An electrical motor spins at a constant 2857.0 rev/min. If the armature radius is 2.685 cm, what is the acceleration of the outer edge of the armature? A) 2403 m/s² B) 844.4 m/s² C) 241,100 m/s² D) 84.40 m/s²

Question

Quizplus · Accepted Answer

The acceleration of the outer edge of the armature is the centripetal acceleration, which can be calculated using the formula a = ω²r, where ω is the angular velocity in rad/s and r is the radius in meters. First, convert 2857.0 rev/min to rad/s: ω = 2857.0 * (2π/60) = 299.53 rad/s. Then, convert the radius to meters: r = 2.685 cm = 0.02685 m. Finally, calculate the acceleration: a = (299.53)² * 0.02685 = 2403 m/s².

An Electrical Motor Spins at a Constant 2857