26 DAIRYING 



square of the velocity in feet per second, and dividing by the 

 radius in feet multiplied by 32.2.* 



Centrifugal force equals 



W X V 2 in feet per second 



RX 32.2 



In the above illustration the centrifugal force exerted on a 

 weight of .14 pound when revolved at a speed of 100 revolutions 

 per minute in a circle having a radius of one foot is calculated 

 as follows: 



266. The velocity in feet per second is found by multiplying 

 the diameter of the circle by 3.1416, in this case it is 2 X 3.1416 = 

 6.283, then multiplying this figure by the speed 100, and dividing 

 by 60 to reduce to seconds, gives the velocity as "10.94 ft. per 

 second. The square of 10.49 is 110, which multiplied by the 

 weight .14, gives 15.40, and this divided by the radius, one foot, 

 after multiplying by 32.2, gives 0.48, the centrifugal force in 

 pounds. 



267. The effect of centrifugal force on milk in a cream sep- 

 arator has also been illustrated by Fredericksen.** "When a par- 

 ticle of matter is swinging round a central point, the force by 

 which it presses outward from the center revolution depends upon 

 the gravity, the speed, and the distance from the center. Sup- 

 posing a weight W to revolve around an axis, the distance from 

 the center (the radius) being R feet, and the number of revolu- 

 tions S hundred a minute, then the centrifugal force F=3.4X 

 RXW'XS 2 . Consequently, if R is one foot and W is one pound 

 the centrifugal force will be: 



For 100 revolutions a minute 3.4X 1= 3.4 pounds 



For 200 revolutions a minute 3.4X 4= 13.6 pounds 



For 400 revolutions a minute 3.4X 16= 54.4 pounds 



For 1000 revolutions a minute 3.4X 100= 304. pounds 



For 5000 revolutions a minute 3.4X2500=8500. pounds 



* The velocity of a body falling in vacuum increases in each 

 second by 32.2 feet per second. 



**2 J. D. Fredericksen in The Dairy Messenger. 



