Question

A digital audio compact disc carries data, each bit of which occupies 0.6 µm along a continuous spiral track from the inner circumference of the disc to the outside edge. A CD player turns the disc to carry the track counterclockwise above a lens at a constant speed of 1.40 m/s.
(a) Find the required angular speed at the beginning of the recording, where the spiral has a radius of 2.60 cm. rad/s
(b) Find the angular speed at the end of the recording, where the spiral has a radius of 5.70 cm. rad/s
(c) A full-length recording lasts for 74 min 33 s. Find the average angular acceleration of the disc. rad/s²
(d) Assuming that the acceleration is constant, find the total angular displacement of the disc as it plays. rad
(e) Find the total length of the track.

Answer 1

Answer:

Part a)

$\omega = 53.85 rad/s$

Part b)

$\omega = 24.56 rad/s$

Part c)

$\alpha = -6.55 \times 10^{-3} rad/s^2$

Part d)

$\theta = 1.75\times 10^5 Rad$

Part e)

$L = 6262.2 m$

Explanation:

Part a)

As we know that speed of the lens is 1.40 m/s

radius of the path is R = 2.60 cm

so angular speed is

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

$\omega = \frac{1.40}{0.026}$

$\omega = 53.85 rad/s$

Part b)

radius of the outer path is R = 5.70 cm

so angular speed is

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

$\omega = \frac{1.40}{0.057}$

$\omega = 24.56 rad/s$

Part c)

Average angular acceleration is rate of change in angular speed

so it is given as

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

$\alpha = \frac{24.56 - 53.85}{(74 \times 60) + 33}$

$\alpha = \frac{-29.29}{4473}$

$\alpha = -6.55 \times 10^{-3} rad/s^2$

Part d)

For constant angular acceleration the angular displacement is given as

$\theta = \frac{\omega_f + \omega_i}{2}\Delta t$

$\theta = \frac{53.85 + 24.56}{2}(4473)$

$\theta = 1.75\times 10^5 Rad$

Part e)

Total length of the track is given as

$L = vt$

$L = 1.40 \times 4473$

$L = 6262.2 m$