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

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 24.5 mm and 52.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.36 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?

(b) If the maximum playing time of a CD is 73.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line?
,...........km

(c) What is the average angular acceleration of a maximum-duration CD during its 73.0-min playing time? Take the direction of rotation of the disc to be positive.
........rad/s2

Explanation / Answer

A)There are 2pi(r) lengths of circumference around a circle of radius r and 2pi radians in each revolution
The inner track length is 2pi(0.0245) m/rev = 0.049pi m
1.36m/s / 0.049pi m/rev) = 27.75/pi rev/sec
at 2pi radians per revolution
27.75/pi(2pi) =55.51 rad/sec ANSWER

B) outer track same formulas
2pi(0.052) m/rev = 0.104pi m
1.36 m/s / 0.104pi m/rev) = 13.07/pi rev/sec
at 2pi radians per revolution
13.07/pi(2pi) =26.15 rad/sec ANSWER

C) 73 min (60 sec/min) (1.36 m/sec) = 5956.8m ANSWER

D) as a cd normally starts at zero speed and acceleration and ends at the same condition. the average acceleration is (0-0)/73 or a = 0 rad/s/s ANSWER

Another way of question is

if the time duration is considered as first and last read then

alpha=dw/dt=55.51-26.15/4380=6.703 *10^-3 rad/s^2

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