Angular acceleration of a skater

AI Thread Summary
A skater performing a pirouette at 2.0 revolutions per second stops within 3/4 of a revolution, leading to a calculation of angular deceleration. The initial calculation yielded -18 rad/s², while the textbook states -17 rad/s², prompting a request for clarification. The discussion highlights the use of the formula ω² = ω₀² + 2αΔθ to find angular acceleration. Participants confirm that the textbook's value is a rounded figure. The conversation concludes with appreciation for the assistance provided.
brad sue
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Hi,
I need help with the following problem.

A skater does a pirouette at the rate of 2.0 rev / second and then stop within 3/4 rev. Assume that the angular decceleration is constant and calculate its magnitude.

I tried to compute it but I found -18 rad/s2.
BUt the textbook gives -17 rad/s2.
Please can someone help me with this problem?

Thank you
B
 
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brad sue,
use
\omega^2 \ = \ {\omega^2}_o \ + \ 2\alpha \ \Delta\theta
The rotational equivalent of
v^2 \ = \ {v^2}_o \ + \ 2as
 
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What formula did you use to compute 18 rad/s2?

What is the angular analogy to v2 = vo2 + 2ax?

The 17 rad/s would be rounded.
 
Astronuc said:
What formula did you use to compute 18 rad/s2?
What is the angular analogy to v2 = vo2 + 2ax?
The 17 rad/s would be rounded.

Yes the -17 has been rounded.
Thank you very much you all..

B
 
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