Answer
A point on the rim moves through a distance of $~~83.2~m$
Work Step by Step
We can express the angular speed in units of $rad/s$:
$\omega_f = (2760~rev/min)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s}) = 289~rad/s$
We can find the angle $\theta$ through which the wheel rotates:
$\omega_f^2 = \omega_0^2+2~\alpha~\theta$
$\omega_f^2 = 0+2~\alpha~\theta$
$\theta = \frac{\omega_f^2}{2\alpha}$
$\theta = \frac{(289~rad/s)^2}{(2)(14.2~rad/s^2)}$
$\theta = 2940.9~rad$
We can find the distance that a point on the rim moves:
$d = \theta~r$
$d = (2940.9~rad)(0.0283~m)$
$d = 83.2~m$
A point on the rim moves through a distance of $~~83.2~m$
