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. (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? Take the direction of rotation of the disc to be positive.

Answer 1

a) Angular speed of the innermost part: 50 rad/s, outermost part: 21.6 rad/s

b) The length of the track would be 5,550 m

c) The average angular acceleration is $-6.4\cdot 10^{-3}rad/s$

Explanation:

a)

For an object in uniform circular motion, the relationship between angular speed and linear speed is

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular speed

r is the radius of the circle

Here the linear speed of the track is constant:

v = 1.25 m/s

For the innermost part of the disk,

r = 25.0 mm = 0.025 m

So the angular speed is

$\omega_i = \frac{v}{r_i}=\frac{1.25}{0.025}=50 rad/s$

For the outermost part,

r = 58.0 mm = 0.058 m

So the angular speed is

$\omega_o = \frac{v}{r_o}=\frac{1.25}{0.058}=21.6 rad/s$

b)

The maximum playing time of the disk is

$t=74.0 min \cdot (60 s/min)=4440 s$

If the track was stretched out in a straight line, the motion of the track would be a uniform motion (since it is moving at constant speed), so we can use the equation

$v=dt$

where

v is the linear speed

d is the length coverted in a straight line

t is the time

Substituting v = 1.25 m/s and solving for d,

$d=vt=(1.25)(4440)=5550 m$

c)

The angular acceleration is given by

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

where

$\omega_f$ is the final angular speed

$\omega_i$ is the initial angular speed

t is the time

A CD is read from the innermost part to the outermost part, so we have:

$\omega_i = 50 rad/s$ (at the innermost part)

$\omega_f = 21.6 rad/s$ (at the outermost part)

t = 74.0 min = 4440 s is the time

Substituting, the average angular acceleration is

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

where the negative sign means the CD is decelerating.

Learn more about circular motion:

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