Chapter 4, Problem 16RE

a.

To determine

To find: the angular speed of a record album.

Answer to Problem 16RE

  $66\frac{2}{3}\pi \text{ radian per minute}$

Explanation of Solution

Given: Diameter of record $12\text{ inches}$

No. of revolution per minute = $33\frac{1}{3}=\frac{100}{3}$

Concept Used:

If $\theta$ is the angle (in radian) corresponding to the arc length s , then the angular speed w is:

  $w=\frac{\theta }{t}$

It is known that each revolution corresponds to $2\pi$ radians.

So, in $\frac{100}{3}$ revolutions, the record turns $\theta =\frac{100}{3}\times 2\pi =\frac{200}{3}\pi \text{ radians}$ .

So, the maximum angular speed is,

  $\begin{array}{c}w=\frac{\theta }{t}\\ =\frac{\frac{200\pi }{3}\text{ radian}}{1\text{ minute}}\\ =\frac{200}{3}\pi \text{ radian per minute}\\ =66\frac{2}{3}\pi \text{ radian per minute}\end{array}$

b.

To determine

To find: the linear speed (in inches per minute) of the outer edge of a record album.

Answer to Problem 16RE

Approximately $\text{1,256}\text{.64 inches per minute}$

Explanation of Solution

Given: Diameter of record $12\text{ inches}$

No. of revolution per minute = $33\frac{1}{3}=\frac{100}{3}$

Concept Used:

If $\theta$ is the angle (in radian) corresponding to the arc length s , then the linear speed v is:

  $v=\frac{s}{t}$

The arc length s is given by,

  $s=r\theta$

It is known that each revolution corresponds to $2\pi$ radians.

So, in $\frac{100}{3}$ revolutions, the record turns $\theta =\frac{100}{3}\times 2\pi =\frac{200}{3}\pi \text{ radians}$ .

The radius of the record is,

  $\begin{array}{c}r=\frac{d}{2}\\ =\frac{12}{2}\\ =6\text{ inches}\end{array}$

So, the maximum linear speed is,

  $\begin{array}{c}v=\frac{s}{t}\\ =\frac{r\theta }{t}\\ =\frac{6\left(\frac{200\pi }{3}\right)\text{ inches}}{1\text{ minute}}\end{array}$

  $\begin{array}{c}=400\pi \text{ inches per minute}\\ \approx \text{1,256}\text{.64 inches per minute}\end{array}$