Question

The outer edge of the grooved area of a long-playing record is at a radial distance...

The outer edge of the grooved area of a long-playing record is at a radial distance 39 cm from the center; the inner edge is at a radial distance of 17 cm. The record rotates at 5.0 rev/min. The needle of the pick-up arm takes 2.8 minutes to move uniformly from the outer edge to the inner edge. What is the radial speed of the needle? m/s What is the speed of the outer edge relative to the needle? m/s What is the speed of the inner edge relative to the needle? m/s Suppose the phonograph is turned off, and the record uniformly and stops rotating after 10 s. What is the angular acceleration? rad/s2

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Answer #1
The angular speed of the disk is ? = 5 rev/min
                                                  = (5 rev/min)(2? rad/rev)
                                                  = 31.416 rad/s
The radius of the outer edge is Ro = 39 cm
                                                  = 0.39 m
The radius of the inner edge is Ri = 17 cm
                                                 = 0.17 m
The time taken by the pin to move from outer edge to inner edge radially is t = 2.8 min
                                                                                                               = 168 s

The radial speed of the needle = radial diatnce/time
                                             = (0.39 m - 0.17 m)/168 s
                                             = 1.309*10-3 m/s

The speed of the outer edge relative to the needle is vo = Ro?
                                                                               = (0.39 m)(31.416 rad/s)
                                                                               = 12.252 m/s

The speed of the inner edge relative to the needle is vi = Ri?
                                                                               = (0.17 m)(31.416 rad/s)
                                                                               = 5.34 m/s

The time taken by the record to stop is t = 10 s
The angular acceleration of the disk is
            ? = (0-?)/t
               = -31.416 rad/s / 10 s
               = -3.1416 rad/s2
negative sign indicates the deceleration
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