Answer
The radial acceleration of a point on the rim is $~~2360~m/s^2$
Work Step by Step
We can express the angular speed in units of $rad/s$:
$\omega = (2760~rev/min)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s}) = 289~rad/s$
We can find the radial acceleration of a point on the rim:
$a_r = \omega^2~r$
$a_r = (289~rad/s)^2~(0.0283~m)$
$a_r = 2360~m/s^2$
The radial acceleration of a point on the rim is $~~2360~m/s^2$
