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Angular Acceleration (p.20)

  1. Oct 21, 2014 #1
    1. The problem statement, all variables and given/known data
    A grindstone, initially at rest, is given a constant angular acceleration so that it makes 20.0 rev in the first 8.00 s. What is its angular acceleration?
    (a) 0.313 rad/s2 (c) 2.50 rad/s2 (e) 3.93 rad/s2

    (b) 0.625 rad/s2 (d) 1.97 rad/s2


    2. Relevant equations


    3. The attempt at a solution

    #1. I am changing the rev part to radians

    20rev = 125.66

    #2. I am using the formula

    w^2 = w^2 + 2 alpha tetha (where are the symbols????)
     
  2. jcsd
  3. Oct 21, 2014 #2
    What do you mean when you write

    "w^2 = w^2 + 2 alpha tetha "

    If we subtract w^2 from both sides, doesn't this leave you with 2 alpha tetha = ? Where did you get that last formula?

    By the way, I think you most likely mean theta, rather than tetha.
     
  4. Oct 22, 2014 #3
    The equation you used relates the acceleration, displacement, initial and final velocities, and leaves out the time. If the question asked to find the angular acceleration given : that the wheel starts from rest (ωo = 0), and reaches a velocity of ωf rad/sec when it is stopped after making Θ number of revolutions, you would use ω^2 - ωo^2 = 2∝ Θ
    The equation you want is the one that includes the time t.
    ***** Θ = Vo t + 1/2 (∝) (t^2) ******

    Θ = 20 rev x (2 pi rad/ rev) = 125.6 tad

    125.6 = 0 + 0.5 (or 1/2) ∝ (8)^2 and
    ∝ = 125.6/32
    = 3.925 rad/sec^2
     
    Last edited: Oct 22, 2014
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