An electrical motor spins at a constant 2130 rev/min. If the armature radius is 3.412 cm,} what is the acceleration of the outer edge of the armature?

 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 =

                               = 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.

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