Question

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m 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. Part A What is the angular speed of the CD when scanning the innermost part of the track? B What is the angular speed of the CD when scanning the outermost part of the track? C 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? D What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time? Take the direction of rotation of the disc to be positive.

Answer 1

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

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

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

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

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

Answer 2

Answer:  (a) 50 rad/s, (b) 21.6 rad/s, (c) 5550 m, (d) $-6.4\cdot 10^{-3} rad/s^2$

Explanation:

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m 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

Part

A What is the angular speed of the CD when scanning the innermost part of the track?

B What is the angular speed of the CD when scanning the outermost part of the track?

C 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?

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

(a) 50 rad/s

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

where

$\omega$ is the angular speed

v is the linear speed,  v = 1.25 m/s

r is the distance from the centre of the CD, r = 25.0 mm = 0.025 m

Therefore, the angular speed

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

The angular speed of the CD is

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

When scanning the outermost part of the track

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

$t =74.0 min \cdot 60 s/min = 4,440 s$

the linear speed of the track is  v = 1.25 m/s

the total length of the track would be:

d=vt=(1.25 m/s)(4,440 s)=5,550 m

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

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