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Net displacement for a point on a circular saw 
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#1
Nov2209, 05:02 PM

P: 3

1. The problem statement, all variables and given/known data
When a carpenter shuts off his circular saw, the 10.0inch diameter blade slows from 4440 rpm to zero in 2.5 s. A.What is the angular acceleration of the blade? In rev/sec^{2} 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) [tex]\omega[/tex]_{f}=[tex]\omega[/tex]_{o}+2[tex]\alpha[/tex]t 2) [tex]\omega[/tex]_{f}^{2}=[tex]\omega[/tex]_{o}^{2}+2[tex]\alpha\Delta\Theta[/tex] 3. The attempt at a solution A. 4440rev/1min * 1min/60sec=74 rev/sec Using equation 1: 0=74+[tex]\alpha[/tex]*2.5 [tex]\alpha[/tex]=29.6 rev/sec^{2} B. Using equation 2: 0=74^{2}+2*29.4*[tex]\Delta\Theta[/tex] [tex]\Delta\Theta[/tex]=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. 


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