Chapter 10 Problem 48

Problem

A centrifuge rotor rotating at 10300 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.20 m.N. If the mass of the rotor is 3.80 kg and it can be approximated as a solid cylinder of radius 0.0710 m, through how many revolutions will the rotor turn before coming to rest, and how long will it take?

Solution

The torque on the rotor will cause an angular acceleration given by $\alpha =\tau /I$ . The torque and angular acceleration will have the opposite sign of the initial angular velocity because the rotor is being brought to rest. The rotational inertia is that of a solid cylinder.

$\omega ^{2}=\omega _{0}^{2}+2\alpha \theta$ $\rightarrow \theta ={\frac {\omega ^{2}-\omega _{0}^{2}}{2\alpha }}={\frac {0-\omega _{0}^{2}}{2\tau /I}}={\frac {-\omega _{0}^{2}}{2\tau /{\frac {1}{2}}MR^{2}}}={\frac {-MR^{2}\omega _{0}^{2}}{4\tau }}$

$={\frac {-\left(3.80{\text{ kg}}\right)\left(0.0710{\text{ m}}\right)^{2}\left[\left(10300{\text{ rpm}}\right)\left({\frac {2\pi {\text{ rad}}}{1{\text{ rev}}}}\right)\left({\frac {1{\text{ min}}}{60{\text{ s}}}}\right)\right]^{2}}{4\left(-1.20{\text{ N.m}}\right)}}$ $=4643{\text{rad}}=739{\text{rev}}$

$\theta ={\frac {1}{2}}(\omega _{0}+\omega )t$ (constant acceleration)

$\rightarrow t={\frac {2\theta }{\omega _{0}+\omega }}$ $={\frac {2(739{\text{rev}})}{10300{\text{rpm}}}}\left({\frac {60{\text{ s}}}{1{\text{ min}}}}\right)$ $=8.61{\text{s}}$