Chapter 8, Problem 37P

To determine

To find: The revolutions that make the rotor turn.

Answer to Problem 37P

The time taken is $10.9\text{s}$ .

Explanation of Solution

Given information:

A centrifuge rotor rotating at $10300\text{rpm}$ is shut off and is eventually brought uniformly to rest by a frictional torque of $1.20\text{N}\cdot \text{m}$ . If the mass of the rotor is $4.80\text{kg}$ and it can be approximated as a solid cylinder of radius $0.0710\text{m}$ , through how many revolutions

Formula used:

  $I=\frac{1}{2}M{R}^2$

Here,

  $I$ is the moment of inertia,

  $M$ is the mass of the cylinder,

  $R$ is the radius of cylinder.

Calculation:

The speed of the rotor $\left(\omega \right)=10300\text{rpm}$

The torque $\left(\tau \right)=1.20\text{N}\cdot \text{m}$

The mass of the rotor $\left(M\right)=4.80\text{kg}$

The radius of the solid cylinder $\left(R\right)=0.0710\text{m}$

The moment of inertia of the solid cylinder $I=\frac{1}{2}M{R}^2$

The angular acceleration $\alpha =\tau /I$

From the equations of the rotational kinematics,

  $\begin{array}{l}{\omega }^2={\omega }_0^2+2\alpha \theta \\ \theta =\frac{{\omega }^2-{\omega }_0^2}{2\alpha }\\ =\frac{0-{\omega }_0^2}{2\left(\tau /I\right)}\\ \theta =\frac{-{\omega }_0^{2}I}{2\tau }\\ \theta =\frac{-M{R}^2{\omega }_0^2}{4\tau }\\ \theta =\frac{-4.80\times {\left(0.0710\right)}^2{\left[\left(\frac{\text{10300}\text{rev}}{\text{min}}\right)\left(\frac{\text{1min}}{\text{60}\text{s}}\right)\left(\frac{\text{2π}\text{rad}}{\text{1}\text{rev}}\right)\right]}^2}{4\left(-1.20\text{N}\cdot \text{m}\right)}\\ =\left(5.865\times {10}^3\text{rad}\right)\left(\frac{\text{1}\text{rev}}{\text{2π}\text{rad}}\right)\\ \theta =933\text{rev}\end{array}$

The time is found from the equation of motion

  $\begin{array}{l}\theta =\frac{1}{2}\left({\omega }_0+\omega \right)t\\ t=\frac{2\theta }{{\omega }_0+\omega }\\ =\frac{\text{2}\left(\text{933}\text{rev}\right)}{\text{10300}\text{rev/min}}\left(\frac{\text{60}\text{s}}{\text{1}\text{min}}\right)\\ t=10.9\text{s}\end{array}$

Conclusion:

The time taken is $10.9\text{s}$ .