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Why would we ignore initial angular velocity in equation?

  1. Nov 22, 2015 #1
    1. The problem statement, all variables and given/known data
    A 1.15-kg grinding wheel 22.0 cm in diameter is spinning counterclockwise at a rate of 20.0 revolutions per second. When the power to the grinder is turned off, the grinding wheel slows with constant angular acceleration and takes 80.0 s to come to a rest.
    (a) What was the angular acceleration (in rad/s2) of the grinding wheel as it came to rest if we take a counterclockwise rotation as positive?
    A: -1.57 rad/s2
    (b) How many revolutions did the wheel make during the time it was coming to rest?
    A: 800 revolutions

    2. Relevant equations
    ωfo+αt
    θ=θ0ot+0.5αt2

    3. The attempt at a solution
    (a) [tex]
    \omega_f=\omega_0+\alpha t\\
    0 = (20 * 2\pi) + 80\alpha\\
    \alpha = -\frac{20 * 2\pi}{80}\\
    \alpha = -1.57 \frac{\texttt{rad}}{\texttt{s}^2}
    [/tex]
    (b) This is where I am confused. Why do we assume w0=0 to get the answer?[tex]
    \theta = \omega_0 t + \frac{1}{2} t^2 \alpha\\
    \theta = \frac{1}{2} t^2 \alpha\\
    \theta = \frac{1}{2} (80)^2 (\frac{-1.57}{2\pi})\\
    \theta = 800 \ \texttt{revolutions}
    [/tex]
     
  2. jcsd
  3. Nov 22, 2015 #2

    haruspex

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    There are two forms of that equation, ##\theta=\omega_0 t+\frac 12 \alpha t^2## or ##\theta=\omega_f t-\frac 12 \alpha t^2##.
    In the present case, it would have been natural to use the second form. Note that alpha as calculated is negative.
    But by considering time in reverse you can swap ##\omega_0## and ##\omega_f## and negate the signs on theta and alpha.
     
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