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S E C T I O N 4 . 4 • Uniform Circular Motion 93 where v " v i " v f and r " r i " r f . This equation can be solved for "!v" and the expres- sion so obtained can be substituted into to give the magnitude of the aver- age acceleration over the time interval for the particle to move from ! to ": Now imagine that points ! and " in Figure 4.17b become extremely close together. As ! and " approach each other, !t approaches zero, and the ratio "!r"/!t approaches the speed v. In addition, the average acceleration becomes the instantaneous accelera- Thus, in uniform circular motion the acceleration is directed inward toward the center 2 /r. In many situations it is convenient to describe the motion of a particle moving with constant speed in a circle of radius r in terms of the period T, which is defined as the time required for one complete revolution. In the time interval T the particle moves a r, which is equal to the circumference of the particle’s circular path. Therefore, because its speed is equal to the circumference of the circular path divided (4.16) T ! 2,r v a c " v 2 r " a " " " !v " ! t " v r
" !r " ! t a " !v/!t Quick Quiz 4.7 Which of the following correctly describes the centripetal ac- celeration vector for a particle moving in a circular path? (a) constant and always per- Quick Quiz 4.8 A particle moves in a circular path of radius r with speed v. It then increases its speed to 2v while traveling along the same circular path. The cen- Example 4.8 The Centripetal Acceleration of the Earth What is the centripetal acceleration of the Earth as it moves Solution We conceptualize this problem by bringing forth To finalize this problem, note that this acceleration is much " 5.93 * 10 # 3 m/s 2 " 4, 2 (1.496 * 10 11 m) (1 yr) 2
# 1 yr 3.156 * 10 7 s $ 2 a c " v 2 r " # 2,r T $ 2 r " 4, 2 r T 2 ▲ PITFALL PREVENTION 4.5 Centripetal Acceleration is not We derived the magnitude of the Period of circular motion |