How Do You Calculate Constant Angular Acceleration?

AI Thread Summary
To calculate constant angular acceleration, the discussion revolves around using the correct equations related to angular motion. One participant initially attempts to use the equation Wfinal² = Winitial² + 2(angular acceleration)(angular displacement) but finds it cumbersome. Another suggests a simpler approach using Wf = Wo + (angular acceleration)t, which leads to confusion due to the lack of angular displacement in the equation. The conversation highlights that while both methods can yield the correct answer, the simpler method is more efficient if one knows what to isolate. Ultimately, understanding the relationships between angular velocity, time, and displacement is key to solving for angular acceleration effectively.
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Homework Statement


A rotating wheel requires 3.00 s to rotate 232.5 rads. It's angular velocity at the end of the 3.00 s interval is 98 rad/s. whatis the constant angular acceleration of the wheel?


Homework Equations


Not sure

The Attempt at a Solution


Is this question more complex than it looks or what am I missing, I have solved it by useing the equation
Wfinal2 = Winitial2 +2(angular accel)(angular displacement)

and subbing in the value of angular accel with the formula (Wfinal - Winitial/T)

then useing the quadratic formula to solve ,
I get the right answer but it seems cumbersome, what are your thoughts?
 
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Constant acceleration. What you're doing is correct.

I would have used the following equation though.

Wf = Wo + (angular accel)t
 
when I use the equation you recommend

Wf = Wo + (angular accel)t

and sub in angular accel = (wf - w0 / t ) everything cancels and I am left with wf = wf

Been studying non stop my brain may be fried, what am I missing?

is it because that equation does not invole angular displacement?
 
Sorry i misinterpreted the question. Wo = Wf - (a)t ; Also,

Wo = sqr(wf^2-2(a)d); So,

Wf - (a)t = sqr(wf^2 - 2(a)d)

Solve (a).
 
Thanks for clarifying that !
 
Note that if you isolate (a) in both equations then solve Wo and use Wo to solve (a) it will still work.

It's just that method requires back substitution, which takes longer (and more algebra ;) ).

If you know what you're looking for you can usually shorten the process up!
 
Found the easy way... Wave = (Wf + W0)/2
so (Wave * 2) - Wf = W0

(Wf - W0 )/ T = Angular Accel
 
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