A small button placed on a horizontal rotating platform with diameter 0.330 m wi

Question

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 36.0 rev/min , provided the button is no more than 0.149 m from the axis.

Part A

What is the coefficient of static friction between the button and the platform?

Express your answer to three significant figures.

Part B

How far from the axis can the button be placed, without slipping, if the platform rotates at 63.0 rev/min ?

Express your answer to three significant figures.

Explanation / Answer

here,

1 rev/min = 0.105 rad/s

diameter of button, d = 0.330 m

angular speed of button, w = 36 rev/min = 3.77 rad/s

radius of path, r = 0.149 m

Part A:
From Newton Second law, Fnet = 0
-Ff + Fc = 0 (Ff = Force of friction , Fc = Centripital force )

mv^2/r = us*mg

m*w^2*r = us*mg ( Since, Linear speed, v = w*r)

solving for coefficinet of static friction, us

us = w^2*r/g --------------------------(1)

us = 3.77^2 * 0.149 / 9.81

us = 0.216

Part B:

angular speed, w = 63 rev/min = 6.6 rad/s

rewriting eqn 1 for path distance, r

r = us*g/w^2

r = 0.216*9.81/(6.6)^2

r = 0.049 m

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