- #1

BHFCBabe

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## Homework Statement

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/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.

## Homework 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]

## 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.