Chapter 4.1, Problem 65E

a.

To determine

Find an interval for the angular speed of the DVD.

Answer to Problem 65E

The angular speed of the carousel is $25.133\text{radians/minute}$ .

Explanation of Solution

Given:

The DVD player rotates between $200$ and $500$ revolutions/minute.

Calculation:

Find the angular speed for $200$ revolutions/minute.

  $\begin{array}{l}1\text{revolution}=\text{2}\pi \text{radians}\\ \theta =200\text{revolutions}\times \text{2}\pi \\ \omega =\frac{\theta }{t}=\frac{200\text{revolutions}\times \text{2}\pi }{\text{minute}}\approx 1256.637\text{radians/minute}\end{array}$

Find the angular speed for $500$ revolutions/minute.

  $\begin{array}{l}1\text{revolution}=\text{2}\pi \text{radians}\\ \theta =500\text{revolutions}\times \text{2}\pi \\ \omega =\frac{\theta }{t}=\frac{500\text{revolutions}\times \text{2}\pi }{\text{minute}}\approx 3141.593\text{radians/minute}\end{array}$

Hence theDVD rotates in betweenthe interval of angular speed $1256.637\text{radians/minute}$ and $3141.593\text{radians/minute}$ .

b.

To determine

Find the interval for the linear speed of the DVD.

Answer to Problem 65E

The DVD rotates in between the interval of linear speed $7539.822\text{centimeters}/\text{minute}$ and $18849.556\text{centimeters}/\text{minute}$ .

Explanation of Solution

Given:

The DVD player rotates between $200$ and $500$ revolutions/minute. The diameter of the DVD is $12$ centimeters.

Calculation:

For $200$ revolutions/minute.

  $\begin{array}{l}1\text{revolution}=\text{2}\pi \text{radians}\\ \theta =200\text{revolutions}\times \text{2}\pi \end{array}$

Radius of the DVD $=\frac{\text{Diameter}}{2}=\frac{12}{2}=6$ centimeters

  $\begin{array}{l}\text{Linear}\text{speed}=\frac{s}{t}=\frac{r\theta }{t}\\ =\frac{6\text{centimeters}\times 200\text{revolutions}\times \text{2}\pi }{\text{minute}}\\ \approx 7539.822\text{centimeters}/\text{minute}\end{array}$

For $500$ revolutions/minute.

  $\begin{array}{l}1\text{revolution}=\text{2}\pi \text{radians}\\ \theta =500\text{revolutions}\times \text{2}\pi \end{array}$

Radius of the DVD $=\frac{\text{Diameter}}{2}=\frac{12}{2}=6$ centimeters

  $\begin{array}{l}\text{Linear}\text{speed}=\frac{s}{t}=\frac{r\theta }{t}\\ =\frac{6\text{centimeters}\times 500\text{revolutions}\times \text{2}\pi }{\text{minute}}\\ \approx 18849.556\text{centimeters}/\text{minute}\end{array}$

Hence the DVD rotates in between the interval of linear speed $7539.822\text{centimeters}/\text{minute}$ and $18849.556\text{centimeters}/\text{minute}$ .