An electrical motor spins at a constant 2662.0 rpm. If the armature radius is 6.725 cm, what is the acceleration of the edge of the rotor? O 524,200 m/s O 29.30 m/s O292.7 m/s O 5226 m/s2

Question

Physics

Hudson Barrett

14 December, 12:40

An electrical motor spins at a constant 2662.0 rpm. If the armature radius is 6.725 cm, what is the acceleration of the edge of the rotor? O 524,200 m/s O 29.30 m/s O292.7 m/s O 5226 m/s2

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Answers (1)

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Agustin Caldwell · Answer 1

18.73 m/s^2

Explanation:

f = 2662 rpm = 2662 / 60 rps

r = 6.725 cm = 0.06725 m

Acceleration, a = r w

a = r x 2 x pi x F

a = 0.06725 * 2 * 3.14 * 2662 / 60

a = 18.73 m/s^2