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 37 cm from the center; the inner edge is at a radial distance of 18 cm. The record rotates at 3.5 rev/min. The needle of the pick-up arm takes 1.5 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

a) radial speed = speed along the radius = d / t = (0.37-0.18)m / 1.5*60s = 0.00211 m/s

b) w = 3.5*2pi rad / 60s = 0.367 rad/s
So,
Outer edge speed is v = r*w = 0.37m * 0.367 rad/s = 0.1358 m/s

c) Inner edge speed is v = r*w = 0.18m * 0.367 rad/s = 0.0661 m/s

d) accel = dw / dt = 0.367 rad/s / 10s = 0.0367 rad/s^2

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