An record player rotates on a turntable at a rate of 45 revolutions per minute (
ID: 1441650 • Letter: A
Question
An record player rotates on a turntable at a rate of 45 revolutions per minute (rpm). What is the period of motion? If the record has a radius of 9 cm, what how fast is the edge of the record travelling? A coin is placed at the edge of the record in the previous problem. The record starts spinning from rest. When the record spins up to 36 rpm, the coin falls off. What was the coefficient of static friction? A 200g block spins in a horizontal circle with a radius of 50 cm and a frequency of 50 rpm. What is the speed of the block?Explanation / Answer
Here ,
1)
angular velocity , wi = 45 rev/mi
wi = 4.715 rad/s
r = 0.09 cm
period of motion = 2pi/wi
period of motion = 2pi/4.715
period of motion = 1.33 s
----
speed of edge = r * w
speed of edge = 0.09 * 4.715 m/s
speed of edge = 0.424 m/s
2)
let the static friction coefficinet is us
for the coin to be placed at the edge without slipping
centripetal force = frictional force
us * m * g= m * v^2/r
us * 9.8 = 0.424^2/0.09
us = 0.204
the static friction coefficinet is 0.204
3)
m = 0.20 Kg
r= 0.50 m
f = 50 rpm = 5.235 rad/s
speed of block = f * r
speed of block = 5.235 * 0.50 m/s
speed of block = 2.62 m/s
the speed of block is 2.62 m/s
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