(from Ohanian) The outer edge of the grooved area of a long-playing record is at

Question

(from Ohanian) The outer edge of the grooved area of a long-playing record is at a radial distance 35 cm from the center; the inner edge is at a radial distance of 17 cm. The record rotates at 3.8 rev/min. The needle of the pick-up arm takes 1.6 minutes to move uniformly from the outer edge to the inner edge.

What is the radial speed of the needle?
What is the speed of the outer edge relative to the needle?
What is the speed of the inner edge relative to the needle?
Suppose the phonograph is turned off, and the record uniformly and stops rotating after 10 s. What is the angular acceleration?

Explanation / Answer

a) radial speed = speed along the radius = d / t = (0.34-0.14)m / 2.7*60s = 0.00123 m/s

b) w = 5.7*2pi rad / 60s = 0.597 rad/s
(LP records actually rotate at 33 rpm, not 5.7 rpm, so Qu. is unrealistic)

Outer edge speed is v = r*w = 0.34m * 0.597 rad/s = 0.203 m/s

c) Inner edge speed is v = r*w = 0.14m * 0.597 rad/s = 0.0836 m/s

d) accel = dw / dt = 0.597 rad/s / 10s = 0.0597 rad/s^2

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