An electrical motor spins at a constant 2695.0 rpm.If the armature radius is 7.165 cm,what is the acceleration of the edge of the rotor? A)5707 m/s² B)281.6 m/s² C)572,400 m/s² D)28.20 m/s²

Question

Quizplus · Accepted Answer

First, we need to convert the rotational speed from rpm to rad/s:
2695.0 rpm = 282.2 rad/s

Then, we can use the following formula for centripetal acceleration:

a = r * ω^2

where:
a = centripetal acceleration
r = armature radius = 0.07165 m
ω = angular velocity = 282.2 rad/s

Plugging in these values:

a = 0.07165 m * (282.2 rad/s)^2
a ≈ 5707 m/s^2

Therefore, the acceleration of the edge of the rotor is approximately 5707 m/s^2, which is answer choice A.

An Electrical Motor Spins at a Constant 2695