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Angluar Velocity and Revolutions

  1. Sep 8, 2008 #1
    1. The problem statement, all variables and given/known data
    You switch a food blender from its high to its low setting; the blade speed drops from 3100 rpm to 1300 rpm in 2.1 s. How many revolutions does it make during this time?

    2. Relevant equations
    [tex]\omega[/tex]i = [tex]\frac{3100 rpm * (2 * pi)}{60}[/tex] = 324.6 rad/sec
    [tex]\omega[/tex]f = [tex]\frac{1300 rpm * (2 * pi)}{60}[/tex] = 136.1 rad/sec

    [tex]\omega[/tex]f = [tex]\omega[/tex]i + [tex]\alpha[/tex]t

    [tex]\omega[/tex]f[tex]^{2}[/tex] = [tex]\omega[/tex]i[tex]^{2}[/tex] + 2[tex]\alpha[/tex]([tex]\Delta[/tex][tex]\Theta[/tex])

    3. The attempt at a solution

    136.1 = 324.6 + 2.1 [tex]\alpha[/tex]

    [tex]\alpha[/tex] = 89.8 rad/sec[tex]^{2}[/tex]

    136.1[tex]^{2}[/tex] = 324.4[tex]^{2}[/tex] + 2 * 89 ([tex]\Delta[/tex][tex]\Theta[/tex])

    ([tex]\Delta[/tex][tex]\Theta[/tex]) = 482.8 rad / (2 * pi) = 76.8 rev

    Want to make sure if this is correct.
     
  2. jcsd
  3. Sep 8, 2008 #2

    mezarashi

    User Avatar
    Homework Helper

    Looks good to me :)
     
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