Question

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?

Answer 1

The outer edge of the armature has an acceleration is 1969.45 m/
`s^2`.

Step-by-step explanation:

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

=
`2883.5 * (2\pi)/(60) = 301.96 rad/ s`

There are 2
`\pi` 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
`(2\pi)/(60)`.

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 ×
`\omega^2`

= 0.0216 ×
`(301.96)^2`

= 1969.45 m/
`s^2`

Therefore the acceleration of the outer edge of the armature is 1969.45 m/
`s^2`