Question

The rotor of an electric motor has an angular velocity of 3600 rpm when the load and power are cut off. the 125-lb rotor, which has a centroidal radius of gyration of 9 in., then coasts to rest. knowing that kinetic friction results in a couple of magnitude 2.5 lb·ft exerted on the rotor, determine the number of revolutions that the rotor executes before coming to rest. (round the final answer to its nearest whole number.)

Answer 1

Answer:

      θ '= 4975 rev

Explanation:

For this exercise let's use the relationship between work and the change in kinetic energy

         W = ΔK

the expression for work is

         W = - τ Δθ

where the negative sign indicates that the torque is in the opposite direction to rotation

kinetic energy

           K = ½ I w²

         

we substitute

           - τ Δθ = 0 - ½ I w²

           θ = $\frac{1}{2} \ \frac{I w^2}{\tau}$

if we approach the rotor to a cylinder with an axis of rotation through its center

          I = ½ m r²

we substitute

          θ = ½ (½ m r²)   $\frac{w^2}{\tau}$

How the measurements are in the English system, the weight

         W = m g

          m = W / g

let's reduce to the english system

          w = 3600 rev / min (2pi rad / 1rev) (1 min / 60s) = 376.99 rad / s

          r = 9 in (1 ft / 12in) = 0.75 ft

let's calculate

           m = 125/32 = 3.91 slug

          θ = ¼ 3.91 0.75² 376.99² / 2.5

          θ = 3.126 10⁴ rad

let's reduce to revolutions

          θ’= 3.126 10⁴ rad (1rev / 2π rad)

          θ’= 4974.9 rev

          θ '= 4975 rev