# Angular Acceleration formula

1. Dec 6, 2009

### rambo5330

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Not sure

3. 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 6, 2009

### jegues

Constant acceleration. What you're doing is correct.

I would have used the following equation though.

Wf = Wo + (angular accel)t

3. Dec 6, 2009

### rambo5330

when I use the equation you recomend

Wf = Wo + (angular accel)t

and sub in angular accel = (wf - w0 / t ) everything cancels and im 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?

4. Dec 7, 2009

### jegues

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).

5. Dec 7, 2009

### rambo5330

Thanks for clarifying that !

6. Dec 7, 2009

### jegues

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!

7. Dec 7, 2009

### rambo5330

Found the easy way... Wave = (Wf + W0)/2
so (Wave * 2) - Wf = W0

(Wf - W0 )/ T = Angular Accel