An electrical motor spins at a constant 2695.0 rpm. If the armature radius is 7.165 cm, what is the acceleration of the edge of the rotor?

A) 28.20 m/s^2

Angular velocity, w = 2695 rpm = 2695*2ft/60 = 282.22 rad/s,

Radius, r = 7.165 cm = 0.07165 meters,
So,
linear velocity, v = rw = 0.07165*282.22 = 20.22 m/s,
Therefore,
Acceleration, a = v^2/r = (20.22)^2/0.07165 = 5706.78 m/s^2 >========< ANSER