Why would we ignore initial angular velocity in equation?

[bob tran]

Homework Statement

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

Homework Equations

ω_f=ω_o+αt
θ=θ₀+ω_ot+0.5αt²

The Attempt at a Solution

(a) <br /> \omega_f=\omega_0+\alpha t\\<br /> 0 = (20 * 2\pi) + 80\alpha\\<br /> \alpha = -\frac{20 * 2\pi}{80}\\<br /> \alpha = -1.57 \frac{\texttt{rad}}{\texttt{s}^2}<br />
(b) This is where I am confused. Why do we assume w₀=0 to get the answer?<br /> \theta = \omega_0 t + \frac{1}{2} t^2 \alpha\\<br /> \theta = \frac{1}{2} t^2 \alpha\\<br /> \theta = \frac{1}{2} (80)^2 (\frac{-1.57}{2\pi})\\<br /> \theta = 800 \ \texttt{revolutions}<br />

 

[haruspex]

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.