ACCELERATION 15 



velocities at the two instants, the line CD will represent the change in 

 velocity between these two instants. 



Now let the two instants differ only by an infinitesimal interval dt, so 

 that the points A , B coincide except for an infinitesimal arc Vdt. In the 

 figure, CD passes through 

 P wherever A , B are on the 

 circle, so that when B is 

 made to coincide with A, 

 CD coincides with the ra- 

 dius through A. But if F 

 is the acceleration of the 

 moving point, the change P ^ 



in velocity produced in 



time dt must be Fdt. Thus CD represents the change of velocity Fdt in 

 direction and magnitude, so that the change of velocity, and hence the 

 acceleration at A , is along the radius at A . 



Here, then, we have a case in which the acceleration is at right angles 

 to the velocity. 



To find the magnitude of the acceleration, we notice that CD = 2 CE, 

 and that, by similar triangles, 



EC : CB = BE : BP. 



Now EC, or I CD, represents the velocity Fdt, while CB on the same 

 scale represents the velocity V. 



Thus i Fdt : V = BE : BP. 



In the limit when BA is very small, BE, or %BA, becomes identical 

 with half of the arc BA of the circle, and therefore with i Vdt. Thus, if a 

 is the radius of the circle, 



iFdt: V= 



V 2 



giving F = as the amount of the acceleration. 

 a 



EXAMPLES 



1. A windmill has sails 20 feet in length, and turns once in ten seconds. 

 Find the acceleration of a point at the end of a sail. 



2. A wheel of radius 3 feet spins at the rate of 10 revolutions a second and 

 is at the same time falling freely with an acceleration of 32 feet per second 

 per second due to gravity. Find the resultant accelerations of the different 

 points on the rim of the wheel. 



3. Taking the earth to have an equatorial diameter of 7927 miles, find the 

 acceleration towards the earth's center of (a) a point at rest, relative to the 

 earth's surface, on the equator; (6) a body falling under gravity at the equator, 



