A compact disc stores music in a coded pattern of tinypits 10^ (-7) m deep. the pits are arranged in a track thenspirals outward toward the rim of the disc; the inner and theouter radii of this spiral are 25.0 mm and 58.0 mmrespectively. as the disc spins inside a CD player, the trackis scanned at a constant linear speed of 1.25 m/s

a) What is the angular speed of the CD when scanning theinnermost part of the track? the outtermost part of thetrack?

b) the maximum playing time of a CD is 74.0 minutes. whatwould be the length of the track on such a maximum-duration CD ifit were stretched out in a straight line?

c) what is the average angular acceleration of amaximim-duration CD during its 74.0 minutes playing time? take the direction of rotation of the disc to be positive.

Question

Physics

Alissa Harper

12 July, 07:46

A compact disc stores music in a coded pattern of tinypits 10^ (-7) m deep. the pits are arranged in a track thenspirals outward toward the rim of the disc; the inner and theouter radii of this spiral are 25.0 mm and 58.0 mmrespectively. as the disc spins inside a CD player, the trackis scanned at a constant linear speed of 1.25 m/s

a) What is the angular speed of the CD when scanning theinnermost part of the track? the outtermost part of thetrack?

b) the maximum playing time of a CD is 74.0 minutes. whatwould be the length of the track on such a maximum-duration CD ifit were stretched out in a straight line?

c) what is the average angular acceleration of amaximim-duration CD during its 74.0 minutes playing time? take the direction of rotation of the disc to be positive.

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Answers (1)

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Cheese · Answer 1

v = ω R

v is linear speed and ω is angular speed

ω = v / R

a) Inner radius = 25 x 10⁻³ m

speed v = 1.25 m/s

ω = 1.25 / (25 x 10⁻³)

=.05 x 10⁻³

= 5 x 10⁻⁵ rad / s

outer radius = 58 x 10⁻³ m

speed v = 1.25 m/s

ω = 1.25 / (58 x 10⁻³)

=.0215 x 10⁻³

= 2.15 x 10⁻⁵ rad / s

b)

linear constant speed v = 1.25 m / s

time = 74 min = 74 x 60 s

distance tracked = speed x time

= 1.25 x 74 x 60

= 5550 m

c) time given

= 74 min = 74 x 60 s

angular acceleration

= (2.15 - 5) x 10⁻⁵ / (74 x 60)

= - 6.42 x 10⁻⁹ rad / s²