A compact disc (CD) stores music in a coded pattern of tiny pits10 -7 m deep. Th

Question

A compact disc (CD) stores music in a coded pattern of tiny pits10-7 m deep. The pits are arranged in a track thatspirals outward toward the rim of the disc; suppose the inner andouter radii of this spiral are 26.0 mm and52.0 mm, respectively. As the disc spinsinside a CD player, the track is scanned at a constantlinear speed of 1.19 m/s. (a) What is the angular speed of the CD whenthe innermost part of the track is scanned? The outermost part ofthe track? innermost = 1 rad/s outermost = 2 rad/s
(b) If the maximum playing time of a CD is 71.0 min. What would be the length of the track onsuch a maximum-duration CD if it were stretched out in a straightline?
3 km

(c) What is the average angular acceleration of a maximum-durationCD during its 71.0-min playing time? Takethe direction of rotation of the disc to be positive.
4 rad/s2
(a) What is the angular speed of the CD whenthe innermost part of the track is scanned? The outermost part ofthe track? innermost = 1 rad/s outermost = 2 rad/s
(b) If the maximum playing time of a CD is 71.0 min. What would be the length of the track onsuch a maximum-duration CD if it were stretched out in a straightline?
3 km

(c) What is the average angular acceleration of a maximum-durationCD during its 71.0-min playing time? Takethe direction of rotation of the disc to be positive.
4 rad/s2 innermost = 1 rad/s outermost = 2 rad/s

Explanation / Answer

Linear speed v = 1.19 m / s Inner radius r = 26 mm =0.026 m Outer radius of spiral r ' = 52.0 mm= 0.052 m (a). the angular speed of the CD when the innermost part ofthe track is scanned w = v / r                                w = 58.46 rad / s    The angular speed of the CD when outermost partof the track is scanned w ' = v / r '                      w ' = 22.88 rad / s (b). time t= 71 min = 71 * 60 s = 4260 s Length of the track L = v t                                  = 5069.4 m                                 = 5.0694 km (c). the average angular acceleration of a maximum-duration CDduring its 71.0-min playing time = ( w + w ' ) / 2t                              plug the values we get = 0.00954 rad / s ^ 2          

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