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

Answer:

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

Explanation:

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

                                                        =  $2883.5 \times \frac{2\pi}{60} = 301.96 rad/ s$

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 $\frac{2\pi}{60}$.

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/$s^2$