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soldi70 [24.7K]
2 years ago
9

A small button placed on a horizontal rotating platform with diameter 0.320 m will revolve with the platform when it is brought

up to a speed of 40.0 rev min, provided the button is no more than 0.150 m from the axis. (a) What is the coefficient of static friction between the button and the platform
Physics
1 answer:
Otrada [13]2 years ago
6 0

Answer:

0.2687 approximately 0.27

Explanation:

Diameter = 0.320

Speed = 40.0 rev/min

We are required to find coefficient of static friction between friction and button

The radius can be calculated as

0.320/2

= 0.160m

Then we have the rotational speed w = 40rev/min x 2pi/60

= 4.19 rad/s

umg = mrw²

u = mrw²/mg

u = rw²/g -------(1)

g = 9.8

When we put values into equation 1

0.150m x 4.19² / 9.8

= 0.150m x 17.5561 /9.8

= 0.2689

This is approximately 0.27

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