Part A A small button placed on a horizontal rotating platform with diameter 0.3

Question

Part A A small button placed on a horizontal rotating platform with diameter 0.330 m will revolve with the platform when it is brought up to a rotational speed of 45.0 rev/min , provided the button is a distance no more than 0.148 m from the axis. What is the coefficient of static friction between the button and the platform? SubmitPr Prevlous Answers Request Answer X Incorrect; Try Again; 6 attempts remaining Part B How far from the axis can the button be placed, without slipping, if the platform rotates at 65.0 rev/min? Submit Request Answer

Explanation / Answer

part A

the net force on button = 0

Fnet = 0

fs - Fc = 0

fs = Fc

fs = static frictional force = us*m*g

Fc = centripetal force = m*r*w^2

us*m*g = m*r*w^2

us = r*w^2/g

r = 0.148 m

w = 45 rev/min = 45*2*pi/60 = 4.71 rad/s

g = 9.8 m/s^2

therefore coefficient of static friction us = 0.148*4.71^2/9.8 = 0.335 <<<<<------------ANSWER

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part B

for angular speed w = 65 rev/min = 65*2pi/60 = 6.81 rad/s

r = us*g/w^2

r = 0.335*9.8/6.81^2

r = 0.071 m <<<<<------------ANSWER

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