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

Question

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 16 cm. The record rotates at 6.5 rev/min. The needle of the pick-up arm takes 2.9 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/s^2

Explanation / Answer

i)radial v = d / t = (0.39 – 0.16)m / (2.9*60)s = 0.00132 m/s

ii) speed of the outer edge

tang v = r = (6.5*/30 rad/s) * 0.39m = 0.265 m/s

iii) speed of the inner edge

tang v' = (6.5*/30 rad/s) * 0.16m = 0.1089 m/s

iv) = / t = -(6.5*/30 rad/s) / 10s = -0.068 rad/s²

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