- #1

Fernando Rios

- 96

- 10

- Homework Statement
- A compact disc (CD) stores music in a coded pattern of tiny pits 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 theCDwhen scanning the innermost part of the track? 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.

- Relevant Equations
- v = r*omega

v = d/t

alpha = (omega_f - omega_0)/t

a) We use the definition of linear speed in terms of angular speed:

v = r*omega

omega_f = v/r = (1.25 m/s)/(0.025 m) = 50 rad/s

omega_0 = v/r = (1.25 m/s)/(0.025 m) = 21.55 rad/s

b) We use the definition of linear speed:

v = d/t

d = vt = (1.25m/s)(74 min)(60 s/1 min) = 5.55 km

c) We use the definition of average angular acceleration:

alpha = (omega_f - omega_0)/t = (50 rad/s- 21.55 rad/s)/(74 min)(1 min/60 s) = 0.00641 rad/s^2

The answers are correct. I just wonder, why if a = r*alpha and alpha has a numerical value, a is still equal to zero (there is constant linear velocity)?

v = r*omega

omega_f = v/r = (1.25 m/s)/(0.025 m) = 50 rad/s

omega_0 = v/r = (1.25 m/s)/(0.025 m) = 21.55 rad/s

b) We use the definition of linear speed:

v = d/t

d = vt = (1.25m/s)(74 min)(60 s/1 min) = 5.55 km

c) We use the definition of average angular acceleration:

alpha = (omega_f - omega_0)/t = (50 rad/s- 21.55 rad/s)/(74 min)(1 min/60 s) = 0.00641 rad/s^2

The answers are correct. I just wonder, why if a = r*alpha and alpha has a numerical value, a is still equal to zero (there is constant linear velocity)?