Angular deceleration of a circular saw blade

Tikki
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Homework Statement
If a circular saw blade is spinning at a rate of 6.57 revolutions per second, what is the magnitude of the (constant) angular acceleration that needs to be applied on the saw blade to bring it to rest in 17.2 revolutions?
Relevant Equations
theta = no of revs * 2pi
w^2=2a(theta)
I tried the formula I showed up there but got 0.399 instead of 7.88rad/s^2.
 
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What number did you use for w?
 
kuruman said:
What number did you use for w?
6.57rads/s
 
Tikki said:
6.57rads/s
wait a min?
 
Tikki said:
wait a min?
Yup, it is given in revolutions per second.
 

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