Asked by Annie
A rotating wheel requires 9.00 s to rotate 29.0 revolutions. Its angular velocity at the end of the 9.00 s interval is 98.0 rad/s. What is the constant angular acceleration of the wheel in radians/second squared? (Do not assume that the wheel starts at rest.)
Answers
Answered by
bobpursley
displacement=Wi*t+1/2 alpha*t^2
where alpha is the angular acceleration in rev/sec
you know:
29=wi*9+1/2 alpha*81
and you know
wf=98/2PI rev/sec=wi+alpha*9
wi= 98/2PI-alpha*9
putting that in to the first equation
29=(98/2PI-alpha*9)*9 +1/2 alpha*81
solve for alpha. Now this alpha will be in rev/s^2, in radians/sec^2, divide rev by 2PI to get that.
where alpha is the angular acceleration in rev/sec
you know:
29=wi*9+1/2 alpha*81
and you know
wf=98/2PI rev/sec=wi+alpha*9
wi= 98/2PI-alpha*9
putting that in to the first equation
29=(98/2PI-alpha*9)*9 +1/2 alpha*81
solve for alpha. Now this alpha will be in rev/s^2, in radians/sec^2, divide rev by 2PI to get that.
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