except that ϴ, ω and a have replaced S, v and a, respectively. As the other equations of linear motion were obtained by algebraic manipulation of these equations, it follows that analogous equations will also apply to angular motion. Given below are angular equations together with their linear counterparts.

Equations-of-angular-Motion-01

The angular equations 5.9 to 5.11 hold true only in the case when the axis of rotation is fixed, so that all the angular vectors have the same direction. Hence they can be manipulated as scalars.

Do You Know?

tangential-distance

As the wheel turns through an angle 0,it lays out a tangential distance S = r0

Example 5.1: an electric fan rotating at 3 rev S-1 is switched off. It comes to rest in 18.0 s. assuming deceleration to be uniform, find its value. How many revolutions did it turn before coming to rest?

Solution: in this problem we have

Equations-of-angular-Motion-02