Net displacement for a point on a circular saw


by BHFCBabe
Tags: angular acceleration, displacement
BHFCBabe
BHFCBabe is offline
#1
Nov22-09, 05:02 PM
P: 3
1. The problem statement, all variables and given/known data

When a carpenter shuts off his circular saw, the 10.0-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) [tex]\omega[/tex]f=[tex]\omega[/tex]o+2[tex]\alpha[/tex]t
2) [tex]\omega[/tex]f2=[tex]\omega[/tex]o2+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/sec2

B. Using equation 2:
0=742+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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Redbelly98
Redbelly98 is offline
#2
Nov22-09, 07:39 PM
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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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