## Question:

The rotor of an electric motor has an angular velocity of 3600 rpm when the load and power are cut off. The 50-kg rotor, which has a centroidal radius of gyration of 180 mm, then coasts to rest. Knowing that kinetic friction results in a couple of magnitude 3.5 N m⋅ exerted on the rotor, determine the number of revolutions that the rotor executes before coming to rest.

## Step-by-step

$\overline { I } =m{ \overline { k } }^{ 2 }$

=$\left( 50 \right) { \left( 0.180 \right) }^{ 2 }$

$=1.62kg.{ m }^{ 2 } M=\overline { I } \alpha :3.5N.$

$m=\left( 1.62kg.{ m }^{ 2 } \right) \alpha$

$\alpha =2.1605rad/{ s }^{ 2 }$ (deceleration)

${ \omega }_{ 0 }=3600rpm\left( \frac { 2\pi }{ 60 } \right)$
=$120\pi \quad rad/s$

${ \omega }^{ 2 }={ \omega }_{ 0 }^{ 2 }+2\alpha \theta$

$0={ \left( 120\pi \quad rad/s \right) }^{ 2 }+2\left(-2.1605rad/{ s }^{ 2 } \right) \theta$

$\theta =32.891\times { 10 }^{ 3 }rad$

$=5235.26rev$

or $\theta=5230rev$