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

AI Thread Summary
The discussion focuses on the use of radians in calculating angular acceleration, specifically addressing the confusion between angular displacement (θ) and time (t). It clarifies that the formula for angular acceleration can be expressed as α = (ω² - ω₀²) / (2θ), where θ represents the angle of rotation, not time. The distinction between the two formulas for angular acceleration is emphasized, noting that one requires time while the other does not. The conversation highlights that θ is derived from the total revolutions, linking it to the wheel's rotation. Ultimately, understanding these relationships is crucial for accurately calculating angular acceleration.
gibson101
Messages
47
Reaction score
0
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?
 

Attachments

  • Screen shot 2011-10-08 at 8.08.52 PM.JPEG
    Screen shot 2011-10-08 at 8.08.52 PM.JPEG
    45.7 KB · Views: 515
  • Screen shot 2011-10-08 at 8.09.35 PM.JPEG
    Screen shot 2011-10-08 at 8.09.35 PM.JPEG
    20.8 KB · Views: 553
Physics news on Phys.org
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))?
 
gibson101 said:
i see that the equation for find angular acceleration was derived from ω^2=ω°^2+2αt.
Not quite, the correct formula is ω2=ω°2+2αθ, where θ is the angle through which the wheel rotates.
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))?
It doesn't. The angle θ through which the wheel rotates is 2π(number of revolutions).
 
I SEE. BUT i still don't see how θ replaces time. because angular acceleration is Δω/Δt, not Δω/Δθ.
 
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.
 
gibson101 said:
I SEE. BUT i still don't see how θ replaces time. because angular acceleration is Δω/Δt, not Δω/Δθ.
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?
gibson101 said:
But why can't the formula ω=ω°+αt be used.
If you're trying to find α, but don't know t yet, then that formula is not useful -- even though it is true.
 
vela said:
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.
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 .
 
Back
Top