# Why is 2theta(radians) being used to find the angular acceleration.

1. Oct 8, 2011

### gibson101

I understand how the angular velocity was gotten, and why the amount of revolutions it took to travel 115 m was just 115/circumference of wheel. But why would radians need to be configured to find the angular acceleration? I thought angular acceleration was Δω/τ? And why is 2(thetasymbol) being substituted for time T? And why is the 2 there?

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2. Oct 8, 2011

### gibson101

i see that the equation for find angular acceleration was derived from ω^2=ω°^2+2αt. But why can't the formula ω=ω°+αt be used. I noticed that when i used the first formula i ended up with -305/((2∏)(53.8)) = -.9 which is correct but when i used the second formula I got
-.073 which is incorrect. And why does t(time) equal ((2∏)(number of revolutions))?

3. Oct 8, 2011

### Redbelly98

Staff Emeritus
Not quite, the correct formula is ω2=ω°2+2αθ, where θ is the angle through which the wheel rotates.
It doesn't. The angle θ through which the wheel rotates is 2π(number of revolutions).

4. Oct 8, 2011

### gibson101

I SEE. BUT i still dont see how θ replaces time. because angular acceleration is Δω/Δt, not Δω/Δθ.

5. Oct 8, 2011

### vela

Staff Emeritus
It's not replacing the time. There are two different formulas:
$$\alpha = \frac{\omega-\omega_0}{t}$$and$$\alpha = \frac{\omega^2-\omega_0^2}{2\theta}$$which allow you to solve for the angular acceleration when the acceleration is constant. The solution used the second equation to find the angular acceleration.

6. Oct 8, 2011

### Redbelly98

Staff Emeritus
It doesn't replace time. θ and time are different concepts, I never said they were the same.

You're correct that angular acceleration is Δω/Δt. But I never said or implied that it is Δω/Δθ, what makes you think I did?
If you're trying to find α, but don't know t yet, then that formula is not useful -- even though it is true.

7. Oct 8, 2011

### SammyS

Staff Emeritus
In fact, the second equation can be written as:

$$\alpha = \frac{(\omega-\omega_0)(\omega+\omega_0)}{2\theta}$$

$$\phantom{x}\ \ \ = \frac{(\omega-\omega_0)}{\left(\frac{2\theta}{\omega+\omega_0} \right)}$$

So it;s more like t is replaced by θ/((ω+ω0)/2), which comes from, average (angular) velocity = the total (angular) displacement / elapsed time .

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