Set 8 Problem number 5

• Important Note on Notation

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

An object travels around a circular path of radius 5 meters in 1.899 seconds.

• Through how many revolutions per second does the object travel?
• Through how many radians per second does it travel?
• At what speed does the object travel?

Solution

In 1.899 seconds this is one revolution, so in 1 second there would be 1/ 1.899 = .5265 revolutions.

Since a revolution corresponds to 2 `pi radians:

• .5265 revolutions per second is 2 `pi ( .5265 ) radians per second = 3.308 radians per second.

On a circle of radius 5 meters:

• 1 radian would correspond to a distance of 5 meters, so
• 3.308 radians per second would be 5 ( 3.308 ) meters per second = 16.54 meters per second.

Generalized Solution

If an object moves around a circle with period Tseconds  (i.e., completing a revolution every T seconds), then it completes 1 / T revolutions per second.  The number of revoutions per second is the frequency of the motion.

• We therefore say that period T implies frequency f = 1 / T.
• The period T is therefore equal to   T = 1 / f.

Period T seconds  implies that the angle changes at a rate of 2 `pi radians / T seconds = 2 `pi / T rad / sec.

• Just as the rate at which position changes is called velocity, the rate at which angle changes is called angular velocity.
• Angular velocity is designated by the Greek letter `omega.

So period T implies angular velocity 2 `pi / T.

We can write the angular velocity in terms of frequency:

• Since f = 1 / T, frequency f implies angular velocity 2 `pi f.
• We summarize by saying that  

• `omega = rate of change of angular position

= `d`theta / `dt

= 2 `pi / T = 2 `pi f.