Chapter 4.1, Problem 66E

a.

To determine

To find: the maximum angular speed (in radians per second) of a Blu-ray disc as it rotates.

Answer to Problem 66E

  $333.33\pi \text{ radian per second}$

Explanation of Solution

Given: Diameter of disc $12cm$

Maximum No. of revolution per minute = 10,000

Concept Used:

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

  $w=\frac{\theta }{t}$

Maximum angular speed is obtained when the disc rotates maximum number of revolution per minute.

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

So, in 10,000 revolutions, the disc turns $\theta =10,000\times 2\pi =20,000\pi \text{ radians}$ .

So, the maximum angular speed is,

  $\begin{array}{c}w=\frac{\theta }{t}\\ =\frac{20,000\pi \text{ radian}}{1\text{ minute}}\\ =\frac{20,000\pi \text{ radian}}{\text{60 second}}\\ =333.33\pi \text{ radian per second}\end{array}$

b.

To determine

To find: the maximum linear speed (in meters per second).

Answer to Problem 66E

Approximately $62.83\text{ meters per second}$

Explanation of Solution

Given: Diameter of disc $d=12cm$

Maximum No. of revolution per minute = 10,000

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,

  $s=r\theta$

Maximum linear speed is obtained when the disc rotates maximum number of revolution per minute.

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

So, in 10,000 revolutions, the disc turns $\theta =10,000\times 2\pi =20,000\pi \text{ radians}$ .

The radius of the disc is,

  $\begin{array}{c}r=\frac{d}{2}\\ =\frac{12}{2}\\ =6cm\\ =0.06m\end{array}$

So, the maximum linear speed is,

  $\begin{array}{c}v=\frac{s}{t}\\ =\frac{r\theta }{t}\\ =\frac{0.06\left(20,000\pi \right)\text{ meters}}{1\text{ minute}}\end{array}$

  $\begin{array}{c}=\frac{0.06\left(20,000\pi \right)\text{ meters}}{\text{60 second}}\\ \approx 62.83\text{ meters per second}\end{array}$