A Classical Approach of Newtonian Mechanics

7 CIRCULAR MOTION 7.3 Centripetal acceleration

seconds. Here, T is the repetition period of the circular motion. If the object

executes a complete cycle (i.e., turns through 360 ◦) in T seconds, then the number
of cycles executed per second is

1

f = =

T

ω

. (7.8)
2 π

Here, the repetition frequency, f, of the motion is measured in cycles per second—
otherwise known as hertz (symbol Hz).

As an example, suppose that an object executes uniform circular motion, ra-

dius r = 1.2 m, at a frequency of f = 50 Hz (i.e., the object executes a complete
rotation 50 times a second). The repetition period of this motion is simply

1

T = = 0.02 s. (7.9)

f

Furthermore, the angular frequency of the motion is given by

ω = 2π f = 314.16 rad./s. (7.10)

Finally, the tangential velocity of the object is

v = r ω = 1.2 × 314.16 = 376.99 m/s. (7.11)

7.3 Centripetal acceleration

An object executing a circular orbit of radius r with uniform tangential speed v

possesses a velocity vector v whose magnitude is constant, but whose direction
is continuously changing. It follows that the object must be accelerating, since

(vector) acceleration is the rate of change of (vector) velocity, and the (vector)

velocity is indeed varying in time.

Suppose that the object moves from point P to point Q between times t and

t + δt, as shown in Fig. 58. Suppose, further, that the object rotates through δθ

radians in this time interval. The vector P

→

X, shown in the diagram, is identical

to the vector Q

→

Y. Moreover, the angle subtended between vectors P

→

Z and P

→

X is