Question

A centrifuge rotor rotating at 9200 rpm is shut off and eventually brought uniformly to rest by a frictional torque of \({\bf{1}}{\bf{.20}}\;{\bf{m}} \cdot {\bf{N}}\). If the mass of the rotor is 3.10 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?

Short answer

The rotor moves \(481\;{\rm{rev}}\) before coming to rest.

The rotor takes 6.27 s before coming to rest.

Step-by-step solution

Concepts

Torque is the product of the moment of inertia and the square of the distance. For this problem, first, find the relationship between angular acceleration and moment of inertia. Then, find the relationship between angular displacement and angular acceleration.

Given data

The initial angular velocity of the rotor is \({\omega _1} = 9200\;{\rm{rpm}} = \frac{{9200 \times 2\pi }}{{60}}\;{\rm{rad/s}}\).

The final angular velocity is \({\omega _2} = 0\).

The frictional torque is \(\tau = 1.20\;{\rm{m}} \cdot {\rm{N}}\).

The mass of the rotor is \(m = 3.10\;{\rm{kg}}\).

The radius of the rotor is \(r = 0.0710\;{\rm{m}}\).

You can assume that the centrifuge rotor is a solid cylinder.

Let the centrifuge rotor take t time before come to rest.

Calculation for the total revolution before coming to rest

The moment of inertia of the rotor is \(I = \frac{1}{2}m{r^2}\).

Now, the torque can be written as

\(\begin{align}\tau &= I\alpha \\\tau &= \frac{1}{2}m{r^2}\alpha \\\alpha &= \frac{{2\tau }}{{m{r^2}}}\end{align}\).

Now, for this angular motion of the rotor,

\(\begin{align}\omega _2^2 &= \omega _1^2 + 2\alpha \theta \\\theta &= \frac{{\omega _2^2 - \omega _1^2}}{{2\alpha }}\\\theta &= \frac{{\omega _2^2 - \omega _1^2}}{{2 \times \frac{{2\tau }}{{m{r^2}}}}}\\\theta &= \frac{{m{r^2}\left( {\omega _2^2 - \omega _1^2} \right)}}{{4\tau }}\end{align}\).

Now, substituting all the values,

\(\begin{align}\theta &= \frac{{\left\{ {\left( {3.10\;{\rm{kg}}} \right) \times {{\left( {0.0710\;{\rm{m}}} \right)}^2}} \right\}\left\{ {0 - {{\left( {\frac{{9200 \times 2\pi }}{{60}}\;{\rm{rad/s}}} \right)}^2}} \right\}}}{{4 \times \left( {1.20\;{\rm{m}} \cdot {\rm{N}}} \right)}}\\ &= 3018.76\;{\rm{rad}}\\ &= \left( {3018.75\;{\rm{rad}}} \right) \times \left( {\frac{{1\;{\rm{rev}}}}{{2\pi \;{\rm{rad}}}}} \right)\\ \ &approx 481\;{\rm{rev}}\end{align}\).

Hence, the rotor moves \(481\;{\rm{rev}}\)before coming to rest.

Calculation for the time to come to rest

Now, the average angular velocity is \(\frac{1}{2}\left( {{\omega _1} + {\omega _2}} \right)\).

\(\begin{align}\theta &= \frac{1}{2}\left( {{\omega _1} + {\omega _2}} \right)t\\t &= \frac{{2\theta }}{{{\omega _1} + {\omega _2}}}\\ &= \frac{{2 \times 3018.76\;{\rm{rad}}}}{{\frac{{9200 \times 2\pi }}{{60}}\;{\rm{rad/s}} + 0}}\\ &= 6.27\;s\end{align}\).

Hence, the rotor takes 6.27 s before coming to rest.