A compact disc (CD) stores music in a coded pattern of tiny pits 10-7 m deep. Th

Question

A compact disc (CD) stores music in a coded pattern of tiny pits 10-7 m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; suppose the inner and outer radii of this spiral are 20.5 mm and 56.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.11 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? innermost rad/s 0utermost- rad/s (b) If the maximum playing time of a CD is 80.0 min. What would be the length of the track on such a maximum-duration CD if it were stretchec out in a straight line? km (c) What is the average angular acceleration of a maximum-duration CD during its 80.0-min playing time? Take the direction of rotation of the disc to be positive. rad/s2

Explanation / Answer

Given,

ri = 20.5 mm ; ro = 56 mm

v = 1.11 m/s

we know that,

v = r w => w = v/r

w(innermost) = v/ri = 1.11/20.5 x 10^-3 = 54.15 rad/s

w(outermost) = v/ro = 1.11/56 x 10^-3 = 19.82 rad/s

Hence, w(innermost) = 54.15 rad/s ; w(outermost) = 19.82 rad/s

b)t = 80 min = 4800 sec

d = v t = 1.11 x 4800 = 5328 m = 5.238 km

Hence, d = 5.238 km

c)Average angular acceleration should be zero. since it beigns and stops with the same speed. so average is zero.

alpha = 0 rad/s^2

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