Answer:
a = 1697.56 m/s²
Explanation:
Given,
Electric motor Spin= 2130 rev/min
armature radius, r = 3.412 cm = 0.03412 m
acceleration of the outer edge = ?
angular speed = [tex]2130\times \dfrac{2\pi}{60}[/tex]
= 223.05 rad/s
acceleration of the motor
a = ω²r
a = 223.05² x 0.03412
a = 1697.56 m/s²
The acceleration of the outer edge of the armature is equal to 1697.56 m/s².
