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Net displacement for a point on a circular saw
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Nov22-09, 05:02 PM
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
3. The attempt at a solution
A. 4440rev/1min * 1min/60sec=74 rev/sec
Using equation 1:
B. Using equation 2:
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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