An electic motor spins in uniform circular motion at a constant 2,883.5 rev/min. If the armature radius is 2.16 cm, what is the acceleration of the outer edge of the armature?

The outer edge of the armature has an acceleration is 1969.45 m/.

Explanation:

The speed of an electric motor, = 2883.5 rev / min

                                                        =  

There are 2 in one revolution and there 60 seconds in one minute. So to convert form revs/min to rad/sec we multiply the value in revs/min by .

The radius of the armature, r  = 2.16 cm = 0.0216m

The acceleration of the outer edge of the armature  is given by the formula

= r ×

=  0.0216 ×

= 1969.45 m/

Therefore the acceleration of the outer edge of the armature is 1969.45 m/

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