Question

A compact disc (CD) stores music in a coded pattern of tiny pits 10-7 m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s.

(a) What is the angular speed of the CD when the innermost part of the track is scanned

? The outermost part of the track?

(b) The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line?

(c) What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time?

d) Take the direction of rotation of the disc to be positive.

Short answer

The outermost part of the track is $50\frac{rad}{s}$.

(b) Thus, the angular speed.

(c) The direction of rotation of the disc5550m.

(d) Angular acceleration is $-6.4\times {10}^{-3}\frac{rad}{s^2}$.

Step-by-step solution

Step:-1 Expression 

The angular acceleration speed of the CD is related to the linear speed $\omega =\frac{V}{r}$.

Where,$\omega$ is the angular speed.

r is the linear speed.

v is the distance from the center of the CD.

Step:-2 Concept

The angular speed of the CD when the innermost part of the track is scanned.

We know that the angular speed is,

$\omega =\frac{v}{r}$

Step:-3 calculation

We have $v=1.25\frac{m}{s}$.

role="math" localid="1667995843373" $r=25.0mm$

$$\begin{gathered} =\frac{25}{1000} \\ =0.025m \\ =\frac{1.25{\displaystyle \frac{m}{s}}}{0.025m} \\ =50\frac{rad}{s} \end{gathered}$$

(b) Step:-4 explanation 

We know that the angular speed of the CD is given by $\omega =\frac{V}{r}$.

role="math" localid="1667996101874" $v=1.25\frac{m}{s}$and

$$\begin{gathered} r=58.0m \\ =0.058m \end{gathered}$$

(b) Step:-5 Concept 

the length of the track on such a maximum-duration CD if it were stretched out in a straight.

We know that the angular speed is,

$\omega =\frac{v}{r}$

(b) Step:-6 calculation 

$$\begin{gathered} \omega =\frac{1.25{\displaystyle \frac{m}{s}}}{0.058m} \\ =21.6\frac{rad}{s} \end{gathered}$$

(c) Step:-7 Expression 

the maximum playing CD is,

$$\begin{gathered} t=74.0min\times 60\frac{s}{\mathrm{mm}} \\ =440s \end{gathered}$$.

We know that the linear speed of the track is

$v=1.25\frac{m}{s}$

(c) Step:-8 concept  

The average angular acceleration of a maximum-duration CD during its 74.0-min playing time.

we would have a uniform motion total length of the track is,

d = vt------(1)

(c) Step:-9 solution 

Here put the vale v and t in equation (1)

$$\begin{gathered} =\frac{\left(125\frac{m}{s}\right)}{\left(4440s\right)} \\ =5550m \end{gathered}$$

(d) Step:-10 expression 

We know that the angular acceleration of the CD is,

$\alpha =\frac{{\omega }_s-{\omega }_t}{t}$________(1)

(d) Step:11 concept  

We know that$\alpha =\frac{{\omega }_s-{\omega }_t}{t}$

Put the value ${\omega }_f=21.6\frac{rad}{s}$.

This is the final angular speed.

${\omega }_i=50.0\frac{rad}{s}$.

Initial angular speed.

$t=4440s$

(d) Step:-12 Solution 

Put the value in the equation (1)

$$\begin{gathered} \alpha =\frac{21.6\frac{rad}{s}-50.0\frac{rad}{s}}{4440s} \\ =-6.4\times {10}^{-3}\frac{rad}{s^2} \end{gathered}$$

Hence, Angular acceleration is $-6.4\times {10}^{-3}\frac{rad}{s^2}$.