Relating the linear and angular characteristics

Earlier, we discussed a uniform circular motion, in which a particle travels at a constant linear speed υ along a circle and around an axis of rotation. When a rigid body, such as a merry-go-round, turns around an axis of rotation, each particle in the body moves in its own circle around that axis. Since the body is rigid, all the particles make one revolution in the same amount of time; that is, they all have the same angular speed ω.

However, the farther a particle is from the axis, the greater the circumference of its circle is, and so the faster its linear speed umust be. You can notice this on a merry-go-round. You turn with the same angular speed ωregardless of your distance from the center, but your linear speed υ increases noticeably if you move to the outside edge of the merry-go-round.

We often need to relate the linear variables s, ,and a for a particular point in a rotating body to the angular variables , ,and for that body.

The Position

If a rigid body is rotated over an elementary angle d a point within the body is moved a distance ds along a circular arc:

Fig. 1.16

ds = rd j.

If the position of a material point is stated by position-vetor then

The Speed

Differentiating equation for a distance with respect to time leads to

In the algebraic form it is , where is the angle between vectors and . Often (the vectors and are at right angles, ), in this case . Otherwise . Fig. 1.16. In this case is the position vector of a rotating point and is the radius of the circle.

Differentiating the equation for linear acceleration with respect to time we get

where

;

Then

and

,

if the angle .

Thus

.

Table 1.4.

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