# Net displacement for a point on a circular saw

by BHFCBabe
Tags: angular acceleration, displacement
 P: 3 1. The problem statement, all variables and given/known data When a carpenter shuts off his circular saw, the 10.0-inch diameter blade slows from 4440 rpm to zero in 2.5 s. A.What is the angular acceleration of the blade? In rev/sec2 B.What is the distance traveled by a point on the rim of the blade during the deceleration? In feet. What is the magnitude of the net displacement of a point on the rim of the blade during the deceleration? in inches I have solved A and B, but cannot get C. The answers are in the back of the book so I know A and B are correct. 2. Relevant equations 1) $$\omega$$f=$$\omega$$o+2$$\alpha$$t 2) $$\omega$$f2=$$\omega$$o2+2$$\alpha\Delta\Theta$$ 3. The attempt at a solution A. 4440rev/1min * 1min/60sec=74 rev/sec Using equation 1: 0=74+$$\alpha$$*2.5 $$\alpha$$=-29.6 rev/sec2 B. Using equation 2: 0=742+2*-29.4*$$\Delta\Theta$$ $$\Delta\Theta$$=92.5 Rev 1ft/12in*10pi in./1 rev * 92.5rev = 242.16 feet. C. I know that it completed 92.5 revolutions. There are .5 revolutions left over. 10pi in/ 1 rev *.5 rev = 15.7 inches. However, this answer is wrong. Why? The proper answer is 10 inches.
 Mentor P: 11,904 Displacement is determined by the straight-line distance between two points, not the distance around the circumference. p.s since nobody said it to you before: welcome to Physics Forums

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